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

[0, 3, 5, ]