# Program to find sum of k non-overlapping sublists whose sum is maximum in C++

Suppose we have a list of numbers called nums, and another value k, we have to find the k non-overlapping, non-empty sublists such that the sum of their sums is maximum. We can consider k is less than or equal to the size of nums.

So, if the input is like nums = [11, -1, 2, 1, 6, -24, 11, -9, 6] k = 3, then the output will be 36, as we can select the sublists [11, -1, 2, 1, 6], [11], and [6] to get sums of [19, 11, 6] = 36.

To solve this, we will follow these steps −

• n := size of nums
• if n is same as 0 or k is same as 0, then −
• return 0
• Define an array hi of size k + 1 and fill with -inf,
• Define another array open of size k + 1 and fill with -inf
• hi[0] := 0
• for each num in nums −
• Define an array nopen of size k + 1 and fill with -inf
• for initialize i := 1, when i <= k, update (increase i by 1), do
• if open[i] > -inf, then −
• nopen[i] := open[i] + num
• if hi[i - 1] > -inf, then −
• nopen[i] := maximum of nopen[i] and hi[i - 1] + num
• open := move(nopen)
• for initialize i := 1, when i <= k, update (increase i by 1), do
• hi[i] := maximum of hi[i] and open[i]
• return hi[k]

## Example (C++)

Let us see the following implementation to get better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
int solve(vector<int>& nums, int k) {
int n = nums.size();
if (n == 0 || k == 0)
return 0;
vector<int> hi(k + 1, INT_MIN), open(k + 1, INT_MIN);
hi[0] = 0;
for (int num : nums) {
vector<int> nopen(k + 1, INT_MIN);
for (int i = 1; i <= k; ++i) {
if (open[i] > INT_MIN)
nopen[i] = open[i] + num;
if (hi[i - 1] > INT_MIN)
nopen[i] = max(nopen[i], hi[i - 1] + num);
}
open = move(nopen);
for (int i = 1; i <= k; ++i)
hi[i] = max(hi[i], open[i]);
}
return hi[k];
}
int main(){
vector<int> v = {11, -1, 2, 1, 6, -24, 11, -9, 6};
int k = 3;
cout << solve(v, 3);
}

## Input

{11, -1, 2, 1, 6, -24, 11, -9, 6}, 3

## Output

36

Updated on: 12-Dec-2020

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