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Maximize product of subarray sum with its maximum element
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Maximize the maximum subarray sum after removing atmost one element

Last Updated : 13 Jul, 2024
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Given an array arr[] of N integers, the task is to find the maximum sum of a subarray with at most one deletion allowed. The size of subarray to be considered should be at least 1.


Examples: 
 

Input: arr[] = {1, 2, 3, -2, 3} 
Output: 9 
The maximum sub-array sum is given by the sub-array {2, 3, -2, 3} 
Hence, we can remove -2 to further maximize the sub-array sum. 
Input: arr[] = {-1, -2} 
Output: -1 
The maximum sub-array sum is from the sub-array {-1} and no removal is required. 
 


Approach: Idea is based upon Kadane’s algorithm to find the maximum subarray sum. In the standard Kadane’s algorithm, we traverse the array from left to right and keep track of maximum sum ending with every element. We finally return the maximum of all max ending here sums. Here we modify the algorithm to keep track of the maximum sum ending here with one deletion allowed.
 

C++ 
#include <iostream> using namespace std;  int maximizeSum(int arr[], int n) {              // Max sum ending here with           // at most one deletion         int maxEHereDel = arr[0];                    // Max sum ending here          // without any deletion         int mEhere = arr[0];                    // To store result         int res = arr[0];                   for (int i=1; i<n; i++) {                      int x = arr[i];                          // a) Start a new subarray             // b) Exclude current element             // c) Include current element             maxEHereDel = max(x, max(mEhere,                              maxEHereDel+x));                        // Updating maxEHere (without any deletion)             // a) Begin a new subarray             // b) Continue with the prev one             mEhere = max(x, mEhere + x);                           res = max(res, maxEHereDel);           }         return res; }  // Driver code int main() { 	int a[] = { 10, -100, -1, 10, -5, 4 }; 	int n = sizeof(a) / sizeof(a[0]); 	cout << maximizeSum(a, n); 	return 0; }

Output
19

Time Complexity: O(n)
Auxiliary Space: O(1)



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Maximize product of subarray sum with its maximum element

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Article Tags :
  • Arrays
  • DSA
  • Mathematical
  • subarray
  • subarray-sum
Practice Tags :
  • Arrays
  • Mathematical

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