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 sumint 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 Codeint main(){ int arr[] = {1,2,31,2,1}; int n = sizeof(arr) / sizeof(arr[0]); cout << maxValue(arr, n); return 0;} |
Java
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
# Python3 implementation of the above approachfrom collections import defaultdict# Function to find the maximum sumdef 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 Codeif __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 approachusing 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 |
Javascript
<script>// JavaScript implementation of the above approach// Function to find the maximum sumfunction 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 Codelet 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> |
Output
37
Time Complexity: O(N)
Auxiliary Space: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
