Subarreglo de suma máxima tal que los valores inicial y final son los mismos

Dado un arreglo de N números positivos, la tarea es encontrar un subarreglo contiguo (LR) tal que a[L]=a[R] y la suma de a[L] + a[L+1] +…+ a[R ] es máximo.
Ejemplos: 
 

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

Enfoque: para cada elemento de la array, busquemos 2 valores: la primera aparición (más a la izquierda) en la array y la última aparición (más a la derecha) en la array. Como todos los números son positivos, aumentar el número de términos solo puede aumentar la suma. Por lo tanto, para cada número en la array, encontramos la suma entre su ocurrencia más a la izquierda y la más a la derecha, lo que se puede hacer rápidamente usando sumas de prefijos. Podemos realizar un seguimiento del valor máximo encontrado hasta el momento e imprimirlo al final.
A continuación se muestra la implementación del enfoque anterior:
 

// 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];
 
    for (int i = 0; i < n; i++) {
 
        // Build prefix sum array
        if (i)
            pr[i] = pr[i - 1] + a[i];
        else
            pr[i] = a[i];
        // If the value hasn't been encountered before,
        // It is the first occurrence
        if (first[a[i]] == 0)
            first[a[i]] = i + 1;
 
        // Keep updating the last occurrence
        last[a[i]] = i + 1;
    }
 
    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 == 1) ans = max(ans, pr[end - 1]);
        else ans = max(ans, pr[end - 1] - pr[start - 2]);
    }
    return ans;
}
 
// Driver Code
int main()
{
    int arr[] = {1,2,31,2,1};
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << maxValue(arr, n);
 
    return 0;
}

import java.util.HashMap;
 
class GFG {
    static int maxValue(int[] a, int n)
    {
        HashMap<Integer, Integer> first = new HashMap<>();
        HashMap<Integer, Integer> last = new HashMap<>();
 
        int[] prefix = new int[n];
 
        for (int i = 0; i < n; i++) {
 
            // Build prefix sum array
            if (i != 0)
                prefix[i] = prefix[i - 1] + a[i];
            else
                prefix[i] = a[i];
            // If the value hasn't been encountered before,
            // It is the first occurrence
            if (!first.containsKey(a[i]))
                first.put(a[i], i);
 
            // Keep updating the last occurrence
            last.put(a[i], i);
        }
 
        int ans = -1;
 
        // Find the maximum sum with same first and last
        // value
        for (int i = 0; i < n; i++) {
            int start = first.get(a[i]);
            int end = last.get(a[i]);
            int sum = 0;
            if(start == 0)
                sum = prefix[end];
            else
                sum = prefix[end] - prefix[start - 1];
            if(sum > ans)
                ans = sum;
        }
 
        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));
    }
}

# 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]
    first[a[0]] = 1
    last[a[0]] = 1
    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+1
 
        # Keep updating the last occurrence
        last[a[i]] = i+1
 
    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]]
        if start != 1:
            ans = max(ans, pr[end-1] - pr[start - 2])
 
    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# 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];
        first[a[0]] = 1;
        last[a[0]] = 1;
        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 + 1;
 
            // Keep updating the last occurrence
            last[a[i]] = i + 1;
        }
 
        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 != 1)
                ans = Math.Max(ans,
                               pr[end - 1] - pr[start - 2]);
        }
        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

<script>
 
// JavaScript implementation of the above approach
 
// Function to find the maximum sum
function maxValue(a,n)
{
    let first = new Map();
        let last = new Map();
        for (let i = 0; i < n; i++) {
            first.set(a[i], 0);
            last.set(a[i], 0);
        }
   
        let pr = new Array(n);
        pr[0] = a[0];
        first[a[0]] = 0;
        last[a[0]] = 0;
        for (let 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 (parseInt((first.get(a[i]))) == 0)
                first.set(a[i], i+1);
   
            // Keep updating the last occurrence
            last.set(a[i], i+1);
        }
   
        let ans = 0;
   
        // Find the maximum sum with same first and last value
        for (let i = 0; i < n; i++) {
            let start = parseInt(first.get(a[i]));
            let end = parseInt(last.get(a[i]));
            if (start != 1)
                ans = Math.max(ans, pr[end-1] - pr[start - 2]);
        }
   
        return ans;
}
 
// Driver Code
let arr=[1, 3, 5, 2, 4, 18, 2, 3 ];
let n = arr.length;
document.write(maxValue(arr, n));   
 
     
 
// This code is contributed by patel2127
 
</script>

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Complejidad de tiempo: O(N)