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Maximum sum subarray such that start and end values are same
  • Difficulty Level : Basic
  • Last Updated : 09 May, 2020

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

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 <bits/stdc++.h>
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[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 (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[0]);
  
    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<Integer, Integer> first = new HashMap<>();
        HashMap<Integer, Integer> 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] *
    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]] =
    
        # Keep updating the last occurrence 
        last[a[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[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 (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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