Find maximum sum taking every Kth element in the array

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Given an array arr[] of integers and an integer K, the task is to find the maximum sum taking every K^th 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 K^th 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:

Time Complexity: O(N²) where N is the number of elements in array.
Auxiliary Space: O(1), no extra space required, so it is a constant.

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:

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

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