# Maximum sum subarray such that start and end values are same

Given an array of N positive numbers, the task is to find a contiguous subarray (L-R) such that a[L]=a[R] and sum of a[L] + a[L+1] +…+ a[R] is maximum.

Examples:

```Input: arr[] = {1, 3, 2, 2, 3}
Output: 10
Subarray [3, 2, 2, 3] starts and ends with 3 and has sum = 10

Input: arr[] = {1, 3, 2, 2, 3}
Output: 10
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: For every element in the array, let’s find 2 values: First (Leftmost) occurrence in the array and Last (Rightmost) occurrence in the array. Since all numbers are positive, increasing the number of terms can only increase the sum. Hence for every number in the array, we find the sum between it’s leftmost and rightmost occurrence, which can be done quickly using prefix sums. We can keep track of the maximum value found so far and print it in the end.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the maximum sum ` `int` `maxValue(``int` `a[], ``int` `n) ` `{ ` `    ``unordered_map<``int``, ``int``> first, last; ` `    ``int` `pr[n]; ` `    ``pr = a; ` ` `  `    ``for` `(``int` `i = 1; i < n; i++) { ` ` `  `        ``// Build prefix sum array ` `        ``pr[i] = pr[i - 1] + a[i]; ` ` `  `        ``// If the value hasn't been encountered before, ` `        ``// It is the first occurrence ` `        ``if` `(first[a[i]] == 0) ` `            ``first[a[i]] = i; ` ` `  `        ``// Keep updating the last occurrence ` `        ``last[a[i]] = i; ` `    ``} ` ` `  `    ``int` `ans = 0; ` ` `  `    ``// Find the maximum sum with same first and last value ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``int` `start = first[a[i]]; ` `        ``int` `end = last[a[i]]; ` `        ``ans = max(ans, pr[end] - pr[start - 1]); ` `    ``} ` `    ``return` `ans; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 3, 5, 2, 4, 18, 2, 3 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << maxValue(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach ` `import` `java.io.*; ` `import` `java.util.*; ` ` `  `class` `GFG { ` ` `  `    ``// Function to find the maximum sum ` `    ``static` `int` `maxValue(``int` `a[], ``int` `n) ` `    ``{ ` `        ``HashMap first = ``new` `HashMap<>(); ` `        ``HashMap last = ``new` `HashMap<>(); ` `        ``for` `(``int` `i = ``0``; i < n; i++) { ` `            ``first.put(a[i], ``0``); ` `            ``last.put(a[i], ``0``); ` `        ``} ` ` `  `        ``int``[] pr = ``new` `int``[n]; ` `        ``pr[``0``] = a[``0``]; ` ` `  `        ``for` `(``int` `i = ``1``; i < n; i++) { ` ` `  `            ``// Build prefix sum array ` `            ``pr[i] = pr[i - ``1``] + a[i]; ` ` `  `            ``// If the value hasn't been encountered before, ` `            ``// It is the first occurrence ` `            ``if` `(Integer.parseInt(String.valueOf(first.get(a[i]))) == ``0``) ` `                ``first.put(a[i], i); ` ` `  `            ``// Keep updating the last occurrence ` `            ``last.put(a[i], i); ` `        ``} ` ` `  `        ``int` `ans = ``0``; ` ` `  `        ``// Find the maximum sum with same first and last value ` `        ``for` `(``int` `i = ``0``; i < n; i++) { ` `            ``int` `start = Integer.parseInt(String.valueOf(first.get(a[i]))); ` `            ``int` `end = Integer.parseInt(String.valueOf(last.get(a[i]))); ` `            ``if` `(start != ``0``) ` `                ``ans = Math.max(ans, pr[end] - pr[start - ``1``]); ` `        ``} ` ` `  `        ``return` `ans; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int``[] arr = { ``1``, ``3``, ``5``, ``2``, ``4``, ``18``, ``2``, ``3` `}; ` `        ``int` `n = arr.length; ` `        ``System.out.print(maxValue(arr, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by rachana soma `

## Python3

 `# Python3 implementation of the above approach  ` `from` `collections ``import` `defaultdict ` ` `  `# Function to find the maximum sum  ` `def` `maxValue(a, n):  ` `  `  `    ``first ``=` `defaultdict(``lambda``:``0``) ` `    ``last ``=` `defaultdict(``lambda``:``0``) ` `     `  `    ``pr ``=` `[``None``] ``*` `n  ` `    ``pr[``0``] ``=` `a[``0``]  ` `   `  `    ``for` `i ``in` `range``(``1``, n):   ` `   `  `        ``# Build prefix sum array  ` `        ``pr[i] ``=` `pr[i ``-` `1``] ``+` `a[i]  ` `   `  `        ``# If the value hasn't been encountered before,  ` `        ``# It is the first occurrence  ` `        ``if` `first[a[i]] ``=``=` `0``:  ` `            ``first[a[i]] ``=` `i  ` `   `  `        ``# Keep updating the last occurrence  ` `        ``last[a[i]] ``=` `i  ` `      `  `   `  `    ``ans ``=` `0`  `   `  `    ``# Find the maximum sum with same first and last value  ` `    ``for` `i ``in` `range``(``0``, n):   ` `        ``start ``=` `first[a[i]]  ` `        ``end ``=` `last[a[i]]  ` `        ``ans ``=` `max``(ans, pr[end] ``-` `pr[start ``-` `1``])  ` `      `  `    ``return` `ans  ` `  `  `   `  `# Driver Code  ` `if` `__name__ ``=``=` `"__main__"``:  ` `  `  `    ``arr ``=`  `[``1``, ``3``, ``5``, ``2``, ``4``, ``18``, ``2``, ``3``]   ` `    ``n ``=` `len``(arr)  ` `   `  `    ``print``(maxValue(arr, n))  ` `   `  `# This code is contributed by Rituraj Jain `

## C#

 `// C# implementation of the above approach ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// Function to find the maximum sum ` `    ``static` `int` `maxValue(``int` `[]a, ``int` `n) ` `    ``{ ` `        ``Dictionary<``int``,  ` `                   ``int``> first = ``new` `Dictionary<``int``,  ` `                                               ``int``>(); ` `        ``Dictionary<``int``,  ` `                   ``int``> last = ``new` `Dictionary<``int``,   ` `                                              ``int``>(); ` `         `  `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{ ` `            ``first[a[i]] = 0; ` `            ``last[a[i]] = 0; ` `        ``} ` ` `  `        ``int``[] pr = ``new` `int``[n]; ` `        ``pr = a; ` ` `  `        ``for` `(``int` `i = 1; i < n; i++)  ` `        ``{ ` ` `  `            ``// Build prefix sum array ` `            ``pr[i] = pr[i - 1] + a[i]; ` ` `  `            ``// If the value hasn't been encountered before, ` `            ``// It is the first occurrence ` `            ``if` `(first[a[i]] == 0) ` `                ``first[a[i]] = i; ` ` `  `            ``// Keep updating the last occurrence ` `            ``last[a[i]] = i; ` `        ``} ` ` `  `        ``int` `ans = 0; ` ` `  `        ``// Find the maximum sum with  ` `        ``// same first and last value ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{ ` `            ``int` `start = first[a[i]]; ` `            ``int` `end = last[a[i]]; ` `            ``if` `(start != 0) ` `                ``ans = Math.Max(ans, pr[end] - pr[start - 1]); ` `        ``} ` `        ``return` `ans; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``static` `void` `Main() ` `    ``{ ` `        ``int``[] arr = { 1, 3, 5, 2, 4, 18, 2, 3 }; ` `        ``int` `n = arr.Length; ` `        ``Console.Write(maxValue(arr, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by mohit kumar `

Output:

```37
```

Time Complexity: O(N)

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