Given an integer array of N elements, the task is to divide this array into K non-empty subsets such that the sum of elements in every subset is same. All elements of this array should be part of exactly one partition.

Examples:

Input : arr = [2, 1, 4, 5, 6], K = 3 Output : Yes we can divide above array into 3 parts with equal sum as [[2, 4], [1, 5], [6]] Input : arr = [2, 1, 5, 5, 6], K = 3 Output : No It is not possible to divide above array into 3 parts with equal sum

We can solve this problem recursively, we keep an array for sum of each partition and a boolean array to check whether an element is already taken into some partition or not.

First we need to check some base cases,

If K is 1, then we already have our answer, complete array is only subset with same sum.

If N < K, then it is not possible to divide array into subsets with equal sum, because we can’t divide the array into more than N parts.

If sum of array is not divisible by K, then it is not possible to divide the array. We will proceed only if k divides sum. Our goal reduces to divide array into K parts where sum of each part should be array_sum/K

In below code a recursive method is written which tries to add array element into some subset. If sum of this subset reaches required sum, we iterate for next part recursively, otherwise we backtrack for different set of elements. If number of subsets whose sum reaches the required sum is (K-1), we flag that it is possible to partition array into K parts with equal sum, because remaining elements already have a sum equal to required sum.

`// C++ program to check whether an array can be ` `// partitioned into K subsets of equal sum ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Recursive Utility method to check K equal sum ` `// subsetition of array ` `/** ` ` ` `array - given input array ` ` ` `subsetSum array - sum to store each subset of the array ` ` ` `taken - boolean array to check whether element ` ` ` `is taken into sum partition or not ` ` ` `K - number of partitions needed ` ` ` `N - total number of element in array ` ` ` `curIdx - current subsetSum index ` ` ` `limitIdx - lastIdx from where array element should ` ` ` `be taken */` `bool` `isKPartitionPossibleRec(` `int` `arr[], ` `int` `subsetSum[], ` `bool` `taken[], ` ` ` `int` `subset, ` `int` `K, ` `int` `N, ` `int` `curIdx, ` `int` `limitIdx) ` `{ ` ` ` `if` `(subsetSum[curIdx] == subset) ` ` ` `{ ` ` ` `/* current index (K - 2) represents (K - 1) subsets of equal ` ` ` `sum last partition will already remain with sum 'subset'*/` ` ` `if` `(curIdx == K - 2) ` ` ` `return` `true` `; ` ` ` ` ` `// recursive call for next subsetition ` ` ` `return` `isKPartitionPossibleRec(arr, subsetSum, taken, subset, ` ` ` `K, N, curIdx + 1, N - 1); ` ` ` `} ` ` ` ` ` `// start from limitIdx and include elements into current partition ` ` ` `for` `(` `int` `i = limitIdx; i >= 0; i--) ` ` ` `{ ` ` ` `// if already taken, continue ` ` ` `if` `(taken[i]) ` ` ` `continue` `; ` ` ` `int` `tmp = subsetSum[curIdx] + arr[i]; ` ` ` ` ` `// if temp is less than subset then only include the element ` ` ` `// and call recursively ` ` ` `if` `(tmp <= subset) ` ` ` `{ ` ` ` `// mark the element and include into current partition sum ` ` ` `taken[i] = ` `true` `; ` ` ` `subsetSum[curIdx] += arr[i]; ` ` ` `bool` `nxt = isKPartitionPossibleRec(arr, subsetSum, taken, ` ` ` `subset, K, N, curIdx, i - 1); ` ` ` ` ` `// after recursive call unmark the element and remove from ` ` ` `// subsetition sum ` ` ` `taken[i] = ` `false` `; ` ` ` `subsetSum[curIdx] -= arr[i]; ` ` ` `if` `(nxt) ` ` ` `return` `true` `; ` ` ` `} ` ` ` `} ` ` ` `return` `false` `; ` `} ` ` ` `// Method returns true if arr can be partitioned into K subsets ` `// with equal sum ` `bool` `isKPartitionPossible(` `int` `arr[], ` `int` `N, ` `int` `K) ` `{ ` ` ` `// If K is 1, then complete array will be our answer ` ` ` `if` `(K == 1) ` ` ` `return` `true` `; ` ` ` ` ` `// If total number of partitions are more than N, then ` ` ` `// division is not possible ` ` ` `if` `(N < K) ` ` ` `return` `false` `; ` ` ` ` ` `// if array sum is not divisible by K then we can't divide ` ` ` `// array into K partitions ` ` ` `int` `sum = 0; ` ` ` `for` `(` `int` `i = 0; i < N; i++) ` ` ` `sum += arr[i]; ` ` ` `if` `(sum % K != 0) ` ` ` `return` `false` `; ` ` ` ` ` `// the sum of each subset should be subset (= sum / K) ` ` ` `int` `subset = sum / K; ` ` ` `int` `subsetSum[K]; ` ` ` `bool` `taken[N]; ` ` ` ` ` `// Initialize sum of each subset from 0 ` ` ` `for` `(` `int` `i = 0; i < K; i++) ` ` ` `subsetSum[i] = 0; ` ` ` ` ` `// mark all elements as not taken ` ` ` `for` `(` `int` `i = 0; i < N; i++) ` ` ` `taken[i] = ` `false` `; ` ` ` ` ` `// initialize first subsubset sum as last element of ` ` ` `// array and mark that as taken ` ` ` `subsetSum[0] = arr[N - 1]; ` ` ` `taken[N - 1] = ` `true` `; ` ` ` ` ` `// call recursive method to check K-substitution condition ` ` ` `return` `isKPartitionPossibleRec(arr, subsetSum, taken, ` ` ` `subset, K, N, 0, N - 1); ` `} ` ` ` `// Driver code to test above methods ` `int` `main() ` `{ ` ` ` `int` `arr[] = {2, 1, 4, 5, 3, 3}; ` ` ` `int` `N = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` `int` `K = 3; ` ` ` ` ` `if` `(isKPartitionPossible(arr, N, K)) ` ` ` `cout << ` ```
"Partitions into equal sum is possible.
"
``` `; ` ` ` `else` ` ` `cout << ` ```
"Partitions into equal sum is not possible.
"
``` `; ` `} ` |

Output:

Partitions into equal sum is possible.

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