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