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Partition to K Equal Sum Subsets in C++
Suppose we have an array of integers called nums and a positive integer k, check whether it's possible to divide this array into k non-empty subsets whose sums are all same. So if the array is like [4,3,2,3,5,2,1] and k = 4, then the result will be True, as the given array can be divided into four subarray like [[5], [1,4], [2,3], [2,3]] with equal sums.
To solve this, we will follow these steps −
- define two table called dp and total of size 2^n,
- sort the given array nums, set sum := sum of all elements in the nums array
- if sum mod k is non 0 OR last element of nums > sum / k, then return false
- set dp[0] := true and sum := sum / k
- for i in range 0 to 2^n
- if dp[i] is non zero, then
- for j in range 0 to ,
- temp := i OR 2 ^ j
- if temp is not same as i, then
- if nums[j] <= sum – total[i] mod sum, then dp[temp] := true
- total[temp] := total[i] + nums[j]
- else come out from the loop
- for j in range 0 to ,
- if dp[i] is non zero, then
- return dp[(2^n) - 1]
Example(C++)
Let us see the following implementation to get a better understanding −
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
bool canPartitionKSubsets(vector<int>& nums, int k) {
int n = nums.size();
vector <bool> dp(1 << n);
vector <int> total(1 << n);
sort(nums.begin(), nums.end());
int sum = 0;
for(int i = 0; i < nums.size(); i++)sum += nums[i];
if(sum % k || nums[nums.size() - 1] > sum / k) return false;
dp[0] = true;
sum /= k;
for(int i = 0; i < (1 << n); i++){
if(dp[i]){
for(int j = 0; j < n; j++){
int temp = i | (1 << j);
if(temp != i){
if(nums[j] <= sum - (total[i] % sum)){
dp[temp] = true;
total[temp] = total[i] + nums[j];
}
else{
break;
}
}
}
}
}
return dp[(1 << n) - 1];
}
};
main(){
Solution ob;
vector<int> v = {4,3,2,3,5,2,1};
cout << (ob.canPartitionKSubsets(v, 4));
}Input
[4,3,2,3,5,2,1] 4
Output
1
Updated on: 29-Apr-2020
301 Views
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