# Program to check whether we can partition a list with k-partitions of equal sum in C++

Suppose we have a list of numbers called nums and another value k, we have to check whether it is possible to partition nums into k different subsets where the sum of each subset are same.

So, if the input is like nums = [4, 2, 6, 5, 1, 6, 3] k = 3, then the output will be True, as we can partition them like: [6, 3], [6, 2, 1], and [4, 5].

To solve this, we will follow these steps −

• Define a function check(), this will take an array v,
• for initialize i := 1, when i < size of v, update (increase i by 1), do −
• if v[i] is not equal to v, then −
• return false
• return true
• Define a function dfs(), this will take idx, an array nums, an array temp,
• if idx is same as size of nums, then −
• return check(temp)
• ret := false
• for initialize i := 0, when i < size of temp, update (increase i by 1), do −
• temp[i] := temp[i] + nums[idx]
• ret := dfs(idx + 1, nums, temp)
• if ret is true, then −
• return true
• temp[i] := temp[i] - nums[idx]
• return false
• From the main method do the following −
• Define an array temp of size k
• return dfs(0, nums, temp)

## Example (C++)

Let us see the following implementation to get better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
bool check(vector<int>& v) {
for (int i = 1; i < v.size(); i++) {
if (v[i] != v)
return false;
}
return true;
}
bool dfs(int idx, vector<int>& nums, vector<int>& temp) {
if (idx == nums.size()) {
return check(temp);
}
bool ret = false;
for (int i = 0; i < temp.size(); i++) {
temp[i] += nums[idx];
ret = dfs(idx + 1, nums, temp);
if (ret)
return true;
temp[i] -= nums[idx];
}
return false;
}
bool solve(vector<int>& nums, int k) {
vector<int> temp(k);
return dfs(0, nums, temp);
}
};
bool solve(vector<int>& nums, int k) {
return (new Solution())->solve(nums, k);
}
int main(){
vector<int> v = {4, 2, 6, 5, 1, 6, 3};
int k = 3;
cout << solve(v, 3);
}

## Input

{4, 2, 6, 5, 1, 6, 3}, 3

## Output

1