Max sum of Subarray of length N/2 where sum is divisible by N
Last Updated :
25 Sep, 2023
Given an array, arr[] of size N where N is even, the task is to find the maximum possible sum of the subarray of length N/2 where the sum is divisible by N.
Examples:
Input: arr[] = {2, 3, 4, 5, 6, 7}, N = 6
Output:18
Explanation: Here, N = 6, The maximum possible sum of a sublist of length 3 (N/2) is 18 (divisible by 6), which can be obtained by selecting the subarray [5, 6, 7].
Input: arr[] = {3, 5, 6, 11, 7, 8}, N = 6
Output: 24
Explanation: The maximum possible sum of a sublist of length 3 (N/2) is 24 (divisible by 6), which can be obtained by selecting the subarray [6, 11, 7].
Approach: This can be solved with the following idea:
One way to solve this problem is to use Dynamic programming. We can create a 2D array DP with dimensions (N/2 + 1) x N, where DP[i][j] represents the maximum possible sum of an arr[] of length i (i.e., i = N/2) ending at index j.
Steps involved in the implementation of code:
- The base case is DP[0][j] = 0 for all j.
- For the recursive case, we can compute DP[i][j] by considering two options:
- If we include the jth element in the subarray, the sum will be divisible by N only if the sum of the previous i - 1 elements is congruent to the j % N. Therefore, we can compute DP[i][j] as DP[i - 1][(j - K + N) % N] + A[j], where K ranges from 0 to N - 1. The maximum of these values is the maximum possible sum of a subarray of length i that ends at index j.
- If we do not include the jth element in the subarray, then DP[i][j] = DP[i][j-1].
- The final answer is the maximum value in DP[N/2][j] for all j.
Below is the implementation of the above approach:
C++
// CPP approach
#include <bits/stdc++.h>
using namespace std;
int max_sum_sublist(vector<int> arr)
{
// Takes in a list of n integers,
// where n is even, and returns the
// maximum possible sum of a sublist
// such that the sum of the elements
// in the sublist is divisible by n
// and the length of the sublist is n/2.
int n = arr.size();
vector<vector<int> > DP((n / 2 + 1),
vector<int>(n + 1, 0));
// Initialize the base case
for (int j = 0; j < n; j++)
DP[0][j] = 0;
// Compute DP[i][j] for
// i > 0 and j > 0
for (int i = 1; i < n / 2 + 1; i++) {
for (int j = 0; j < n; j++) {
DP[i][j] = 0;
for (int k = 0; k < n; k++) {
if ((j - k + n) % n < j)
DP[i][j] = max(
DP[i][j], DP[i - 1][(j - k + n) % n]
+ arr[j]);
}
DP[i][j] = max(DP[i][j], DP[i][j - 1]);
}
}
// Find the maximum value
// in the last row of DP
int max_sum = 0;
for (int j = 0; j < n; j++)
max_sum = max(max_sum, DP[n / 2][j]);
// Return the maximum possible sum
return max_sum;
}
// Driver code
int main()
{
vector<int> arr = { 2, 3, 4, 5, 6, 7 };
// Function call
cout << max_sum_sublist(arr);
return 0;
}
// java approach
import java.util.Arrays;
import java.util.List;
public class GFG {
public static int maxSumSublist(List<Integer> arr)
{
// Takes in a list of n integers,
// where n is even, and returns the
// maximum possible sum of a sublist
// such that the sum of the elements
// in the sublist is divisible by n
// and the length of the sublist is n/2.
int n = arr.size();
int[][] DP = new int[n / 2 + 1][n + 1];
// Initialize the base case
for (int j = 0; j < n + 1; j++)
DP[0][j] = 0;
// Compute DP[i][j] for i > 0 and j > 0
for (int i = 1; i < n / 2 + 1; i++) {
for (int j = 1; j < n + 1; j++) {
DP[i][j] = 0;
for (int k = 0; k < n; k++) {
if ((j - k + n) % n < j)
DP[i][j] = Math.max(
DP[i][j],
DP[i - 1][(j - k + n) % n]
+ arr.get(j - 1));
}
DP[i][j] = Math.max(DP[i][j], DP[i][j - 1]);
}
}
// Find the maximum value in the last row of DP
int maxSum = 0;
for (int j = 1; j < n + 1; j++)
maxSum = Math.max(maxSum, DP[n / 2][j]);
// return maximum possible sum
return maxSum;
}
// driver code
public static void main(String[] args)
{
List<Integer> arr = Arrays.asList(2, 3, 4, 5, 6, 7);
// function call
System.out.println(maxSumSublist(arr));
}
}
# Python Implementation of code
import sys
def max_sum_sublist(arr):
# Takes in a list of n integers,
# where n is even, and returns the
# maximum possible sum of a sublist
# such that the sum of the elements
# in the sublist is divisible by n
# and the length of the sublist is n/2.
n = len(arr)
DP = [[0 for j in range(n + 1)] for i in range(n // 2 + 1)]
# Initialize the base case
for j in range(n):
DP[0][j] = 0
# Compute DP[i][j] for
# i > 0 and j > 0
for i in range(1, n // 2 + 1):
for j in range(n):
DP[i][j] = 0
for k in range(n):
if (j - k + n) % n < j:
DP[i][j] = max(DP[i][j], DP[i - 1][(j - k + n) % n] + arr[j])
DP[i][j] = max(DP[i][j], DP[i][j - 1])
# Find the maximum value
# in the last row of DP
max_sum = 0
for j in range(n):
max_sum = max(max_sum, DP[n // 2][j])
# Return the maximum possible sum
return max_sum
# Driver code
if __name__ == "__main__":
arr = [2, 3, 4, 5, 6, 7]
# Function call
print(max_sum_sublist(arr))
# This code is contributed by Susobhan Akhuli
using System;
using System.Collections.Generic;
public class GFG
{
public static int MaxSumSublist(List<int> arr)
{
// Takes in a list of n integers,
// where n is even, and returns the
// maximum possible sum of a sublist
// such that the sum of the elements
// in the sublist is divisible by n
// and the length of the sublist is n/2.
int n = arr.Count;
int[][] DP = new int[n / 2 + 1][];
for (int i = 0; i <= n / 2; i++)
{
DP[i] = new int[n + 1];
}
// Initialize the base case
for (int j = 0; j < n + 1; j++)
{
DP[0][j] = 0;
}
// Compute DP[i][j] for i > 0 and j > 0
for (int i = 1; i <= n / 2; i++)
{
for (int j = 1; j < n + 1; j++)
{
DP[i][j] = 0;
for (int k = 0; k < n; k++)
{
if ((j - k + n) % n < j)
{
DP[i][j] = Math.Max(DP[i][j], DP[i - 1][(j - k + n) % n] + arr[j - 1]);
}
}
DP[i][j] = Math.Max(DP[i][j], DP[i][j - 1]);
}
}
// Find the maximum value in the last row of DP
int maxSum = 0;
for (int j = 1; j < n + 1; j++)
{
maxSum = Math.Max(maxSum, DP[n / 2][j]);
}
// Return the maximum possible sum
return maxSum;
}
// Driver code
public static void Main(string[] args)
{
List<int> arr = new List<int> { 2, 3, 4, 5, 6, 7 };
// Function call
Console.WriteLine(MaxSumSublist(arr));
}
}
function maxSumSublist(arr) {
const n = arr.length;
const DP = new Array(Math.floor(n / 2) + 1).fill(null).map(() => new Array(n + 1).fill(0));
// Initialize the base case
for (let j = 0; j < n + 1; j++) {
DP[0][j] = 0;
}
// Compute DP[i][j] for i > 0 and j > 0
for (let i = 1; i < Math.floor(n / 2) + 1; i++) {
for (let j = 1; j < n + 1; j++) {
DP[i][j] = 0;
for (let k = 0; k < n; k++) {
if ((j - k + n) % n < j) {
DP[i][j] = Math.max(
DP[i][j],
DP[i - 1][(j - k + n) % n] + arr[j - 1]
);
}
}
DP[i][j] = Math.max(DP[i][j], DP[i][j - 1]);
}
}
// Find the maximum value in the last row of DP
let maxSum = 0;
for (let j = 1; j < n + 1; j++) {
maxSum = Math.max(maxSum, DP[Math.floor(n / 2)][j]);
}
// Return the maximum possible sum
return maxSum;
}
// Driver code
const arr = [2, 3, 4, 5, 6, 7];
// Function call
console.log(maxSumSublist(arr));
Time Complexity: O(N2)
Auxiliary Space: O(N2)
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