# Subarray with largest sum after excluding its maximum element

Given an array arr[], the task is to find the starting and ending indices of the subarray with the largest sum after excluding its maximum element.
Examples:

Input: arr[] = {5, -2, 10, -1, 4}
Output: 1 5
Explanation:
Subarray[1:5] = {5, -2, 10, -1, 4}
Sum of subarray excluding maximum element = 5 + (-2) + (-1) + 4 = 6

Input:
arr[] = {5, 2, 5, 3, -30, -30, 6, 9}
Output: 1 4
Explanation:
Subarray[1:4] = {5, 2, 5, 3}
Sum of subarray excluding maximum element = 5 + 2 + 3 = 10

Approach: The idea is to use Kadane algorithm to solve this problem.

1. As in this problem we have to choose one element which is the maximum in the subarray.
2. Therefore, we can choose all the positive elements from the array, and each time we can make elements greater than that element to INT_MIN, such that it is not included in the array.
3. Finally, apply the Kadane algorithm to find the maximum sum subarray.
4. If there are no positive elements in the array then we can choose any one element from the array to get the maximum sum as 0.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the  ` `// maximum sum subarray such by  ` `// excluding the maximum element  ` `// from the subarray  ` ` `  `#include   ` `using` `namespace` `std;  ` ` `  `// Function to find the maximum sum  ` `// subarray by excluding the maximum  ` `// element from the array  ` `void` `maximumSumSubarray(``int` `arr[], ``int` `n)  ` `{  ` `    ``unordered_map<``int``, ``int``> mp;  ` ` `  `    ``// Loop to store all the positive  ` `    ``// elements in the map  ` `    ``for` `(``int` `i = 0; i < n; i++) {  ` ` `  `        ``if` `(arr[i] >= 0  ` `            ``&& mp.find(arr[i])  ` `                ``== mp.end())  ` ` `  `            ``mp[arr[i]] = 1;  ` `    ``}  ` ` `  `    ``int` `first = 0;  ` `    ``int` `last = 0;  ` `    ``int` `ans = 0;  ` `    ``int` `INF = 1e6;  ` ` `  `    ``// Loop to iterating over the map  ` `    ``// and considering as the maximum  ` `    ``// element of the current including  ` `    ``// subarray  ` `    ``for` `(``auto` `i : mp) {  ` ` `  `        ``// Make the current  ` `        ``// element maximum  ` `        ``int` `mx = i.first;  ` ` `  `        ``int` `curr = 0;  ` `        ``int` `curr_start;  ` ` `  `        ``// Iterate through array and  ` `        ``// apply kadane's algorithm  ` `        ``for` `(``int` `j = 0; j < n; j++) {  ` `            ``if` `(curr == 0)  ` `                ``curr_start = j;  ` ` `  `            ``// Condition if current element is  ` `            ``// greater than mx then make  ` `            ``// the element -infinity  ` `            ``int` `val = arr[j] > mx  ` `                        ``? -INF  ` `                        ``: arr[j];  ` ` `  `            ``curr += val;  ` ` `  `            ``if` `(curr < 0)  ` `                ``curr = 0;  ` ` `  `            ``if` `(curr > ans) {  ` `                ``ans = curr;  ` ` `  `                ``// Store the indices  ` `                ``// in some variable  ` `                ``first = curr_start;  ` `                ``last = j;  ` `            ``}  ` `        ``}  ` `    ``}  ` ` `  `    ``cout << first + 1  ` `        ``<< ``" "` `<< last + 1;  ` `}  ` ` `  `// Driver Code  ` `int` `main()  ` `{  ` `    ``int` `arr[] = { 5, -2, 10, -1, 4 };  ` ` `  `    ``int` `size = ``sizeof``(arr) / ``sizeof``(arr);  ` ` `  `    ``// Function Call  ` `    ``maximumSumSubarray(arr, size);  ` `    ``return` `0;  ` `}  `

## Java

 `// Java implementation to find the  ` `// maximum sum subarray such by  ` `// excluding the maximum element  ` `// from the subarray  ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `     `  `// Function to find the maximum sum ` `// subarray by excluding the maximum ` `// element from the array ` `static` `void` `maximumSumSubarray(``int` `arr[], ``int` `n) ` `{ ` `    ``Map mp = ``new` `HashMap<>(); ` `     `  `    ``// Loop to store all the positive ` `    ``// elements in the map ` `    ``for``(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `        ``if` `(arr[i] >= ``0``) ` `            ``mp.put(arr[i], ``1``); ` `    ``} ` `     `  `    ``int` `first = ``0``; ` `    ``int` `last = ``0``; ` `    ``int` `ans = ``0``; ` `    ``int` `INF = (``int``)1e6; ` `     `  `    ``// Loop to iterating over the map ` `    ``// and considering as the maximum ` `    ``// element of the current including ` `    ``// subarray ` `    ``for` `(Map.Entry i : mp.entrySet())  ` `    ``{ ` `         `  `        ``// Make the current ` `        ``// element maximum ` `        ``int` `mx = i.getKey(); ` `        ``int` `curr = ``0``; ` `        ``int` `curr_start = -``1``; ` `         `  `        ``// Iterate through array and ` `        ``// apply kadane's algorithm ` `        ``for``(``int` `j = ``0``; j < n; j++) ` `        ``{ ` `            ``if` `(curr == ``0``) ` `                ``curr_start = j; ` `                 `  `            ``// Condition if current element is ` `            ``// greater than mx then make ` `            ``// the element -infinity ` `            ``int` `val = arr[j] > mx ? -INF : arr[j]; ` `            ``curr += val; ` `             `  `            ``if` `(curr < ``0``) ` `                ``curr = ``0``; ` `             `  `            ``if` `(curr > ans) ` `            ``{ ` `                ``ans = curr; ` `                 `  `                ``// Store the indices ` `                ``// in some variable ` `                ``first = curr_start; ` `                ``last = j; ` `            ``} ` `        ``} ` `    ``} ` `    ``System.out.print((first + ``1``) + ``" "` `+ ` `                      ``(last + ``1``)); ` `}  ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``5``, -``2``, ``10``, -``1``, ``4` `}; ` `    ``int` `size = arr.length; ` `     `  `    ``// Function call ` `    ``maximumSumSubarray(arr, size); ` `} ` `} ` ` `  `// This code is contributed by offbeat `

## Python3

 `# Python3 implementation to find  ` `# the maximum sum subarray such  ` `# by excluding the maximum  ` `# element from the subarray  ` ` `  `# Function to find the maximum sum  ` `# subarray by excluding the maximum  ` `# element from the array  ` `def` `maximumSumSubarray(arr, n):  ` `    ``mp ``=` `{}  ` ` `  `    ``# Loop to store all the positive  ` `    ``# elements in the map  ` `    ``for` `i ``in` `range``(n):  ` ` `  `        ``if` `(arr[i] >``=` `0` `and` `            ``arr[i] ``not` `in` `mp):  ` `            ``mp[arr[i]] ``=` `1` ` `  `    ``first ``=` `0` `    ``last ``=` `0` `    ``ans ``=` `0` `    ``INF ``=` `1e6` ` `  `    ``# Loop to iterating over the map  ` `    ``# and considering as the maximum  ` `    ``# element of the current including  ` `    ``# subarray  ` `    ``for` `i ``in` `mp:  ` ` `  `        ``# Make the current  ` `        ``# element maximum  ` `        ``mx ``=` `i  ` ` `  `        ``curr ``=` `0` ` `  `        ``# Iterate through array and  ` `        ``# apply kadane's algorithm  ` `        ``for` `j ``in` `range``(n):  ` `            ``if` `(curr ``=``=` `0``):  ` `                ``curr_start ``=` `j  ` ` `  `            ``# Condition if current element  ` `            ``# is greater than mx then make  ` `            ``# the element -infinity  ` `            ``if` `arr[j] > mx:  ` `                ``val ``=``-` `INF  ` `            ``else``:  ` `                ``val``=` `arr[j];  ` ` `  `            ``curr ``+``=` `val  ` ` `  `            ``if` `(curr < ``0``):  ` `                ``curr ``=` `0` ` `  `            ``if` `(curr > ans):  ` `                ``ans ``=` `curr  ` ` `  `                ``# Store the indices  ` `                ``# in some variable  ` `                ``first ``=` `curr_start  ` `                ``last ``=` `j  ` ` `  `    ``print``(first ``+` `1``, last ``+` `1``)  ` ` `  `# Driver Code  ` `if` `__name__ ``=``=` `"__main__"``:  ` `     `  `    ``arr ``=` `[ ``5``, ``-``2``, ``10``, ``-``1``, ``4` `]  ` `    ``size ``=` `len``(arr)  ` ` `  `    ``# Function Call  ` `    ``maximumSumSubarray(arr, size)  ` ` `  `# This code is contributed by chitranayal  `

Output:

```1 5
```

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