Big O Notation of this BinaryHeap code - c#

I'm trying to determine the Big O notation of two methods that are based on Jim Mischel's BinaryHeap class.
public void Add(T item) {
int i = _items.Count;
_items.Add(item);
while (i > 0 && _items[(i - 1) / 2].CompareTo(item) > 0) {
_items[i] = _items[(i - 1) / 2];
i = (i - 1) / 2;
}
_items[i] = item;
}
public T Remove() {
T firstItem = _items[0];
T tempItem = _items[_items.Count - 1];
_items.RemoveAt(_items.Count - 1);
if (_items.Count > 0) {
int i = 0;
while (i < _items.Count / 2) {
int j = (2 * i) + 1;
if ((j < _items.Count - 1) && (_items[j].CompareTo(_items[j + 1]) > 0)) ++j;
if (_items[j].CompareTo(tempItem) >= 0) break;
_items[i] = _items[j];
i = j;
}
_items[i] = tempItem;
}
return firstItem;
}
For the Add method, I believe it's O(log(n)) since the loop appears to be of the form:
for(int i = n; i > 0; i=(i-1)/2);
And for the Remove method, I believe it's O(log(log(n)) since the loop appears to be of the form:
for(int i=0; i < n/2; i = (2*i)+1);
Is this correct?

Related

Unusual StackOverflowException in ListDictionaryInternal

I have class library with a bunch of static methods. While trying to call one of them, I experience unhandled StackOverflowException somewhere in the ListDictionaryInternal class.
I tried enabling .Net Framework (v 4.5.2) stepping, surrounding call with try/catch block, and executing it step by step. When I place continue statement after Appendix A comment, then comment it while debugging, method works as expected. Otherwise I cannot even hit breakpoint at the start at the method. I also tried to call method with all parameters set to null, but it did not help either.
public static List<CalcSector> Split(List<CalcSector> calibration, List<ProfilePoint> profile, List<MeasurementPoint> additionalPoints)
{
double lengthCorridor = 10d;
double lengthEpsilon = 1d;
if (!(calibration?.Any() ?? false)) throw new ArgumentNullException(nameof(calibration), "Empty calibration table");
if (!(profile?.Any() ?? false)) throw new ArgumentNullException(nameof(profile), "Empty profile points collection");
for (int i = 0; i < calibration.Count - 1; i++)
if (Math.Abs(calibration[i].EndDistance - calibration[i + 1].StartDistance) > lengthEpsilon)
throw new ArgumentException($"calibration[{i}]", "Calibration table integrity is compromised");
List<CalcSector> result = new List<CalcSector>();
List<ProfilePoint> SummitPoints = new List<ProfilePoint>();
calibration.ForEach(x => result.Add(x));
profiles = profile.OrderBy(x => x.Distance).ToList();
//
if (additionalPoints?.Any() ?? false)
foreach (MeasurementPoint mp in additionalPoints.Where(x => x.Id != int.MinValue && x.Id != int.MaxValue))
for (int i = 0; i < result.Count; i++)
if (Math.Abs(mp.Distance - result[i].StartDistance) > lengthEpsilon && Math.Abs(mp.Distance - result[i].EndDistance) > lengthEpsilon && mp.Distance > result[i].StartDistance && mp.Distance < result[i].EndDistance)
{
CalcSector c = new CalcSector()
{
StartDistance = mp.Distance,
StartHeight = BinaryHeightSearch(mp.Distance),
StartPointId = mp.Id,
EndDistance = result[i].EndDistance,
EndHeight = result[i].EndHeight,
Length = result[i].EndDistance - mp.Distance,
Thickness = result[i].Thickness,
};
result[i].EndDistance = mp.Distance;
result[i].EndHeight = c.StartHeight;
result[i].EndPointId = mp.Id;
c.Volume = result[i].Volume / result[i].Length * c.Length;
result[i].Length -= c.Length;
result[i].Volume -= c.Volume;
result.Insert(i + 1, c);
break;
}
else if (Math.Abs(mp.Distance - result[i].StartDistance) < lengthEpsilon)
result[i].StartPointId = mp.Id;
else if (Math.Abs(mp.Distance - result[i].EndDistance) < lengthEpsilon)
result[i].EndPointId = mp.Id;
int start = 0;
int end = 0;
bool hasSpikes = true;
while (hasSpikes)
{
hasSpikes = false;
//Appendix A
for (int j = 0; j < result.Count; j++)
{
result[j].z = -1d * (result[j].StartHeight - result[j].EndHeight) / (result[j].EndDistance - result[j].StartDistance);
result[j].sI = start = BinaryProfileSearch(result[j].StartDistance);
result[j].eI = end = BinaryProfileSearch(result[j].EndDistance);
for (int i = start + 1; i < end; i++)
if (Math.Abs(result[j].z * (profiles[i].Distance - result[j].StartDistance) + result[j].StartHeight - profiles[i].Height) > lengthCorridor)
{
int maxIndex = -1;
double maxH = double.MinValue;
int minIndex = -1;
double minH = double.MaxValue;
for (; start < end; start++)
{
if (Math.Abs(result[j].z * (profiles[start].Distance - result[j].StartDistance) + result[j].StartHeight - profiles[start].Height) <= lengthCorridor)
continue;
if (result[j].z * (profiles[i].Distance - result[j].StartDistance) + result[j].StartHeight - profiles[i].Height > maxH)
{
maxH = profiles[start].Height;
maxIndex = start;
}
if (result[j].z * (profiles[i].Distance - result[j].StartDistance) + result[j].StartHeight - profiles[i].Height < minH)
{
minH = profiles[start].Height;
minIndex = start;
}
}
int target = Math.Min(maxIndex, minIndex);
CalcSector c = new CalcSector()
{
StartDistance = profiles[target].Distance,
StartHeight = profiles[target].Height,
sI = target,
EndDistance = result[j].EndDistance,
EndHeight = result[j].EndHeight,
EndPointId = result[j].EndPointId,
eI = result[j].eI,
Length = result[j].EndDistance - profiles[target].Distance,
Thickness = result[j].Thickness,
};
result[j].EndDistance = c.StartDistance;
result[j].EndHeight = c.StartHeight;
result[j].EndPointId = null;
result[j].eI = target;
result[j].z = -1d * (result[j].StartHeight - result[j].EndHeight) / (result[j].EndDistance - result[j].StartDistance);
c.Volume = result[j].Volume / result[j].Length * c.Length;
result[j].Length -= c.Length;
result[j].Volume -= c.Volume;
result.Insert(j + 1, c);
hasSpikes = true;
break;
}
}
}
for (int j = 0; j < result.Count; j++)
{
result[j].Diameter = 1000d * Math.Sqrt(4d * result[j].Volume / Constants["PI"] / result[j].Length);
result[j].OrdNum = j;
}
result.First().StartPointId = int.MinValue;
result.Last().EndPointId = int.MaxValue;
for (int i = 1; i < profiles.Count - 1; i++)
if (profiles[i - 1].Height < profiles[i].Height && profiles[i].Height > profiles[i + 1].Height)
SummitPoints.Add(profiles[i]);
return result;
}
public class CalcSector
{
public int OrdNum;
public double StartDistance;
public double StartHeight;
public int? StartPointId;
public double EndDistance;
public double EndHeight;
public int? EndPointId;
public double Length;
public double Volume;
public double Diameter;
public double Thickness;
public int sI;
public int eI;
public double z;
}
public class ProfilePoint
{
public double Distance;
public double Height;
}
public class MeasurementPoint
{
public int Id;
public double Distance;
}
I expect this method to split some of the original CalcSectors into smaller ones, but all I have is this unhandled fatal exception.
Added:
private static int BinaryProfileSearch(double distance)
{
if (profiles == null || profiles.Count == 0)
return -1;
//assuming that profile points are already ordered by distance
if (distance <= profiles.First().Distance)
return 0;
if (distance >= profiles.Last().Distance)
return profiles.Count - 1;
int first = 0;
int last = profiles.Count - 1;
while (first + 1 < last)
{
int mid = (first + last) / 2;
if (distance <= profiles[mid].Distance)
last = mid;
else
first = mid + 1;
}
if (distance - profiles[first].Distance > profiles[last].Distance - distance)
return last;
else
return first;
}
The solution came quite unexpected: invalid value of 8087 control word. Next line changes it back.
_controlfp(0x9001F, 0xFFFFF);

Rounding amount with available set of denominations

This isn't regular rounding thing which rounds up or down based of a single value.
I would want to have a function where I pass the amount as integer and denominations as array of integer.
What that function should return to me is a nearest possible integer value achievable with passed array of denominations.
Whether to round up or down will again be sent as a parameter.
Code:
var amount = 61; // for. e.g.
int[] denoms = [20, 50]; // for. e.g.
bool roundUp = true;
amount = RoundAmount(amount, denoms, roundUp);
Expected result :
RoundAmount function should return me the nearest possible amount achievable with denoms that I have passed.
If roundUp = true, The return value should be 70, because 70 = 20+50
and amount 70 can be achieved by one note of 20s and one note of 50s.
If roundUp = false, It should have returned 60, because 60 =
20+20+20 and amount 60 can be achieved by 3 notes of 20s
What I got so far :
I was only reached to the point where I can manage to round the amount up or down based on a single integer (and not the array of integers)
public int RoundAmount(int amount, int value, bool roundUp)
{
if (roundUp)
amount = amount - (amount % value) + value;
else
amount = amount - (amount % value)
return amount;
}
Edit:
I have another recursive function which checks if amount is achievable or not,
Only if amount isn't achievable, RoundAmount function is called.
So in my example, amount = 70 will never be the input because 70 is achievable with available denoms and I won't call the RoundAmount in that case.
Solution: (Thanks to maraca and Koray)
I'm glad its working with long numbers though it wasn't original requirement.
private static long RoundAmount_maraca(long a, long[] d, bool up)
{
d = d.ToArray();
Array.Sort(d);
if (a < d[0])
return up ? d[0] : 0;
long count = 0;
for (long i = 0; i < d.Length; i++)
{
if (d[i] == 0)
continue;
for (long j = i + 1; j < d.Length; j++)
if (d[j] % d[i] == 0)
d[j] = 0;
if (d[i] > a && !up)
break;
d[count++] = d[i];
if (d[i] > a)
break;
}
if (count == 1)
return (!up ? a : (a + d[0] - 1)) / d[0] * d[0];
long gcd = euclid(d[1], d[0]);
for (long i = 2; i < count && gcd > 1; i++)
gcd = euclid(d[i], gcd);
if (up)
a = (a + gcd - 1) / gcd;
else
a /= gcd;
for (long i = 0; i < count; i++)
{
d[i] /= gcd;
if (a % d[i] == 0)
return a * gcd;
}
var set = new HashSet<long>();
set.Add(0);
long last = 0;
for (long n = d[0]; ; n++)
{
if (!up && n > a)
return last * gcd;
for (long i = 0; i < count && n - d[i] >= 0; i++)
{
if (set.Contains(n - d[i]))
{
if (n >= a)
return n * gcd;
if ((a - n) % d[0] == 0)
return a * gcd;
set.Add(n);
last = n;
break;
}
}
}
}
private static long euclid(long a, long b)
{
while (b != 0)
{
long h = a % b;
a = b;
b = h;
}
return a;
}
I am assuming that you are looking for a performant solution with a relatively small amount of denominations b (e.g. less than 100 denominations). While the amount a and the denominations d[i] can be quite large (e.g. less than 10^6).
Sort d ascending and remove duplicates. When rounding down only keep the values smaller or equal than a and when rounding up keep only the smallest value greater or equal than a and discard the greater ones.
(Optional) remove all numbers which are a multiple of some other number O(b^2).
Calculate the greatest common divisor gcd of the denominations. You can use the Euclidean algorithm starting with the first two numbers then calculate the greatest common divisor of the result and the third number and so on. Of course you can stop as soon as you reach one.
Divide a by gcd, round like you want to round the result (using integer division, rounding down: a /= gcd, rounding up: a = (a + gcd - 1) / gcd).
Divide all denominations by gcd (d[i] /= gcd). Now the greatest common divisor of all denominations is one and therefore it is guaranteed that a Frobenius number exists (all amounts greater than that number can be built and require no rounding). While doing so you can also check if the new value leads to a % d[i] == 0 and immediately return a * gcd if so.
Create a hash set for the values which can be built. It is better than an array because the array is potentially wasting a lot of space (remember the Frobenius number). Add zero to the set.
Create a variable n for the current number, initialize with smallest denomination: n = d[0].
If n can be built with any of the available denominations, in other words the set contains any of n - d[i] then proceed with the next step. Otherwise increase n by one and repeat this step unless n == a and you are rounding down, then you can return the last number that could be built multiplied by gcd immediately. You could also remove n - d[b - 1] from the set each time because this value will not be requested any more.
If n >= a return n * gcd (can only be true when rounding up, rounding down would have returned the result in step 8. already). Else if (a - n) % d[0] == 0 return a * gcd. This check is even better than looking for the Frobenius number (the number after which d[0] - 1 consecutive values can be built), it is more or less the equivalent (d[0] - 1 consecutive values means the difference between one of them and a modulo d[0] has to be zero) but could return much faster. Else increase n by one and continue with step 8.
An example with d = {4, 6} and a = 9999 (or any other big odd number) shows the advantages of this algorithm. It is easy to see that odd numbers can never be built and we would fill up the whole set with all even numbers except 2. But if we divide by gcd we get d = {2, 3} and aUp = 5000 and aDown = 4999. The Frobenius number for {2, 3} is 1 (the only number which cannot be built), so after at most 3 (first number where all modulos are covered) steps (instead of 10000) the modulo would be zero and we would return a * gcd which gives 9998 or 10000 depending on rounding direction, which is the correct result.
Here is the code with test included. I did six runs on my crappy notebook and it took 90, 92, 108, 94, 96 and 101 seconds (edit: early loop escape if current denomination greater than current number && n - d[i] >= 0 halves the times and gives an average of about 45s) for 7200 random roundings (3600 in each direction) with combinations of different amounts of denominations (range 2 to 100), dMax (range 100 to 10^6) and aMax (range 10^4 to 10^6), (see the code at the bottom for the exact values). I think the time for the random number generation and output can be neglected, so with this input and the given ranges the algorithm rounds about 160 numbers per second on average (edit: see thirty times faster version below).
public static final int round(int a, int[] d, boolean up) {
d = d.clone(); // otherwise input gets changed
Arrays.sort(d);
if (a < d[0])
return up ? d[0] : 0;
int count = 0;
for (int i = 0; i < d.length; i++) {
if (d[i] == 0)
continue;
for (int j = i + 1; j < d.length; j++)
if (d[j] % d[i] == 0)
d[j] = 0;
if (d[i] > a && !up)
break;
d[count++] = d[i];
if (d[i] > a)
break;
}
if (count == 1)
return (!up ? a : (a + d[0] - 1)) / d[0] * d[0];
int gcd = euclid(d[1], d[0]);
for (int i = 2; i < count && gcd > 1; i++)
gcd = euclid(d[i], gcd);
if (up)
a = (a + gcd - 1) / gcd;
else
a /= gcd;
for (int i = 0; i < count; i++) {
d[i] /= gcd;
if (a % d[i] == 0)
return a * gcd;
}
Set<Integer> set = new HashSet<>();
set.add(0);
int last = 0;
for (int n = d[0];; n++) {
if (!up && n > a)
return last * gcd;
for (int i = 0; i < count && n - d[i] >= 0; i++) {
if (set.contains(n - d[i])) {
if (n >= a)
return n * gcd;
if ((a - n) % d[0] == 0)
return a * gcd;
set.add(n);
last = n;
break;
}
}
}
}
public static final int euclid(int a, int b) {
while (b != 0) {
int h = a % b;
a = b;
b = h;
}
return a;
}
public static final int REPEAT = 100;
public static final int[] D_COUNT = {2, 5, 10, 20, 50, 100};
public static final int[] D_MAX = {100, 10000, 1000000};
public static final int[] A_MAX = {10000, 1000000};
public static void main(String[] args) {
long start = System.currentTimeMillis();
Random r = new Random();
for (int i = 0; i < REPEAT; i++) {
for (int j = 0; j < D_COUNT.length; j++) {
for (int k = 0; k < D_MAX.length; k++) {
for (int l = 0; l < A_MAX.length; l++) {
int[] d = new int[D_COUNT[j]];
for (int m = 0; m < d.length; m++)
d[m] = r.nextInt(D_MAX[k]);
int a = r.nextInt(A_MAX[l]);
System.out.println(round(a, d, false));
System.out.println(round(a, d, true));
}
}
}
}
System.out.println((System.currentTimeMillis() - start) / 1000 + " seconds");
}
As it turns out #Koray's edit 7 is about three times faster for the given input (only for very large gcds my algorithm above is faster). So to get the ultimate algorithm I replaced the dynamic programming part of my algorithm by that of #Koray (with some improvements). It worked, it is roughly ten times faster than edit 7 and thirty times faster than the algorithm above. Which would give about 5000 roundings per second (very rough estimation) on average.
private static int round(int a, int[] d, boolean up) {
d = d.clone();
Arrays.sort(d);
if (a < d[0])
return up ? d[0] : 0;
int count = 0;
for (int i = 0; i < d.length; i++) {
if (d[i] == 0)
continue;
if (a % d[i] == 0)
return a;
for (int j = i + 1; j < d.length; j++)
if (d[j] > 0 && d[j] % d[i] == 0)
d[j] = 0;
if (d[i] > a && !up)
break;
d[count++] = d[i];
if (d[i] > a)
break;
}
if (count == 1)
return (!up ? a : (a + d[0] - 1)) / d[0] * d[0];
int gcd = euclid(d[1], d[0]);
for (int i = 2; i < count && gcd > 1; i++)
gcd = euclid(d[i], gcd);
if (gcd > 1) {
if (up)
a = (a + gcd - 1) / gcd;
else
a /= gcd;
for (int i = 0; i < count; i++) {
d[i] /= gcd;
if (a % d[i] == 0)
return a * gcd;
}
}
int best = !up ? d[count - 1] : ((a + d[0] - 1) / d[0] * d[0]);
if (d[count - 1] > a) {
if (d[count - 1] < best)
best = d[count - 1];
count--;
}
Stack<Integer> st = new Stack<Integer>();
BitSet ba = new BitSet(a + 1);
for (int i = 0; i < count; i++) {
ba.set(d[i]);
st.push(d[i]);
}
while (st.size() > 0) {
int v1 = st.pop();
for (int i = 0; i < count; i++) {
int val = v1 + d[i];
if (val <= a && !ba.get(val)) {
if ((a - val) % d[0] == 0)
return a * gcd;
ba.set(val, true);
st.push(val);
if (!up && val > best)
best = val;
} else if (val > a) {
if (up && val < best)
best = val;
break;
}
}
}
return best * gcd;
}
private static void test()
{
var amount = 61;
int[] denoms = new int[] { 20, 50 };
int up = RoundAmount(amount, denoms, true);//->70
int down = RoundAmount(amount, denoms, false);//->60
}
private static int RoundAmount(int amount, int[] denoms, bool roundUp)
{
HashSet<int> hs = new HashSet<int>(denoms);
bool added = true;
while (added)
{
added = false;
var arr = hs.ToArray();
foreach (int v1 in arr)
foreach (int v2 in arr)
if ((v1 < amount) && (v2 < amount) && (hs.Add(v1 + v2)))
added = true;
}
int retval = roundUp ? int.MaxValue : int.MinValue;
foreach (int v in hs)
{
if (roundUp)
{
if ((v < retval) && (v >= amount))
retval = v;
}
else
{
if ((v > retval) && (v <= amount))
retval = v;
}
}
return retval;
}
Edit 7
Edit 6 had a bug if a "0" denom exists. I examined #maraca's code in detail (its just great I think) and inspired on that, I've tried some optimizations on my code. Here are the performance comparisons. (I've tried to convert maraca's code to c#, I hope I ve done it right.)
private static int REPEAT = 100;
private static int[] D_COUNT = { 2, 5, 10, 20, 50, 100 };
private static int[] D_MAX = { 100, 10000, 1000000 };
private static int[] A_MAX = { 10000, 1000000 };
private static void testR()
{
Random r = new Random();
long wMaraca = 0;
long wKoray = 0;
for (int i = 0; i < REPEAT; i++)
{
for (int j = 0; j < D_COUNT.Length; j++)
{
for (int k = 0; k < D_MAX.Length; k++)
{
for (int l = 0; l < A_MAX.Length; l++)
{
int[] d = new int[D_COUNT[j]];
for (int m = 0; m < d.Length; m++)
d[m] = r.Next(D_MAX[k]);
int a = r.Next(A_MAX[l]);
Stopwatch maraca = Stopwatch.StartNew();
int m1 = RoundAmount_maraca(a, d, false);
int m2 = RoundAmount_maraca(a, d, true);
maraca.Stop();
wMaraca += maraca.ElapsedMilliseconds;
Stopwatch koray = Stopwatch.StartNew();
int k1 = RoundAmount_koray(a, d, false);
int k2 = RoundAmount_koray(a, d, true);
koray.Stop();
wKoray += koray.ElapsedMilliseconds;
if ((m1 != k1) || (m2 != k2))
{
throw new Exception("something is wrong!");
}
}
}
}
}
//some results with debug compile
//try1
//maraca: 50757 msec
//koray: 19188 msec
//try2
//maraca: 52623 msec
//koray: 19102 msec
//try3
//maraca: 57139 msec
//koray: 18952 msec
//try4
//maraca: 64911 msec
//koray: 21070 msec
}
private static int RoundAmount_koray(int amount, int[] denoms, bool roundUp)
{
List<int> lst = denoms.ToList();
lst.Sort();
if (amount < lst[0])
return roundUp ? lst[0] : 0;
HashSet<int> hs = new HashSet<int>();
for (int i = 0, count = lst.Count; i < count; i++)
{
int v = lst[i];
if (v != 0)
{
if (v > amount && !roundUp)
break;
if (hs.Add(v))
{
if (amount % v == 0)
return amount;
else
for (int j = i + 1; j < count; j++)
if (lst[j] != 0)
if (v % lst[j] == 0)
lst[j] = 0;
else if (amount % (v + lst[j]) == 0)
return amount;
}
}
}
denoms = hs.ToArray();
HashSet<int> hsOK = new HashSet<int>(denoms);
Stack<int> st = new Stack<int>(denoms);
BitArray ba = new BitArray(amount + denoms.Max() * 2 + 1);
int minOK = amount - denoms.Min();
while (st.Count > 0)
{
int v1 = st.Pop();
foreach (int v2 in denoms)
{
int val = v1 + v2;
if (!ba.Get(val))
{
if (amount % val == 0)
return amount;
ba.Set(val, true);
if (val < amount)
st.Push(val);
if (val >= minOK)
hsOK.Add(val);
}
}
}
if (!roundUp)
{
int retval = 0;
foreach (int v in hsOK)
if (v > retval && v <= amount)
retval = v;
return retval;
}
else
{
int retval = int.MaxValue;
foreach (int v in hsOK)
if (v < retval && v >= amount)
retval = v;
return retval;
}
}
private static int RoundAmount_maraca(int a, int[] d, bool up)
{
d = d.ToArray();
Array.Sort(d);
if (a < d[0])
return up ? d[0] : 0;
int count = 0;
for (int i = 0; i < d.Length; i++)
{
if (d[i] == 0)
continue;
for (int j = i + 1; j < d.Length; j++)
if (d[j] % d[i] == 0)
d[j] = 0;
if (d[i] > a && !up)
break;
d[count++] = d[i];
if (d[i] > a)
break;
}
if (count == 1)
return (!up ? a : (a + d[0] - 1)) / d[0] * d[0];
int gcd = euclid(d[1], d[0]);
for (int i = 2; i < count && gcd > 1; i++)
gcd = euclid(d[i], gcd);
if (up)
a = (a + gcd - 1) / gcd;
else
a /= gcd;
for (int i = 0; i < count; i++)
{
d[i] /= gcd;
if (a % d[i] == 0)
return a * gcd;
}
var set = new HashSet<int>();
set.Add(0);
int last = 0;
for (int n = d[0]; ; n++)
{
if (!up && n > a)
return last * gcd;
for (int i = 0; i < count && n - d[i] >= 0; i++)
{
if (set.Contains(n - d[i]))
{
if (n >= a)
return n * gcd;
if ((a - n) % d[0] == 0)
return a * gcd;
set.Add(n);
last = n;
break;
}
}
}
}
private static int euclid(int a, int b)
{
while (b != 0)
{
int h = a % b;
a = b;
b = h;
}
return a;
}
Edit - Maraca in c#
Maraca's last edit clearly outperforms all! I have tried to prepare a better c# conversion of his code + added a ulong version. (int version is ~1.6 times faster than the ulong version)
#region maraca int
private static int RoundAmount_maraca(int a, int[] d0, bool up)
{
int[] d = new int[d0.Length];
Buffer.BlockCopy(d0, 0, d, 0, d.Length * sizeof(int));
Array.Sort(d);
if (a < d[0])
return up ? d[0] : 0;
int count = 0;
for (int i = 0; i < d.Length; i++)
{
if (d[i] == 0)
continue;
for (int j = i + 1; j < d.Length; j++)
if (d[j] % d[i] == 0)
d[j] = 0;
if (d[i] > a && !up)
break;
d[count++] = d[i];
if (d[i] > a)
break;
}
if (count == 1)
return (!up ? a : (a + d[0] - 1)) / d[0] * d[0];
int gcd = euclid(d[1], d[0]);
for (int i = 2; i < count && gcd > 1; i++)
gcd = euclid(d[i], gcd);
if (up)
a = (a + gcd - 1) / gcd;
else
a /= gcd;
for (int i = 0; i < count; i++)
{
d[i] /= gcd;
if (a % d[i] == 0)
return a * gcd;
}
int best = !up ? d[count - 1] : ((a + d[0] - 1) / d[0] * d[0]);
if (d[count - 1] > a)
{
if (d[count - 1] < best)
best = d[count - 1];
count--;
}
var st = new Stack<int>();
BitArray ba = new BitArray(a+1);
for (int i = 0; i < count; i++)
{
ba.Set(d[i], true);
st.Push(d[i]);
}
while (st.Count > 0)
{
int v1 = st.Pop();
for (int i = 0; i < count; i++)
{
int val = v1 + d[i];
if (val <= a && !ba.Get(val))
{
if ((a - val) % d[0] == 0)
return a * gcd;
ba.Set(val, true);
st.Push(val);
if (!up && val > best)
best = val;
}
else if (up && val > a && val < best)
best = val;
}
}
return best * gcd;
}
private static int euclid(int a, int b)
{
while (b != 0)
{
int h = a % b;
a = b;
b = h;
}
return a;
}
#endregion
#region maraca ulong
private static ulong RoundAmount_maraca_ulong(ulong a, ulong[] d0, bool up)
{
ulong[] d = new ulong[d0.Length];
Buffer.BlockCopy(d0, 0, d, 0, d.Length * sizeof(ulong));
Array.Sort(d);
if (a < d[0])
return up ? d[0] : 0ul;
int count = 0;
for (int i = 0; i < d.Length; i++)
{
if (d[i] == 0ul)
continue;
for (int j = i + 1; j < d.Length; j++)
if (d[j] % d[i] == 0ul)
d[j] = 0ul;
if (d[i] > a && !up)
break;
d[count++] = d[i];
if (d[i] > a)
break;
}
if (count == 1)
return (!up ? a : (a + d[0] - 1ul)) / d[0] * d[0];
ulong gcd = euclid(d[1], d[0]);
for (int i = 2; i < count && gcd > 1; i++)
gcd = euclid(d[i], gcd);
if (up)
a = (a + gcd - 1ul) / gcd;
else
a /= gcd;
for (int i = 0; i < count; i++)
{
d[i] /= gcd;
if (a % d[i] == 0ul)
return a * gcd;
}
ulong best = !up ? d[count - 1] : ((a + d[0] - 1ul) / d[0] * d[0]);
if (d[count - 1] > a)
{
if (d[count - 1] < best)
best = d[count - 1];
count--;
}
var st = new Stack<ulong>();
UlongBitArray ba = new UlongBitArray(a + 1ul);
for (int i = 0; i < count; i++)
{
ba.Set(d[i], true);
st.Push(d[i]);
}
while (st.Count > 0)
{
ulong v1 = st.Pop();
for (int i = 0; i < count; i++)
{
ulong val = v1 + d[i];
if (val <= a && !ba.Get(val))
{
if ((a - val) % d[0] == 0ul)
return a * gcd;
ba.Set(val, true);
st.Push(val);
if (!up && val > best)
best = val;
}
else if (up && val > a && val < best)
best = val;
}
}
return best * gcd;
}
private static ulong euclid(ulong a, ulong b)
{
while (b != 0)
{
ulong h = a % b;
a = b;
b = h;
}
return a;
}
class UlongBitArray
{
ulong[] bits;
public UlongBitArray(ulong length)
{
this.bits = new ulong[(length - 1ul) / 32ul + 1ul];
}
public bool Get(ulong index)
{
return (this.bits[index / 32ul] & (1ul << (int)(index % 32ul))) > 0ul;
}
public void Set(ulong index, bool val)
{
if (val)
this.bits[index / 32ul] |= 1ul << (int)(index % 32ul);
else
this.bits[index / 32ul] &= ~(1ul << (int)(index % 32ul));
}
}
#endregion
Edit 8
I have made some improvements and in random tests outperformed #maraca's latest update :) If you choose to use my custom stack class, please make measurements in release mode. (This custom stack class is of course much slower in debug mode but %5-15 faster than .NET's in relase mode. In my tests using the .NET Stack class did not change the performance comparison between two, its just an extra boost.)
private delegate int RoundAmountDelegate(int amount, int[] denoms, bool roundUp);
private static int REPEAT = 100;
private static int[] D_COUNT = { 2, 5, 10, 20, 50, 100 };
private static int[] D_MAX = { 100, 10000, 1000000 };
private static int[] A_MAX = { 10000, 1000000 };
private static void testR()
{
#if DEBUG
while (true)
#endif
{
Random r = new Random();
long wT1 = 0; RoundAmountDelegate func1 = RoundAmount_maraca;
long wT2 = 0; RoundAmountDelegate func2 = RoundAmount_koray;
for (int i = 0; i < REPEAT; i++)
{
for (int j = 0; j < D_COUNT.Length; j++)
{
for (int k = 0; k < D_MAX.Length; k++)
{
for (int l = 0; l < A_MAX.Length; l++)
{
int[] d = new int[D_COUNT[j]];
ulong[] dl = new ulong[D_COUNT[j]];
for (int m = 0; m < d.Length; m++)
{
d[m] = r.Next(D_MAX[k]) + 1;
dl[m] = (ulong)d[m];
}
int a = r.Next(A_MAX[l]);
ulong al = (ulong)a;
Stopwatch w1 = Stopwatch.StartNew();
int m1 = func1(a, d, false);
int m2 = func1(a, d, true);
w1.Stop();
wT1 += w1.ElapsedMilliseconds;
Stopwatch w2 = Stopwatch.StartNew();
int k1 = func2(a, d, false);
int k2 = func2(a, d, true);
w2.Stop();
wT2 += w2.ElapsedMilliseconds;
if ((m1 != k1) || (m2 != k2))
{
#if !DEBUG
MessageBox.Show("error");
#else
throw new Exception("something is wrong!");
#endif
}
}
}
}
}
//some results with release compile
//maraca: 1085 msec
//koray(with .NET Stack<int>): 801 msec
//maraca: 1127 msec
//koray(with .NET Stack<int>): 741 msec
//maraca: 989 msec
//koray(with .NET Stack<int>): 736 msec
//maraca: 962 msec
//koray(with .NET Stack<int>): 632 msec
//-------------------------------------------
//maraca: 1045 msec
//koray(with custom stack): 674 msec
//maraca: 1060 msec
//koray(with custom stack): 606 msec
//maraca: 1175 msec
//koray(with custom stack): 711 msec
//maraca: 878 msec
//koray(with custom stack): 699 msec
#if !DEBUG
MessageBox.Show(wT1 + " " + wT2 + " %" + (double)wT2 / (double)wT1 * 100d);
#endif
}
}
#region Koray
private static int RoundAmount_koray(int amount, int[] denoms, bool roundUp)
{
int[] sorted = new int[denoms.Length];
Buffer.BlockCopy(denoms, 0, sorted, 0, sorted.Length * sizeof(int));
Array.Sort(sorted);
int minD = sorted[0];
if (amount < minD)
return roundUp ? minD : 0;
HashSet<int> hs = new HashSet<int>();
for (int i = 0, count = sorted.Length; i < count; i++)
{
int v = sorted[i];
if (v != 0)
{
if (!roundUp && v > amount)
break;
else if (hs.Add(v))
{
if (amount % v == 0)
return amount;
else
for (int j = i + 1; j < count; j++)
if (sorted[j] != 0)
if (v % sorted[j] == 0)
sorted[j] = 0;
else if (amount % (v + sorted[j]) == 0)
return amount;
}
}
}
denoms = new int[hs.Count];
int k = 0;
foreach (var v in hs)
denoms[k++] = v;
HashSet<int> hsOK = new HashSet<int>(denoms);
stack st = new stack(denoms);
//Stack<int> st = new Stack<int>(denoms);
BitArray ba = new BitArray(amount + denoms[denoms.Length - 1] * 2 + 1);
int minOK = roundUp ? amount : amount - minD;
int maxOK = roundUp ? amount + minD : amount;
while (st.Count > 0)
{
int v1 = st.Pop();
foreach (int v2 in denoms)
{
int val = v1 + v2;
if (val <= maxOK)
{
if (!ba.Get(val))
{
if (amount % val == 0)
return amount;
int diff = amount - val;
if (diff % v1 == 0 || diff % v2 == 0)
return amount;
ba.Set(val, true);
if (val < amount)
st.Push(val);
if (val >= minOK)
hsOK.Add(val);
}
}
else
break;
}
}
if (!roundUp)
{
int retval = 0;
foreach (int v in hsOK)
if (v > retval && v <= amount)
retval = v;
return retval;
}
else
{
int retval = int.MaxValue;
foreach (int v in hsOK)
if (v < retval && v >= amount)
retval = v;
return retval;
}
}
private sealed class stack
{
int[] _array;
public int Count;
public stack()
{
this._array = new int[0];
}
public stack(int[] arr)
{
this.Count = arr.Length;
this._array = new int[this.Count*2];
Buffer.BlockCopy(arr, 0, this._array, 0, this.Count * sizeof(int));
}
public void Push(int item)
{
if (this.Count == this._array.Length)
{
int[] destinationArray = new int[2 * this.Count];
Buffer.BlockCopy(this._array, 0, destinationArray, 0, this.Count * sizeof(int));
this._array = destinationArray;
}
this._array[this.Count++] = item;
}
public int Pop()
{
return this._array[--this.Count];
}
}
#endregion
#region Maraca
private static int RoundAmount_maraca(int a, int[] d0, bool up)
{
int[] d = new int[d0.Length];
Buffer.BlockCopy(d0, 0, d, 0, d.Length * sizeof(int));
Array.Sort(d);
if (a < d[0])
return up ? d[0] : 0;
int count = 0;
for (int i = 0; i < d.Length; i++)
{
if (d[i] == 0)
continue;
for (int j = i + 1; j < d.Length; j++)
if (d[j] % d[i] == 0)
d[j] = 0;
if (d[i] > a && !up)
break;
d[count++] = d[i];
if (d[i] > a)
break;
}
if (count == 1)
return (!up ? a : (a + d[0] - 1)) / d[0] * d[0];
int gcd = euclid(d[1], d[0]);
for (int i = 2; i < count && gcd > 1; i++)
gcd = euclid(d[i], gcd);
if (up)
a = (a + gcd - 1) / gcd;
else
a /= gcd;
for (int i = 0; i < count; i++)
{
d[i] /= gcd;
if (a % d[i] == 0)
return a * gcd;
}
int best = !up ? d[count - 1] : ((a + d[0] - 1) / d[0] * d[0]);
if (d[count - 1] > a)
{
if (d[count - 1] < best)
best = d[count - 1];
count--;
}
var st = new Stack<int>();
BitArray ba = new BitArray(a + 1);
for (int i = 0; i < count; i++)
{
ba.Set(d[i], true);
st.Push(d[i]);
}
while (st.Count > 0)
{
int v1 = st.Pop();
for (int i = 0; i < count; i++)
{
int val = v1 + d[i];
if (val <= a && !ba.Get(val))
{
if ((a - val) % d[0] == 0)
return a * gcd;
ba.Set(val, true);
st.Push(val);
if (!up && val > best)
best = val;
}
else if (up && val > a && val < best)
best = val;
}
}
return best * gcd;
}
private static int euclid(int a, int b)
{
while (b != 0)
{
int h = a % b;
a = b;
b = h;
}
return a;
}
#endregion
This is a standard Knapsack problem and you can google it to refer to its wiki page for its concept.
I think your problem can be splitted to two parts.
Do Knapsack for denominations.
Use f[i] to represent the last denomination used to construct amount i, and f[i]==-1 means that i is not able to get.
fill f with -1
f[0] = 0
for i from 0 to target_amount + min(denoms) - 1
for j from 0 to denoms.size()
if f[i - denoms[j]] != -1
{
f[i] = denoms[j]
break
}
Find nearest amount based on roundUp.
roundUp == true
Starting from target_amount, ascendingly find a f[i] which is not -1.
roundUp == false
Starting from target_amount, descendingly find a f[i] which is not -1.
Optional: find which denominations construct your target amount
Backtrack your f[target_amount].
Just fill array of length amount + smallestdenomination + 1 with possible combinations of coins (standard dynamic programming problem).
Then walk this array from amount index in rounding direction.
Delphi example
var
A : array of Integer;
Denoms: array of Integer;
coin, amount, idx, i, Maxx: Integer;
roundUp: Boolean;
s: string;
begin
amount := 29;
SetLength(Denoms, 2);
Denoms[0] := 7;
Denoms[1] := 13;
Maxx := amount + MinIntValue(Denoms);
SetLength(A, Maxx + 1);
A[0] := 1;
for coin in Denoms do begin
for i := 0 to Maxx - coin do
if A[i] <> 0 then
A[i + coin] := coin;
end;
roundUp := True;
idx := amount;
i := 2 * Ord(roundUp) - 1;// 1 for roundUp=true, -1 for false
while A[idx] = 0 do //scan for nonzero entry
idx := idx + i;
s := '';
while idx > 0 do begin //roll back to get components of this sum
s := s + Format('%d ', [A[idx]]);
idx := idx - A[idx];
end;
Memo1.Lines.Add(s);
outputs 13 13 7 combination for roundUp := True; and 7 7 7 7 otherwise.
(Code does not seek for "optimal" solution)
Example for coins 3 and 5:
[0, 0, 0, 3, 0, 5, 3, 0, 5, 3, 5]
To find what coins make cell 8, step down by cell value:by 5 then by 3.
The Coin Problem is a well-researched topic and I would like to reference some papers where you can probably find better solutions:
The Money Changing Problem Revisited
Coin Problem
Also, using C# (statically typed language) will restrict you from having the most efficient algorithm over a dynamically typed language. If you plan to go down that route, you can have a look at this website The Frobenius problem. You can right click and inspect the code (though I really didn't understand much having no experience of javascript)
Anyhow, this is how I would tackle the problem in C#:
private static List<int> _denominations = new List<int>() { 1000, 5000 };
private static int _denominationMin = _denominations[0];
static void Main()
{
bool roundDown = false;
Console.WriteLine("Enter number: ");
int input = Convert.ToInt32(Console.ReadLine());
if(roundDown)
{
for(int i = input; i > _denominationMin; i--)
{
if(Check(0,0,i))
{
Console.WriteLine("Number: {0}", i);
break;
}
}
}
else
{
for (int i = input; i < int.MaxValue; i++)
{
if (Check(0, 0, i))
{
Console.WriteLine("Number: {0}", i);
break;
}
}
}
Console.Read();
}
static bool Check(int highest, int sum, int goal)
{
//Bingo!
if (sum == goal)
{
return true;
}
//Oops! exceeded here
if (sum > goal)
{
return false;
}
// Loop through _denominations.
foreach (int value in _denominations)
{
// Add higher or equal amounts.
if (value >= highest)
{
if(Check(value, sum + value, goal))
{
return true;
}
}
}
return false;
}
Worked well with {4,6} for input 19999, so I don't think it is all that bad. Surely has scope for improvements for not running into Stackoverflow Exception. One could half the input or quarter it. Or subtract a number that has factors whose subset are the denominations. Also, important to have the denominations sorted and contain no multiples of another entry E.x. {4, 6, 8} -> {4, 6}.
Anyhow, if I have time I will try to make it more efficient. Just wanted to provide an alternate solution.

Counting Sort Implementation in C#

I am implementing counting sort But some thing is wrong with my code
I am new in Programming Please help me to find an error.
I am implenting it step by step .
namespace ConsoleApplication1
{
class Program
{
public static int[] a = { 0,0,0,5,4,8,9,9,7,3, 3, 2, 1 };
public static void Sorting()
{
int j = 0, i = 0, smallestvalue = 0, largestvalue = 0, n = a.Length, lengthof_B = 0, temp = 0, anothersmallestvalue;
smallestvalue = largestvalue = a[0];
for (i = 0; i < n; i++)
{
if (smallestvalue > a[i])
{
smallestvalue = a[i];
}
else if (largestvalue < a[i])
{
largestvalue = a[i];
}
}
int x = anothersmallestvalue = smallestvalue;
lengthof_B = largestvalue - smallestvalue + 1;
int[] b = new int[lengthof_B];
for (i = 0; i < lengthof_B && smallestvalue <= largestvalue; i++)
{
for (j = 0; j < n; j++)
{
if (smallestvalue == a[j])
{
b[i] = b[i] + 1;
}
}
b[i] = temp + b[i];
temp = b[i];
smallestvalue++;
}
int[] c = new int[a.Length];
// I think error here
for (i = n - 1; i >= 0; i--)
{
anothersmallestvalue = x;
for (j = 0; j <= lengthof_B ; j++)
{
if (a[i] == anothersmallestvalue)
{
temp = b[j];
c[temp - 1] = anothersmallestvalue;
b[j] = b[j];
}
anothersmallestvalue++;
}
}
for (i = 0; i < c.Length; i++)
{
Console.WriteLine("c[i] : " + c[i]);
}
}
}
class Demo
{
static void Main(string[] args)
{
Program.Sorting();
Console.ReadLine();
}
}
}
Desired Output is
000123457899
But output of my program is
000120457809
This Is Your Code Here I found a mistake.
And your Code is too complex Please Go through your code Once more.
for (i = n - 1; i >= 0; i--)
{
anothersmallestvalue = x;
for (j = 0; j <= lengthof_B ; j++)
{
if (a[i] == anothersmallestvalue)
{
temp = b[j];
c[temp - 1] = anothersmallestvalue;
b[j] = b[j] -1 ;// Possible Mistake I think here
}
anothersmallestvalue++;
}
}
the very simple and stylish way is described and shown here.
en.wikipedia.org/wiki/Counting_sort#The_algorithm
Normal sorting your two loops should look like this
for (i = 0; i < lengthof_B - 1; i++)
{
for (j = i + 1; j < lengthof_B; j++)
{
}
}​

C# InsertionSort 1xforeach 2xfor

I want to create an insertionsort but I can't get any further.
I solved it on my way until I get out of bounds, in the second "for-command" which I can't solve. I'm not sure if I am on the right way, but I want to keep the solution simple.
int arrayzähler = 0;
int[] Speicherarray = new int[randomarray.Length];
//ausgabearray[1] = randomarray[1]; //für vergleich
foreach (int wert in randomarray)
{
if (wert == randomarray[0])
{
Speicherarray[0] = wert;
ausgabearray[0] = wert;
arrayzähler++;
continue; // erster wert = ausgabearray[0]
}
arrayzähler++;
for (int i = arrayzähler - 1; i >= arrayzähler - 1; i--)
{
for (int a = arrayzähler - 2; a >= arrayzähler - 2; a--)
{
if (Speicherarray[i] < Speicherarray[a])
{
Speicherarray[a] = Speicherarray[a + 1];
}
else if (Speicherarray[i] >= Speicherarray[a])
{
Speicherarray[a] = wert;
ausgabearray[i] = Speicherarray[i];
}
}
}
}
This is exact coding for insertion sort,
public void Sort(int[] collection)
{
int inner, temp;
for (int i = 1; i < collection.Length; i++)
{
temp = collection[i];
inner = i;
while (inner > 0 && collection[inner - 1] >= temp)
{
collection[i] = collection[inner - 1];
--inner;
}
collection[inner] = temp;
}
Console.WriteLine("Printing Insertion Sorted Items");
Print();
}

How to find the longest palindrome in a given string? [duplicate]

This question already has answers here:
Write a function that returns the longest palindrome in a given string
(23 answers)
Closed 9 years ago.
Possible Duplicate:
Write a function that returns the longest palindrome in a given string
I know how to do this in O(n^2). But it seems like there exist a better solution.
I've found this, and there is a link to O(n) answer, but it's written in Haskell and not clear for me.
It would be great to get an answer in c# or similar.
I've found clear explanation of the solution here. Thanks to Justin for this link.
There you can find Python and Java implementations of the algorithm (C++ implementation contains errors).
And here is C# implementation that is just a translation of those algorithms.
public static int LongestPalindrome(string seq)
{
int Longest = 0;
List<int> l = new List<int>();
int i = 0;
int palLen = 0;
int s = 0;
int e = 0;
while (i<seq.Length)
{
if (i > palLen && seq[i-palLen-1] == seq[i])
{
palLen += 2;
i += 1;
continue;
}
l.Add(palLen);
Longest = Math.Max(Longest, palLen);
s = l.Count - 2;
e = s - palLen;
bool found = false;
for (int j = s; j > e; j--)
{
int d = j - e - 1;
if (l[j] == d)
{
palLen = d;
found = true;
break;
}
l.Add(Math.Min(d, l[j]));
}
if (!found)
{
palLen = 1;
i += 1;
}
}
l.Add(palLen);
Longest = Math.Max(Longest, palLen);
return Longest;
}
And this is its java version:
public static int LongestPalindrome(String seq) {
int Longest = 0;
List<Integer> l = new ArrayList<Integer>();
int i = 0;
int palLen = 0;
int s = 0;
int e = 0;
while (i < seq.length()) {
if (i > palLen && seq.charAt(i - palLen - 1) == seq.charAt(i)) {
palLen += 2;
i += 1;
continue;
}
l.add(palLen);
Longest = Math.max(Longest, palLen);
s = l.size() - 2;
e = s - palLen;
boolean found = false;
for (int j = s; j > e; j--) {
int d = j - e - 1;
if (l.get(j) == d) {
palLen = d;
found = true;
break;
}
l.add(Math.min(d, l.get(j)));
}
if (!found) {
palLen = 1;
i += 1;
}
}
l.add(palLen);
Longest = Math.max(Longest, palLen);
return Longest;
}
public static string GetMaxPalindromeString(string testingString)
{
int stringLength = testingString.Length;
int maxPalindromeStringLength = 0;
int maxPalindromeStringStartIndex = 0;
for (int i = 0; i < stringLength; i++)
{
int currentCharIndex = i;
for (int lastCharIndex = stringLength - 1; lastCharIndex > currentCharIndex; lastCharIndex--)
{
if (lastCharIndex - currentCharIndex + 1 < maxPalindromeStringLength)
{
break;
}
bool isPalindrome = true;
if (testingString[currentCharIndex] != testingString[lastCharIndex])
{
continue;
}
else
{
int matchedCharIndexFromEnd = lastCharIndex - 1;
for (int nextCharIndex = currentCharIndex + 1; nextCharIndex < matchedCharIndexFromEnd; nextCharIndex++)
{
if (testingString[nextCharIndex] != testingString[matchedCharIndexFromEnd])
{
isPalindrome = false;
break;
}
matchedCharIndexFromEnd--;
}
}
if (isPalindrome)
{
if (lastCharIndex + 1 - currentCharIndex > maxPalindromeStringLength)
{
maxPalindromeStringStartIndex = currentCharIndex;
maxPalindromeStringLength = lastCharIndex + 1 - currentCharIndex;
}
break;
}
}
}
if(maxPalindromeStringLength>0)
{
return testingString.Substring(maxPalindromeStringStartIndex, maxPalindromeStringLength);
}
return null;
}
C#
First I search for even length palindromes. Then I search for odd length palindromes. When it finds a palindrome, it determines the length and sets the max length accordingly. The average case complexity for this is linear.
protected static int LongestPalindrome(string str)
{
int i = 0;
int j = 1;
int oldJ = 1;
int intMax = 1;
int intCount = 0;
if (str.Length == 0) return 0;
if (str.Length == 1) return 1;
int[] intDistance = new int[2] {0,1};
for( int k = 0; k < intDistance.Length; k++ ){
j = 1 + intDistance[k];
oldJ = j;
intCount = 0;
i = 0;
while (j < str.Length)
{
if (str[i].Equals(str[j]))
{
oldJ = j;
intCount = 2 + intDistance[k];
i--;
j++;
while (i >= 0 && j < str.Length)
{
if (str[i].Equals(str[j]))
{
intCount += 2;
i--;
j++;
continue;
}
else
{
break;
}
}
intMax = getMax(intMax, intCount);
j = oldJ + 1;
i = j - 1 - intDistance[k];
}
else
{
i++;
j++;
}
}
}
return intMax;
}
protected static int getMax(int a, int b)
{
if (a > b) return a; return b;
}
Recently I wrote following code during interview...
public string FindMaxLengthPalindrome(string s)
{
string maxLengthPalindrome = "";
if (s == null) return s;
int len = s.Length;
for(int i = 0; i < len; i++)
{
for (int j = 0; j < len - i; j++)
{
bool found = true;
for (int k = j; k < (len - j) / 2; k++)
{
if (s[k] != s[len - (k - j + 1)])
{
found = false;
break;
}
}
if (found)
{
if (len - j > maxLengthPalindrome.Length)
maxLengthPalindrome = s.Substring(j, len - j);
}
if(maxLengthPalindrome.Length >= (len - (i + j)))
break;
}
if (maxLengthPalindrome.Length >= (len - i))
break;
}
return maxLengthPalindrome;
}
I got this question when i took an interview.
I found out when i was back home, unfortunately.
public static string GetMaxPalindromeString(string testingString)
{
int stringLength = testingString.Length;
int maxPalindromeStringLength = 0;
int maxPalindromeStringStartIndex = 0;
for (int i = 0; i < testingString.Length; i++)
{
int currentCharIndex = i;
for (int lastCharIndex = stringLength - 1; lastCharIndex > currentCharIndex; lastCharIndex--)
{
bool isPalindrome = true;
if (testingString[currentCharIndex] != testingString[lastCharIndex])
{
continue;
}
for (int nextCharIndex = currentCharIndex + 1; nextCharIndex < lastCharIndex / 2; nextCharIndex++)
{
if (testingString[nextCharIndex] != testingString[lastCharIndex - 1])
{
isPalindrome = false;
break;
}
}
if (isPalindrome)
{
if (lastCharIndex + 1 - currentCharIndex > maxPalindromeStringLength)
{
maxPalindromeStringStartIndex = currentCharIndex;
maxPalindromeStringLength = lastCharIndex + 1 - currentCharIndex;
}
}
break;
}
}
return testingString.Substring(maxPalindromeStringStartIndex, maxPalindromeStringLength);
}

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