Given an array arr[] of N elements, the task is to find the maximum sum of any subarray of length X such that X > 0 and X % 2 = 0.
Examples:
Input: arr[] = {1, 2, 3}
Output: 5
{2, 3} is the required subarray.Input: arr[] = {8, 9, -8, 9, 10}
Output: 20
{9, -8, 9, 10} is the required subarray.
Even though {8, 9, -8, 9, 10} has the maximum sum
but it is not of even length.
Approach: This problem is a variation of maximum subarray sum problem and can be solved using dynamic programming approach. Create an array dp[] where dp[i] will store the maximum sum of an even length subarray whose first element is arr[i]. Now the recurrence relation will be:
dp[i] = max((arr[i] + arr[i + 1]), (arr[i] + arr[i + 1] + dp[i + 2]))
This is because the maximum sum even length subarray starting with the element arr[i] can either be the sum of arr[i] and arr[i + 1] or it can be arr[i] + arr[i + 1] added with the maximum sum of even length subarray starting with arr[i + 2] i.e. dp[i + 2]. Take the maximum of these two.
In the end, the maximum value from the dp[] array will be the required answer.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Function to return the maximum// subarray sum of even lengthint maxEvenLenSum(int arr[], int n){ // There has to be at // least 2 elements if (n < 2) return 0; // dp[i] will store the maximum // subarray sum of even length // starting at arr[i] int dp[n] = { 0 }; // Valid subarray cannot start from // the last element as its // length has to be even dp[n - 1] = 0; dp[n - 2] = arr[n - 2] + arr[n - 1]; for (int i = n - 3; i >= 0; i--) { // arr[i] and arr[i + 1] can be added // to get an even length subarray // starting at arr[i] dp[i] = arr[i] + arr[i + 1]; // If the sum of the valid subarray starting // from arr[i + 2] is greater than 0 then it // can be added with arr[i] and arr[i + 1] // to maximize the sum of the subarray // starting from arr[i] if (dp[i + 2] > 0) dp[i] += dp[i + 2]; } // Get the sum of the even length // subarray with maximum sum int maxSum = *max_element(dp, dp + n); return maxSum;} // Driver codeint main(){ int arr[] = { 8, 9, -8, 9, 10 }; int n = sizeof(arr) / sizeof(int); cout << maxEvenLenSum(arr, n); return 0;} |
Java
// Java implementation of the approachimport java.util.Arrays; class GFG { // Function to return the maximum// subarray sum of even lengthstatic int maxEvenLenSum(int arr[], int n){ // There has to be at // least 2 elements if (n < 2) return 0; // dp[i] will store the maximum // subarray sum of even length // starting at arr[i] int []dp = new int[n]; // Valid subarray cannot start from // the last element as its // length has to be even dp[n - 1] = 0; dp[n - 2] = arr[n - 2] + arr[n - 1]; for (int i = n - 3; i >= 0; i--) { // arr[i] and arr[i + 1] can be added // to get an even length subarray // starting at arr[i] dp[i] = arr[i] + arr[i + 1]; // If the sum of the valid subarray starting // from arr[i + 2] is greater than 0 then it // can be added with arr[i] and arr[i + 1] // to maximize the sum of the subarray // starting from arr[i] if (dp[i + 2] > 0) dp[i] += dp[i + 2]; } // Get the sum of the even length // subarray with maximum sum int maxSum = Arrays.stream(dp).max().getAsInt(); return maxSum;} // Driver codepublic static void main(String[] args) { int arr[] = { 8, 9, -8, 9, 10 }; int n = arr.length; System.out.println(maxEvenLenSum(arr, n));}} // This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach # Function to return the maximum# subarray sum of even lengthdef maxEvenLenSum(arr, n): # There has to be at # least 2 elements if (n < 2): return 0 # dp[i] will store the maximum # subarray sum of even length # starting at arr[i] dp = [0 for i in range(n)] # Valid subarray cannot start from # the last element as its # length has to be even dp[n - 1] = 0 dp[n - 2] = arr[n - 2] + arr[n - 1] for i in range(n - 3, -1, -1): # arr[i] and arr[i + 1] can be added # to get an even length subarray # starting at arr[i] dp[i] = arr[i] + arr[i + 1] # If the sum of the valid subarray # starting from arr[i + 2] is # greater than 0 then it can be added # with arr[i] and arr[i + 1] # to maximize the sum of the # subarray starting from arr[i] if (dp[i + 2] > 0): dp[i] += dp[i + 2] # Get the sum of the even length # subarray with maximum sum maxSum = max(dp) return maxSum # Driver codearr = [8, 9, -8, 9, 10]n = len(arr) print(maxEvenLenSum(arr, n)) # This code is contributed by Mohit Kumar |
C#
// C# implementation of the approach using System; class GFG { static int MaxSum(int []arr) { // assigning first element to the array int large = arr[0]; // loop to compare value of large // with other elements for (int i = 1; i < arr.Length; i++) { // if large is smaller than other element // assig that element to the large if (large < arr[i]) large = arr[i]; } return large; } // Function to return the maximum // subarray sum of even length static int maxEvenLenSum(int []arr, int n) { // There has to be at // least 2 elements if (n < 2) return 0; // dp[i] will store the maximum // subarray sum of even length // starting at arr[i] int []dp = new int[n]; // Valid subarray cannot start from // the last element as its // length has to be even dp[n - 1] = 0; dp[n - 2] = arr[n - 2] + arr[n - 1]; for (int i = n - 3; i >= 0; i--) { // arr[i] and arr[i + 1] can be added // to get an even length subarray // starting at arr[i] dp[i] = arr[i] + arr[i + 1]; // If the sum of the valid subarray starting // from arr[i + 2] is greater than 0 then it // can be added with arr[i] and arr[i + 1] // to maximize the sum of the subarray // starting from arr[i] if (dp[i + 2] > 0) dp[i] += dp[i + 2]; } // Get the sum of the even length // subarray with maximum sum int maxSum = MaxSum(dp); return maxSum; } // Driver code public static void Main() { int []arr = { 8, 9, -8, 9, 10 }; int n = arr.Length; Console.WriteLine(maxEvenLenSum(arr, n)); } } // This code is contributed by kanugargng |
20
Time complexity: O(n)
Space complexity: O(n)
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