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Maximize the maximum subarray sum after removing at most one element in C++
Problem statement
Given an array arr[] of N integers. The task is to first find the maximum sub-array sum and then remove at most one element from the sub-array. Remove at most a single element such that the maximum sum after removal is maximized.
If given input array is {1, 2, 3, -2, 3} then maximum sub-array sum is {2, 3, -2, 3}. Then we can remove -2. After removing the remaining array becomes−
{1, 2, 3, 3} with sum 9 which is maximum.Algorithm
1. Use Kadane’s algorithm to find the maximum subarray sum. 2. Once the sum has beens find, re-apply Kadane’s algorithm to find the maximum sum again with some minor changes)
Example
#include <bits/stdc++.h>
using namespace std;
int getMaxSubarraySum(int *arr, int n){
int max = INT_MIN;
int currentMax = 0;
for (int i = 0; i < n; ++i) {
currentMax = currentMax + arr[i];
if (max < currentMax) {
max = currentMax;
}
if (currentMax < 0) {
currentMax = 0;
}
}
return max;
}
int getMaxSum(int *arr, int n){
int cnt = 0;
int minVal = INT_MAX;
int minSubarr = INT_MAX;
int sum = getMaxSubarraySum(arr, n);
int max = INT_MIN;
int currentMax = 0;
for (int i = 0; i < n; ++i) {
currentMax = currentMax + arr[i];
++cnt;
minSubarr = min(arr[i], minSubarr);
if (sum == currentMax) {
if (cnt == 1) {
minVal = min(minVal, 0);
} else {
minVal = min(minVal, minSubarr);
}
}
if (currentMax < 0) {
currentMax = 0;
cnt = 0;
minSubarr = INT_MAX;
}
}
return sum - minVal;
}
int main(){
int arr[] = {1, 2, 3, -2, 3};
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Maximum sum = " << getMaxSum(arr, n) << endl;
return 0;
}Output
When you compile and execute the above program. It generates the following output−
Maximum sum = 9
Updated on: 24-Dec-2019
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