Найдите максимальную сумму, взяв каждый K-й элемент в массиве

Опубликовано: 14 Января, 2022

Учитывая массив целых чисел arr [] и целое число K , задача состоит в том, чтобы найти максимальную сумму, взяв каждый K- й элемент, т.е. sum = arr [i] + arr [i + k] + arr [i + 2 * k] + arr [i + 3 * k] + ……. arr [i + q * k], начиная с любого i .
Примеры:

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 
 

Рекомендуется: сначала попробуйте свой подход в {IDE}, прежде чем переходить к решению.

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

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