Maximum Sum of 3 Non-Overlapping Subarrays in C++

Suppose we have one array called nums of positive integers, we have to find three non-overlapping subarrays with maximum sum. Here each subarray will be of size k, and we want to maximize the sum of all 3*k entries.

We have to find the result as a list of indices representing the starting position of each interval. If there are multiple answers, we will return the lexicographically smallest one.

So if the input is like [1,2,1,2,6,8,4,1] and k = 2, then the result will be [0,3,5], so subarrays are [1,2], [2,6], [8,4] correspond to the starting indices [0,3,5].

To solve this, we will follow these steps −

  • n := size of nums
  • Define an array ret of size 3 fill this with inf
  • Define an array sum of size n + 1
  • for initialize i := 0, when i <n, update (increase i by 1), do −
    • sum[i + 1] = sum[i] + nums[i]
  • Define an array posLeft of size n
  • Define an array posRight of size n fill this with n - k
  • for initialize i := k, currMax := sum[k] - sum[0], when i < n, update (increase i by 1), do −
    • newTotal := sum[i + 1] - sum[i + 1 - k]
    • if newTotal > currMax, then −
      • currMax := newTotal
      • posLeft[i] := i + 1 - k
    • Otherwise
      • posLeft[i] := posLeft[i - 1]
  • for initialize i := n - k - 1, currMax := sum[n] - sum[n - k], when i >= 0, update (decrease i by 1), do −
    • newTotal := sum[i + k] - sum[i]
    • if newTotal >= currMax, then −
      • currMax := newTotal
      • posRight[i] := i
    • Otherwise
      • posRight[i] := posRight[i + 1]
  • req := 0
  • for initialize i := k, when i <= n - 2 * k, update (increase i by 1), do −
    • l := posLeft[i - 1], r := posRight[i + k]
    • temp := (sum[l + k] - sum[l]) + (sum[i + k] - sum[i]) + (sum[r + k] - sum[r])
    • if temp > req, then −
      • ret[0] := l, ret[1] := i, ret[2] := r
      • req := temp
  • return ret

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto> v){
   cout << "[";
   for(int i = 0; i<v.size(); i++){
      cout << v[i] << ", ";
   }
   cout << "]"<<endl;
}
class Solution {
public:
   vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
      int n = nums.size();
      vector <int> ret(3, INT_MAX);
      vector <int> sum(n + 1);
      for(int i = 0; i < n; i++){
         sum[i + 1] = sum[i] + nums[i];
      }
      vector <int> posLeft(n);
      vector <int> posRight(n, n - k);
      for(int i = k, currMax = sum[k] - sum[0]; i < n; i++){
         int newTotal = sum[i + 1] - sum[i + 1- k];
         if(newTotal > currMax){
            currMax = newTotal;
            posLeft[i] = i + 1 - k;
         }else{
            posLeft[i] = posLeft[i - 1];
         }
      }
      for(int i = n - k - 1, currMax = sum[n] - sum[n - k]; i >=0 ; i--){
         int newTotal = sum[i + k] - sum[i];
         if(newTotal >= currMax){
            currMax = newTotal;
            posRight[i] = i;
         }else{
            posRight[i] = posRight[i + 1];
         }
      }
      int req = 0;
      for(int i = k; i <= n - 2 * k; i++){
         int l = posLeft[i - 1];
         int r = posRight[i + k];
         int temp = (sum[l + k] - sum[l]) + (sum[i + k] - sum[i]) + (sum[r + k] - sum[r]);
         if(temp > req){
            ret[0] = l;
            ret[1] = i;
            ret[2] = r;
            req = temp;
         }
      }
      return ret;
   }
};
main(){
   Solution ob;
   vector<int> v = {1,2,1,2,6,8,4,1};
   print_vector(ob.maxSumOfThreeSubarrays(v, 2));
}

Input

{1,2,1,2,6,8,4,1}
2

Output

[0, 3, 5, ]

Updated on: 02-Jun-2020

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