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Find maximum sum taking every Kth element in the array
  • Last Updated : 08 Jun, 2021

Given an array arr[] of integers and an integer K, the task is to find the maximum sum taking every Kth element i.e. sum = arr[i] + arr[i + k] + arr[i + 2 * k] + arr[i + 3 * k] + ……. arr[i + q * k] starting with any i.
Examples: 
 

Input: arr[] = {3, -5, 6, 3, 10}, K = 3 
Output: 10 
All possible sequence are: 
3 + 3 = 6 
-5 + 10 = 5 
6 = 6 
3 = 3 
10 = 10
Input: arr[] = {3, 6, 4, 7, 2}, K = 2 
Output: 13 
 

 

Naive Approach: The idea to solve this by using two nested loops and find the sum of every sequence starting from index i and sum every Kth element up to n, and find the maximum from all of these. The time complexity of this method will be O(N2)
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 sum for
// every possible sequence such that
// a[i] + a[i+k] + a[i+2k] + ... + a[i+qk]
// is maximized
int maxSum(int arr[], int n, int K)
{
 
    // Initialize the maximum with
    // the smallest value
    int maximum = INT_MIN;
 
    // Find maximum from all sequences
    for (int i = 0; i < n; i++) {
 
        int sumk = 0;
 
        // Sum of the sequence
        // starting from index i
        for (int j = i; j < n; j += K)
            sumk = sumk + arr[j];
 
        // Update maximum
        maximum = max(maximum, sumk);
    }
 
    return maximum;
}
 
// Driver code
int main()
{
    int arr[] = { 3, 6, 4, 7, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int K = 2;
 
    cout << maxSum(arr, n, K);
 
    return (0);
}

Java




// Java implementation of the approach
class GFG
{
     
// Function to return the maximum sum for
// every possible sequence such that
// a[i] + a[i+k] + a[i+2k] + ... + a[i+qk]
// is maximized
static int maxSum(int arr[], int n, int K)
{
 
    // Initialize the maximum with
    // the smallest value
    int maximum = Integer.MIN_VALUE;
 
    // Find maximum from all sequences
    for (int i = 0; i < n; i++)
    {
 
        int sumk = 0;
 
        // Sum of the sequence
        // starting from index i
        for (int j = i; j < n; j += K)
            sumk = sumk + arr[j];
 
        // Update maximum
        maximum = Math.max(maximum, sumk);
    }
 
    return maximum;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 3, 6, 4, 7, 2 };
    int n = arr.length;
    int K = 2;
 
    System.out.println(maxSum(arr, n, K));
}
}
 
// This code is contributed by Code_Mech

Python3




# Python 3 implementation of the approach
import sys
 
# Function to return the maximum sum for
# every possible sequence such that
# a[i] + a[i+k] + a[i+2k] + ... + a[i+qk]
# is maximized
def maxSum(arr, n, K):
     
    # Initialize the maximum with
    # the smallest value
    maximum = -sys.maxsize - 1
 
    # Find maximum from all sequences
    for i in range(n):
        sumk = 0
 
        # Sum of the sequence
        # starting from index i
        for j in range(i, n, K):
            sumk = sumk + arr[j]
 
        # Update maximum
        maximum = max(maximum, sumk)
 
    return maximum
 
# Driver code
if __name__ == '__main__':
    arr = [3, 6, 4, 7, 2]
    n = len(arr)
    K = 2
    print(maxSum(arr, n, K))
     
# This code is contributed by
# Surendra_Gangwar

C#




// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return the maximum sum for
// every possible sequence such that
// a[i] + a[i+k] + a[i+2k] + ... + a[i+qk]
// is maximized
static int maxSum(int []arr, int n, int K)
{
 
    // Initialize the maximum with
    // the smallest value
    int maximum = int.MinValue;
 
    // Find maximum from all sequences
    for (int i = 0; i < n; i++)
    {
 
        int sumk = 0;
 
        // Sum of the sequence
        // starting from index i
        for (int j = i; j < n; j += K)
            sumk = sumk + arr[j];
 
        // Update maximum
        maximum = Math.Max(maximum, sumk);
    }
 
    return maximum;
}
 
// Driver code
public static void Main()
{
    int []arr = { 3, 6, 4, 7, 2 };
    int n = arr.Length;
    int K = 2;
 
    Console.WriteLine(maxSum(arr, n, K));
}
}
 
// This code is contributed by Akanksha Rai

PHP




<?php
// PHP implementation of the approach
 
// Function to return the maximum sum for
// every possible sequence such that
// a[i] + a[i+k] + a[i+2k] + ... + a[i+qk]
// is maximized
function maxSum($arr, $n, $K)
{
 
    // Initialize the maximum with
    // the smallest value
    $maximum = PHP_INT_MIN;
 
    // Find maximum from all sequences
    for ($i = 0; $i < $n; $i++)
    {
        $sumk = 0;
 
        // Sum of the sequence
        // starting from index i
        for ($j = $i; $j < $n; $j += $K)
            $sumk = $sumk + $arr[$j];
 
        // Update maximum
        $maximum = max($maximum, $sumk);
    }
 
    return $maximum;
}
 
// Driver code
$arr = array(3, 6, 4, 7, 2);
$n = sizeof($arr);
$K = 2;
 
echo maxSum($arr, $n, $K);
 
// This code is contributed by Akanksha Rai
?>

Javascript




<script>
 
 
// JavaScript implementation of the approach
 
// Function to return the maximum sum for
// every possible sequence such that
// a[i] + a[i+k] + a[i+2k] + ... + a[i+qk]
// is maximized
function maxSum(arr, n, K)
{
 
    // Initialize the maximum with
    // the smallest value
    var maximum = -1000000000;
 
    // Find maximum from all sequences
    for (var i = 0; i < n; i++) {
 
        var sumk = 0;
 
        // Sum of the sequence
        // starting from index i
        for (var j = i; j < n; j += K)
            sumk = sumk + arr[j];
 
        // Update maximum
        maximum = Math.max(maximum, sumk);
    }
 
    return maximum;
}
 
// Driver code
var arr = [3, 6, 4, 7, 2];
var n = arr.length;
var K = 2;
document.write( maxSum(arr, n, K));
 
 
</script>
Output: 



13

 

Efficient Approach: This problem can be solved by using the concept of Suffix Arrays, we iterate the array from right side and store the suffix sum for each (i+k)’th element (ie., i+k < n) , and find the maximum sum. The time complexity of this method will be O(N). 
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 sum for
// every possible sequence such that
// a[i] + a[i+k] + a[i+2k] + ... + a[i+qk]
// is maximized
int maxSum(int arr[], int n, int K)
{
 
    // Initialize the maximum with
    // the smallest value
    int maximum = INT_MIN;
 
    // Initialize the sum array with zero
    int sum[n] = { 0 };
 
    // Iterate from the right
    for (int i = n - 1; i >= 0; i--) {
 
        // Update the sum starting at
        // the current element
        if (i + K < n)
            sum[i] = sum[i + K] + arr[i];
        else
            sum[i] = arr[i];
 
        // Update the maximum so far
        maximum = max(maximum, sum[i]);
    }
 
    return maximum;
}
 
// Driver code
int main()
{
    int arr[] = { 3, 6, 4, 7, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int K = 2;
 
    cout << maxSum(arr, n, K);
 
    return (0);
}

Java




// Java implementation of the approach
class GFG {
 
    // Function to return the maximum sum for
    // every possible sequence such that
    // a[i] + a[i+k] + a[i+2k] + ... + a[i+qk]
    // is maximized
    static int maxSum(int arr[], int n, int K)
    {
 
        // Initialize the maximum with
        // the smallest value
        int maximum = Integer.MIN_VALUE;
 
        // Initialize the sum array with zero
        int[] sum = new int[n];
 
        // Iterate from the right
        for (int i = n - 1; i >= 0; i--) {
 
            // Update the sum starting at
            // the current element
            if (i + K < n)
                sum[i] = sum[i + K] + arr[i];
            else
                sum[i] = arr[i];
 
            // Update the maximum so far
            maximum = Math.max(maximum, sum[i]);
        }
 
        return maximum;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 3, 6, 4, 7, 2 };
        int n = arr.length;
        int K = 2;
 
        System.out.print(maxSum(arr, n, K));
    }
}

Python




# Python implementation of the approach
 
# Function to return the maximum sum for
# every possible sequence such that
# a[i] + a[i + k] + a[i + 2k] + ... + a[i + qk]
# is maximized
def maxSum(arr, n, K):
     
    # Initialize the maximum with
    # the smallest value
    maximum = -2**32;
 
    # Initialize the sum array with zero
    sum = [0]*n
 
    # Iterate from the right
    for i in range(n-1, -1, -1):
         
        # Update the sum starting at
        # the current element
        if( i + K < n ):
            sum[i] = sum[i + K] + arr[i]
        else:
            sum[i] = arr[i];
     
        # Update the maximum so far
        maximum = max( maximum, sum[i] )
     
    return maximum;
 
# Driver code
arr = [3, 6, 4, 7, 2]
n = len(arr);
K = 2
print(maxSum(arr, n, K))

C#




// C# implementation of the approach
using System;
class GFG {
 
    // Function to return the maximum sum for
    // every possible sequence such that
    // a[i] + a[i+k] + a[i+2k] + ... + a[i+qk]
    // is maximized
    static int maxSum(int[] arr, int n, int K)
    {
 
        // Initialize the maximum with
        // the smallest value
        int maximum = int.MinValue;
 
        // Initialize the sum array with zero
        int[] sum = new int[n];
 
        // Iterate from the right
        for (int i = n - 1; i >= 0; i--) {
 
            // Update the sum starting at
            // the current element
            if (i + K < n)
                sum[i] = sum[i + K] + arr[i];
            else
                sum[i] = arr[i];
 
            // Update the maximum so far
            maximum = Math.Max(maximum, sum[i]);
        }
 
        return maximum;
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 3, 6, 4, 7, 2 };
        int n = arr.Length;
        int K = 2;
 
        Console.Write(maxSum(arr, n, K));
    }
}

PHP




<?php
// PHP implementation of the approach
// Function to return the maximum sum for
// every possible sequence such that
// a[i] + a[i+k] + a[i+2k] + ... + a[i+qk]
// is maximized
function maxSum($arr, $n, $K)
{
 
    // Initialize the maximum with
    // the smallest value
    $maximum = PHP_INT_MIN;
 
    // Initialize the sum array with zero
    $sum = array($n);
 
    // Iterate from the right
    for ($i = $n - 1; $i >= 0; $i--)
    {
 
        // Update the sum starting at
        // the current element
        if ($i + $K < $n)
            $sum[$i] = $sum[$i + $K] + $arr[$i];
        else
            $sum[$i] = $arr[$i];
 
        // Update the maximum so far
        $maximum = max($maximum, $sum[$i]);
    }
 
    return $maximum;
}
 
// Driver code
{
    $arr = array(3, 6, 4, 7, 2 );
    $n = sizeof($arr);
    $K = 2;
 
    echo(maxSum($arr, $n, $K));
}
 
// This code is contributed by Learner_

Javascript




<script>
 
// JavaScript implementation of the approach
 
// Function to return the maximum sum for
// every possible sequence such that
// a[i] + a[i+k] + a[i+2k] + ... + a[i+qk]
// is maximized
function maxSum(arr, n, K)
{
 
    // Initialize the maximum with
    // the smallest value
    var maximum = -1000000000;
 
    // Initialize the sum array with zero
    var sum = Array(n).fill(0);
 
    // Iterate from the right
    for (var i = n - 1; i >= 0; i--) {
 
        // Update the sum starting at
        // the current element
        if (i + K < n)
            sum[i] = sum[i + K] + arr[i];
        else
            sum[i] = arr[i];
 
        // Update the maximum so far
        maximum = Math.max(maximum, sum[i]);
    }
 
    return maximum;
}
 
// Driver code
var arr = [3, 6, 4, 7, 2 ];
var n = arr.length;
var K = 2;
document.write( maxSum(arr, n, K));
 
 
</script>
Output: 
13

 

Time Complexity: O(N) where N is the number of elements in array.
 

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