I need to "Find the minimal positive integer not occurring in a given sequence. "
A[0] = 1
A[1] = 3
A[2] = 6
A[3] = 4
A[4] = 1
A[5] = 2, the function should return 5.
Assume that:
N is an integer within the range [1..100,000];
each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].
I wrote the code in codility, but for many cases it did not worked and the performance test gives 0 %. Please help me out, where I am wrong.
class Solution {
public int solution(int[] A) {
if(A.Length ==0) return -1;
int value = A[0];
int min = A.Min();
int max = A.Max();
for (int j = min+1; j < max; j++)
{
if (!A.Contains(j))
{
value = j;
if(value > 0)
{
break;
}
}
}
if(value > 0)
{
return value;
}
else return 1;
}
}
The codility gives error with all except the example, positive and negative only values.
Edit: Added detail to answer your actual question more directly.
"Please help me out, where I am wrong."
In terms of correctness: Consider A = {7,2,5,6,3}. The correct output, given the contents of A, is 1, but our algorithm would fail to detect this since A.Min() would return 2 and we would start looping from 3 onward. In this case, we would return 4 instead; since it's the next missing value.
Same goes for something like A = {14,15,13}. The minimal missing positive integer here is again 1 and, since all the values from 13-15 are present, the value variable will retain its initial value of value=A[0] which would be 14.
In terms of performance: Consider what A.Min(), A.Max() and A.Contains() are doing behind the scenes; each one of these is looping through A in its entirety and in the case of Contains, we are calling it repeatedly for every value between the Min() and the lowest positive integer we can find. This will take us far beyond the specified O(N) performance that Codility is looking for.
By contrast, here's the simplest version I can think of that should score 100% on Codility. Notice that we only loop through A once and that we take advantage of a Dictionary which lets us use ContainsKey; a much faster method that does not require looping through the whole collection to find a value.
using System;
using System.Collections.Generic;
class Solution {
public int solution(int[] A) {
// the minimum possible answer is 1
int result = 1;
// let's keep track of what we find
Dictionary<int,bool> found = new Dictionary<int,bool>();
// loop through the given array
for(int i=0;i<A.Length;i++) {
// if we have a positive integer that we haven't found before
if(A[i] > 0 && !found.ContainsKey(A[i])) {
// record the fact that we found it
found.Add(A[i], true);
}
}
// crawl through what we found starting at 1
while(found.ContainsKey(result)) {
// look for the next number
result++;
}
// return the smallest positive number that we couldn't find.
return result;
}
}
The simplest solution that scored perfect score was:
public int solution(int[] A)
{
int flag = 1;
A = A.OrderBy(x => x).ToArray();
for (int i = 0; i < A.Length; i++)
{
if (A[i] <= 0)
continue;
else if (A[i] == flag)
{
flag++;
}
}
return flag;
}
Fastest C# solution so far for [-1,000,000...1,000,000].
public int solution(int[] array)
{
HashSet<int> found = new HashSet<int>();
for (int i = 0; i < array.Length; i++)
{
if (array[i] > 0)
{
found.Add(array[i]);
}
}
int result = 1;
while (found.Contains(result))
{
result++;
}
return result;
}
A tiny version of another 100% with C#
using System.Linq;
class Solution
{
public int solution(int[] A)
{
// write your code in C# 6.0 with .NET 4.5 (Mono)
var i = 0;
return A.Where(a => a > 0).Distinct().OrderBy(a => a).Any(a => a != (i = i + 1)) ? i : i + 1;
}
}
A simple solution that scored 100% with C#
int Solution(int[] A)
{
var A2 = Enumerable.Range(1, A.Length + 1);
return A2.Except(A).First();
}
public class Solution {
public int solution( int[] A ) {
return Arrays.stream( A )
.filter( n -> n > 0 )
.sorted()
.reduce( 0, ( a, b ) -> ( ( b - a ) > 1 ) ? a : b ) + 1;
}
}
It seemed easiest to just filter out the negative numbers. Then sort the stream. And then reduce it to come to an answer. It's a bit of a functional approach, but it got a 100/100 test score.
Got an 100% score with this solution:
https://app.codility.com/demo/results/trainingUFKJSB-T8P/
public int MissingInteger(int[] A)
{
A = A.Where(a => a > 0).Distinct().OrderBy(c => c).ToArray();
if (A.Length== 0)
{
return 1;
}
for (int i = 0; i < A.Length; i++)
{
//Console.WriteLine(i + "=>" + A[i]);
if (i + 1 != A[i])
{
return i + 1;
}
}
return A.Max() + 1;
}
JavaScript solution using Hash Table with O(n) time complexity.
function solution(A) {
let hashTable = {}
for (let item of A) {
hashTable[item] = true
}
let answer = 1
while(true) {
if(!hashTable[answer]) {
return answer
}
answer++
}
}
The Simplest solution for C# would be:
int value = 1;
int min = A.Min();
int max = A.Max();
if (A.Length == 0) return value = 1;
if (min < 0 && max < 0) return value = 1;
List<int> range = Enumerable.Range(1, max).ToList();
List<int> current = A.ToList();
List<int> valid = range.Except(current).ToList();
if (valid.Count() == 0)
{
max++;
return value = max;
}
else
{
return value = valid.Min();
}
Considering that the array should start from 1 or if it needs to start from the minimum value than the Enumerable.range should start from Min
MissingInteger solution in C
int solution(int A[], int N) {
int i=0,r[N];
memset(r,0,(sizeof(r)));
for(i=0;i<N;i++)
{
if(( A[i] > 0) && (A[i] <= N)) r[A[i]-1]=A[i];
}
for(i=0;i<N;i++)
{
if( r[i] != (i+1)) return (i+1);
}
return (N+1);
}
My solution for it:
public static int solution()
{
var A = new[] { -1000000, 1000000 }; // You can try with different integers
A = A.OrderBy(i => i).ToArray(); // We sort the array first
if (A.Length == 1) // if there is only one item in the array
{
if (A[0]<0 || A[0] > 1)
return 1;
if (A[0] == 1)
return 2;
}
else // if there are more than one item in the array
{
for (var i = 0; i < A.Length - 1; i++)
{
if (A[i] >= 1000000) continue; // if it's bigger than 1M
if (A[i] < 0 || (A[i] + 1) >= (A[i + 1])) continue; //if it's smaller than 0, if the next integer is bigger or equal to next integer in the sequence continue searching.
if (1 < A[0]) return 1;
return A[i] + 1;
}
}
if (1 < A[0] || A[A.Length - 1] + 1 == 0 || A[A.Length - 1] + 1 > 1000000)
return 1;
return A[A.Length-1] +1;
}
class Solution {
public int solution(int[] A) {
int size=A.length;
int small,big,temp;
for (int i=0;i<size;i++){
for(int j=0;j<size;j++){
if(A[i]<A[j]){
temp=A[j];
A[j]=A[i];
A[i]=temp;
}
}
}
int z=1;
for(int i=0;i<size;i++){
if(z==A[i]){
z++;
}
//System.out.println(a[i]);
}
return z;
}
enter code here
}
In C# you can solve the problem by making use of built in library functions. How ever the performance is low for very large integers
public int solution(int[] A)
{
var numbers = Enumerable.Range(1, Math.Abs(A.Max())+1).ToArray();
return numbers.Except(A).ToArray()[0];
}
Let me know if you find a better solution performance wise
C# - MissingInteger
Find the smallest missing integer between 1 - 1000.000.
Assumptions of the OP take place
TaskScore/Correctness/Performance: 100%
using System;
using System.Linq;
namespace TestConsole
{
class Program
{
static void Main(string[] args)
{
var A = new int[] { -122, -5, 1, 2, 3, 4, 5, 6, 7 }; // 8
var B = new int[] { 1, 3, 6, 4, 1, 2 }; // 5
var C = new int[] { -1, -3 }; // 1
var D = new int[] { -3 }; // 1
var E = new int[] { 1 }; // 2
var F = new int[] { 1000000 }; // 1
var x = new int[][] { A, B, C, D, E, F };
x.ToList().ForEach((arr) =>
{
var s = new Solution();
Console.WriteLine(s.solution(arr));
});
Console.ReadLine();
}
}
// ANSWER/SOLUTION
class Solution
{
public int solution(int[] A)
{
// clean up array for negatives and duplicates, do sort
A = A.Where(entry => entry > 0).Distinct().OrderBy(it => it).ToArray();
int lowest = 1, aLength = A.Length, highestIndex = aLength - 1;
for (int i = 0; i < aLength; i++)
{
var currInt = A[i];
if (currInt > lowest) return lowest;
if (i == highestIndex) return ++lowest;
lowest++;
}
return 1;
}
}
}
Got 100% - C# Efficient Solution
public int solution (int [] A){
int len = A.Length;
HashSet<int> realSet = new HashSet<int>();
HashSet<int> perfectSet = new HashSet<int>();
int i = 0;
while ( i < len)
{
realSet.Add(A[i]); //convert array to set to get rid of duplicates, order int's
perfectSet.Add(i + 1); //create perfect set so can find missing int
i++;
}
perfectSet.Add(i + 1);
if (realSet.All(item => item < 0))
return 1;
int notContains =
perfectSet.Except(realSet).Where(item=>item!=0).FirstOrDefault();
return notContains;
}
class Solution {
public int solution(int[] a) {
int smallestPositive = 1;
while(a.Contains(smallestPositive)) {
smallestPositive++;
}
return smallestPositive;
}
}
Well, this is a new winner now. At least on C# and my laptop. It's 1.5-2 times faster than the previous champion and 3-10 times faster, than most of the other solutions. The feature (or a bug?) of this solution is that it uses only basic data types. Also 100/100 on Codility.
public int Solution(int[] A)
{
bool[] B = new bool[(A.Length + 1)];
for (int i = 0; i < A.Length; i++)
{
if ((A[i] > 0) && (A[i] <= A.Length))
B[A[i]] = true;
}
for (int i = 1; i < B.Length; i++)
{
if (!B[i])
return i;
}
return A.Length + 1;
}
Simple C++ solution. No additional memory need, time execution order O(N*log(N)):
int solution(vector<int> &A) {
sort (A.begin(), A.end());
int prev = 0; // the biggest integer greater than 0 found until now
for( auto it = std::begin(A); it != std::end(A); it++ ) {
if( *it > prev+1 ) break;// gap found
if( *it > 0 ) prev = *it; // ignore integers smaller than 1
}
return prev+1;
}
int[] A = {1, 3, 6, 4, 1, 2};
Set<Integer> integers = new TreeSet<>();
for (int i = 0; i < A.length; i++) {
if (A[i] > 0) {
integers.add(A[i]);
}
}
Integer[] arr = integers.toArray(new Integer[0]);
final int[] result = {Integer.MAX_VALUE};
final int[] prev = {0};
final int[] curr2 = {1};
integers.stream().forEach(integer -> {
if (prev[0] + curr2[0] == integer) {
prev[0] = integer;
} else {
result[0] = prev[0] + curr2[0];
}
});
if (Integer.MAX_VALUE == result[0]) result[0] = arr[arr.length-1] + 1;
System.out.println(result[0]);
I was surprised but this was a good lesson. LINQ IS SLOW. my answer below got me 11%
public int solution (int [] A){
if (Array.FindAll(A, x => x >= 0).Length == 0) {
return 1;
} else {
var lowestValue = A.Where(x => Array.IndexOf(A, (x+1)) == -1).Min();
return lowestValue + 1;
}
}
I think I kinda look at this a bit differently but gets a 100% evaluation. Also, I used no library:
public static int Solution(int[] A)
{
var arrPos = new int[1_000_001];
for (int i = 0; i < A.Length; i++)
{
if (A[i] >= 0)
arrPos[A[i]] = 1;
}
for (int i = 1; i < arrPos.Length; i++)
{
if (arrPos[i] == 0)
return i;
}
return 1;
}
public int solution(int[] A) {
// write your code in Java SE 8
Set<Integer> elements = new TreeSet<Integer>();
long lookFor = 1;
for (int i = 0; i < A.length; i++) {
elements.add(A[i]);
}
for (Integer integer : elements) {
if (integer == lookFor)
lookFor += 1;
}
return (int) lookFor;
}
I tried to use recursion in C# instead of sorting, because I thought it would show more coding skill to do it that way, but on the scaling tests it didn't preform well on large performance tests. Suppose it's best to just do the easy way.
class Solution {
public int lowest=1;
public int solution(int[] A) {
// write your code in C# 6.0 with .NET 4.5 (Mono)
if (A.Length < 1)
return 1;
for (int i=0; i < A.Length; i++){
if (A[i]==lowest){
lowest++;
solution(A);
}
}
return lowest;
}
}
Here is my solution in javascript
function solution(A) {
// write your code in JavaScript (Node.js 8.9.4)
let result = 1;
let haveFound = {}
let len = A.length
for (let i=0;i<len;i++) {
haveFound[`${A[i]}`] = true
}
while(haveFound[`${result}`]) {
result++
}
return result
}
class Solution {
public int solution(int[] A) {
var sortedList = A.Where(x => x > 0).Distinct().OrderBy(x => x).ToArray();
var output = 1;
for (int i = 0; i < sortedList.Length; i++)
{
if (sortedList[i] != output)
{
return output;
}
output++;
}
return output;
}
}
You should just use a HashSet as its look up time is also constant instead of a dictionary. The code is less and cleaner.
public int solution (int [] A){
int answer = 1;
var set = new HashSet<int>(A);
while (set.Contains(answer)){
answer++;
}
return answer;
}
This snippet should work correctly.
using System;
using System.Collections.Generic;
public class Program
{
public static void Main()
{
int result = 1;
List<int> lst = new List<int>();
lst.Add(1);
lst.Add(2);
lst.Add(3);
lst.Add(18);
lst.Add(4);
lst.Add(1000);
lst.Add(-1);
lst.Add(-1000);
lst.Sort();
foreach(int curVal in lst)
{
if(curVal <=0)
result=1;
else if(!lst.Contains(curVal+1))
{
result = curVal + 1 ;
}
Console.WriteLine(result);
}
}
}
I'm trying to write code that returns whether or not a given integer is or is not divisible evenly by 1 to 20, but i keep receiving
error CS0161: 'ProblemFive.isTwenty(int)': not all code paths return a value"
Here is my code:
public static bool isTwenty(int num)
{
for(int j = 1; j <= 20; j++)
{
if(num % j != 0)
{
return false;
}
else if(num % j == 0 && num == 20)
{
return true;
}
}
}
The fact you see that method always returns something does not mean compiler can see it.
Compiler does only simple checks and in your example logic is too complicated for compiler to see that.
It's not even smart enough to allow that kind of method:
public static bool isTwenty(int num)
{
for (int j = 1; j <= 20; j++)
{
return true;
}
}
Solution
Add another return at the end of the method:
public static bool isTwenty(int num)
{
for(int j = 1; j <= 20; j++)
{
if(num % j != 0)
{
return false;
}
else if(num % j == 0 && num == 20)
{
return true;
}
}
return true;
}
But in this case you can just make your for loop a little different and delete else if part:
public static bool isTwenty(int num)
{
for(int j = 1; j <= 20; j++)
{
if(num % j != 0)
{
return false;
}
}
return true;
}
You have an if and an else if but no else. It is not guaranteed that either one of the if or else if will always be evaluated.
You have to return a value from the method also after the loop.
In fact, your else if condition is the same as if you just finished the iterations through the loop, so it's simpler just to do this:
public static bool isDivisibleByAllLessThanOrEqualToTwenty(int num) {
for (int j = 1; j <= 20; j++) {
if(num % j != 0) {
return false;
}
}
return true;
}
or just a little one-liner (useless of course), to avoid forgetting a return
public static bool isTwenty(int num)
{
return Enumerable.Range(1, 20).All(x => num % x == 0);
}
You need to consider what happens if neither of the two conditions are met for all iterations of j.
Edit: A more efficient way of determining whether a number is exactly divisible by all numbers in a fixed range is to take the least common multiple (LCM) of your range, and simply check whether the said number is divisible by the LCM.
public static bool isTwenty(int num)
{
// compile-time computation
const int lcm = 2 * 2 * 2 * 2 * 3 * 3 * 5 * 7 * 11 * 13 * 17 * 19;
return num % lcm == 0;
}
Others have answered why in terms of code, but perhaps an example will help?
What is the response from your code for the following inputs?
isTwenty(0);
isTwenty(21);
isTwenty(-1);
Another solution if you want to keep your logic - replace for with well-known infinite loop since you know that it will always exit. Compiler will understand that and let you keep return just inside if:
public static bool isTwenty(int num)
{
for(int j = 1; true; j++)
{
if(num % j != 0)
{
return false;
}
else if(num % j == 0 && num == 20)
{
return true;
}
}
}
Or if while works better:
var j = 1;
while(true)
{
if (...){...}
j++;
}
I have a code here written in C# that finds the smallest multiple by all numbers from 1 to 20. However, I find it very inefficient since the execution took awhile before producing the correct answer. I would like to know what are the different ways that I can do to improve the code. Thank You.
public static void SmallestMultiple()
{
const ushort ARRAY_SIZE = 21;
ushort[] array = new ushort[ARRAY_SIZE];
ushort check = 0;
for (uint value = 1; value < uint.MaxValue; value++)
{
for (ushort j = 1; j < ARRAY_SIZE; j++)
{
array[j] = j;
if (value % array[j] == 0)
{
check++;
}
}
if (check == 20)
{
Console.WriteLine("The value is {0}", value);
}
else
{
check = 0;
}
}
}
static void Main(string[] args)
{
int[] nums = Enumerable.Range(1, 20).ToArray();
int lcm = 1;
for (int i = 0; i < nums.Length; i++)
{
lcm = LCM(lcm, nums[i]);
}
Console.WriteLine("LCM = {0}", lcm);
}
public static int LCM(int value1, int value2)
{
int a = Math.Abs(value1);
int b = Math.Abs(value2);
// perform division first to avoid potential overflow
a = checked((a / GCD(a, b)));
return checked((a * b));
}
public static int GCD(int value1, int value2)
{
int gcd = 1; // Greates Common Divisor
// throw exception if any value=0
if (value1 == 0 || value2 == 0)
{
throw new ArgumentOutOfRangeException();
}
// assign absolute values to local vars
int a = Math.Abs(value1); // local var1
int b = Math.Abs(value2); // local var2
// if numbers are equal return the first
if (a == b) { return a; }
// if var "b" is GCD return "b"
if (a > b && a % b == 0) { return b; }
// if var "a" is GCD return "a"
if (b > a && b % a == 0) { return a; }
// Euclid algorithm to find GCD (a,b):
// estimated maximum iterations:
// 5* (number of dec digits in smallest number)
while (b != 0)
{
gcd = b;
b = a % b;
a = gcd;
}
return gcd;
}
}
Source : Fast Integer Algorithms: Greatest Common Divisor and Least Common Multiple, .NET solution
Since the result must also be divisible by 19 (which is the greatest prime number) up to 20, you might only cycle through multiples of 19.
This should get to to the result about 19 times faster.
Here's the code that does this:
public static void SmallestMultiple()
{
const ushort ARRAY_SIZE = 21;
ushort[] array = new ushort[ARRAY_SIZE];
ushort check = 0;
for (uint value = 19; value < uint.MaxValue; value += 19)
{
for (ushort j = 1; j < ARRAY_SIZE; j++)
{
array[j] = j;
if (value % array[j] == 0)
{
check++;
}
}
if (check == 20)
{
Console.WriteLine("The value is {0}", value);
return;
}
else
{
check = 0;
}
}
}
On my machine, this finds the result 232792560 in a little over 2 seconds.
Update
Also, please note that the initial program did not stop when reaching a solution; I have added a return statement to make it stop.
You're just looking for the LCM of the numbers from 1 to 20:
Where the GCD can be efficiently calculated with the Euclidean algorithm.
I don't know C#, but this Python solution shouldn't be hard to translate:
def gcd(a, b):
while b != 0:
a, b = b, a % b
return a
def lcm(a, b):
return (a * b) / gcd(a, b)
numbers = range(1, 20 + 1)
print reduce(numbers, lcm)
It's pretty fast too:
>>> %timeit reduce(lcm, range(1, 20000))
1 loops, best of 3: 258 ms per loop
EDIT: v2.0 - Major speed improvement
Building on w0lf's solution. A faster solution:
public static void SmallestMultiple()
{
// this is a bit quick and dirty
// (not too difficult to change to generate primeProduct dynamically for any range)
int primeProduct = 2*3*5*7*11*13*17*19;
for (int value = primeProduct; ; value += primeProduct)
{
bool success = true;
for (int j = 11; j < 21; j++)
{
if (value % j != 0)
{
success = false;
break;
}
}
if (success)
{
Console.WriteLine("The value is {0}", value);
break;
}
}
}
You needn't check 1-10 since if something is divisible by x (e.g. 12), it is divisible by x/n (e.g. 12/2 = 6). The smallest multiple will always be a multiple of a product of all the primes involved.
Didn't benchmark C# solution, but equivalent Java solution runs in about 0.0000006 seconds.
Well I'm not sure what exactly you are trying to accomplish here but your out side for loop will run approximately 4,294,967,295 time (uint.MaxValue). So that will take some time...
If you have a way to keep from going to uint.MaxValue - like breaking your loop when you have accomplished what you need to - that will speed it up.
Also, since you are setting array[j] equal to j and then apparently never using the array again why not just do:
value % j
instead of
value % array[j]
Using also code written by W0lf (sorry but i cannot comment on your post) I would improve it (only a little) deleting the array that I think is useless..
public static void SmallestMultiple()
{
const ushort ARRAY_SIZE = 21;
ushort check = 0;
for (uint value = 1; value < uint.MaxValue; value++)
{
for (ushort j = 1; j < ARRAY_SIZE; j++)
{
if (value % j == 0)
{
check++;
}
}
if (check == 20)
{
Console.WriteLine("The value is {0}", value);
}
else
{
check = 0;
}
}
}
I have a 5x5 table of values from 0 to 3 inclusive with all values unknown. I know both the sum of the values and the number of zeros for each row and column. How would I go about solving this 0-1 knapsack problem using C# and retrieving the possible solutions that satisfy the known sums and number of zeros? The tables will always be five rows and five columns, so it's not quite a traditional knapsack.
For example, say we input:
Row[0]: Sum=4, Zeros=1
[1]: Sum=5, Zeros=1
[2]: Sum=4, Zeros=2
[3]: Sum=8, Zeros=0
[4]: Sum=3, Zeros=2
Col[0]: Sum=5, Zeros=1
[1]: Sum=3, Zeros=2
[2]: Sum=4, Zeros=2
[3]: Sum=5, Zeros=1
[4]: Sum=7, Zeros=0
We would get this as a possible solution:
[[ 0 1 1 1 1 ]
[ 1 0 2 1 1 ]
[ 2 1 0 0 1 ]
[ 1 1 1 2 3 ]
[ 1 0 0 1 1 ]]
What type of algorithm should I employ in this rather strange situation? Would I also have to write a class just to enumerate the permutations?
Edit for clarification: the problem isn't that I can't enumerate the possibilities; it's that I have no clue how to go about efficiently determining the permutations adding to an arbitrary sum while containing the specified number of zeros and a maximum of 5 items.
Here there is the code. If you need any comment feel free to ask:
using System;
using System.Diagnostics;
namespace ConsoleApplication15
{
class Program
{
static void Main(string[] args)
{
RowOrCol[] rows = new RowOrCol[] {
new RowOrCol(4, 1),
new RowOrCol(5, 1),
new RowOrCol(4, 2),
new RowOrCol(8, 0),
new RowOrCol(3, 2),
};
RowOrCol[] cols = new RowOrCol[] {
new RowOrCol(5, 1),
new RowOrCol(3, 2),
new RowOrCol(4, 2),
new RowOrCol(5, 1),
new RowOrCol(7, 0),
};
int[,] table = new int[5, 5];
Stopwatch sw = Stopwatch.StartNew();
int solutions = Do(table, rows, cols, 0, 0);
sw.Stop();
Console.WriteLine();
Console.WriteLine("Found {0} solutions in {1}ms", solutions, sw.ElapsedMilliseconds);
Console.ReadKey();
}
public static int Do(int[,] table, RowOrCol[] rows, RowOrCol[] cols, int row, int col)
{
int solutions = 0;
int oldValueRowSum = rows[row].Sum;
int oldValueRowZero = rows[row].Zeros;
int oldValueColSum = cols[col].Sum;
int oldValueColZero = cols[col].Zeros;
int nextCol = col + 1;
int nextRow;
bool last = false;
if (nextCol == cols.Length)
{
nextCol = 0;
nextRow = row + 1;
if (nextRow == rows.Length)
{
last = true;
}
}
else
{
nextRow = row;
}
int i;
for (i = 0; i <= 3; i++)
{
table[row, col] = i;
if (i == 0)
{
rows[row].Zeros--;
cols[col].Zeros--;
if (rows[row].Zeros < 0)
{
continue;
}
if (cols[col].Zeros < 0)
{
continue;
}
}
else
{
if (i == 1)
{
rows[row].Zeros++;
cols[col].Zeros++;
}
rows[row].Sum--;
cols[col].Sum--;
if (rows[row].Sum < 0)
{
break;
}
else if (cols[col].Sum < 0)
{
break;
}
}
if (col == cols.Length - 1)
{
if (rows[row].Sum != 0 || rows[row].Zeros != 0)
{
continue;
}
}
if (row == rows.Length - 1)
{
if (cols[col].Sum != 0 || cols[col].Zeros != 0)
{
continue;
}
}
if (!last)
{
solutions += Do(table, rows, cols, nextRow, nextCol);
}
else
{
solutions++;
Console.WriteLine("Found solution:");
var sums = new int[cols.Length];
var zeross = new int[cols.Length];
for (int j = 0; j < rows.Length; j++)
{
int sum = 0;
int zeros = 0;
for (int k = 0; k < cols.Length; k++)
{
Console.Write("{0,2} ", table[j, k]);
if (table[j, k] == 0)
{
zeros++;
zeross[k]++;
}
else
{
sum += table[j, k];
sums[k] += table[j, k];
}
}
Console.WriteLine("| Sum {0,2} | Zeros {1}", sum, zeros);
Debug.Assert(sum == rows[j].OriginalSum);
Debug.Assert(zeros == rows[j].OriginalZeros);
}
Console.WriteLine("---------------");
for (int j = 0; j < cols.Length; j++)
{
Console.Write("{0,2} ", sums[j]);
Debug.Assert(sums[j] == cols[j].OriginalSum);
}
Console.WriteLine();
for (int j = 0; j < cols.Length; j++)
{
Console.Write("{0,2} ", zeross[j]);
Debug.Assert(zeross[j] == cols[j].OriginalZeros);
}
Console.WriteLine();
}
}
// The for cycle was broken at 0. We have to "readjust" the zeros.
if (i == 0)
{
rows[row].Zeros++;
cols[col].Zeros++;
}
// The for cycle exited "normally". i is too much big because the true last cycle was at 3.
if (i == 4)
{
i = 3;
}
// We readjust the sums.
rows[row].Sum += i;
cols[col].Sum += i;
Debug.Assert(oldValueRowSum == rows[row].Sum);
Debug.Assert(oldValueRowZero == rows[row].Zeros);
Debug.Assert(oldValueColSum == cols[col].Sum);
Debug.Assert(oldValueColZero == cols[col].Zeros);
return solutions;
}
}
public class RowOrCol
{
public readonly int OriginalSum;
public readonly int OriginalZeros;
public int Sum;
public int Zeros;
public RowOrCol(int sum, int zeros)
{
this.Sum = this.OriginalSum = sum;
this.Zeros = this.OriginalZeros = zeros;
}
}
}
How fast does it have to be? I just tested a naive "try pretty much anything" with some early aborts but less than would be possible, and it was pretty fast (less than a millisecond). It gave the solution:
[[ 0 1 1 1 1 ]
[ 1 0 1 1 2 ]
[ 1 0 0 1 2 ]
[ 2 1 2 2 1 ]
[ 1 1 0 0 1 ]]
If that's an acceptable solution to you, I can post the code (or just discuss it, it's quite verbose but the underlying idea is trivial)
edit: it is also trivially extendable to enumerating all solutions. It found 400 of them in 15 milliseconds, and claims that there are no more than that. Is that correct?
What I did, was start at 0,0 and try all values I could fill in at that place (0 though min(3, rowsum[0])), fill it it (subtracting it from rowsum[y] and colsum[x] and subtracting one from rowzero[y] and colzero[x] if the value was zero), then recursively do this for 0,1; 0,2; 0,3; then at 0,4 I have a special case where I just fill in the remaining rowsum if it is non-negative (otherwise, abort the current try - ie go up in the recursion tree), and something similar for when y=4. In the mean time, I abort when any rowsum colsum colzero or rowzero becomes negative.
The current board is a solution if and only if all remaining rowsums columnsums colzero's and rowzero's are zero. So I just test for that, and add it to the solutions if it is one. It won't have any negative entries by construction.