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Pick maximum sum M elements such that contiguous repetitions do not exceed K

  • Difficulty Level : Medium
  • Last Updated : 25 Apr, 2022

Given an array arr[] of distinct elements and two integers M and K, the task is to generate an array from the given array elements (elements can repeat in the generated array) such that the size of the generated array is M and the length of any sub-array with all same elements must not exceed K. Print the maximum sum of the elements among all the possible arrays that can be generated.
Examples: 
 

Input: arr[] = {1, 3, 6, 7, 4, 5}, M = 9, K = 2 
Output: 60 
The maxim sum arrangement is 7 7 6 7 7 6 7 7 6. Note that there is no subarray of size more than 2 with all same elements.
Input: arr[] = {8, 13, 9, 17, 4, 12}, M = 5, K = 1 
Output: 77 
The maxim sum arrangement is 17, 13, 17, 13, 17 
 

 

Approach: If we want the maximum sum we have to take maximum value from the array but we can repeat this maximum value at most K times so we have to separate it by second maximum value only once and after that we again take first maximum value up to K times and this cycle goes on until we take total M values.
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 required sum
long int maxSum(int arr[], int n, int m, int k)
{
 
    int max1 = -1, max2 = -1;
 
    // All the elements in the array are distinct
    // Finding the maximum and the second maximum
    // element from the array
    for (int i = 0; i < n; i++) {
        if (arr[i] > max1) {
            max2 = max1;
            max1 = arr[i];
        }
        else if (arr[i] > max2)
            max2 = arr[i];
    }
 
    // Total times the second maximum element
    // will appear in the generated array
    int counter = m / (k + 1);
 
    long int sum = counter * max2 + (m - counter) * max1;
 
    // Return the required sum
    return sum;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 3, 6, 7, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int m = 9, k = 2;
    cout << maxSum(arr, n, m, k);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function to return the maximum required sum
static int maxSum(int arr[], int n, int m, int k)
{
 
    int max1 = -1, max2 = -1;
 
    // All the elements in the array are distinct
    // Finding the maximum and the second maximum
    // element from the array
    for (int i = 0; i < n; i++)
    {
        if (arr[i] > max1)
        {
            max2 = max1;
            max1 = arr[i];
        }
        else if (arr[i] > max2)
            max2 = arr[i];
    }
 
    // Total times the second maximum element
    // will appear in the generated array
    int counter = m / (k + 1);
 
    int sum = counter * max2 + (m - counter) * max1;
 
    // Return the required sum
    return sum;
}
 
// Driver code
public static void main(String args[])
{
    int arr[] = { 1, 3, 6, 7, 4, 5 };
    int n = arr.length;
    int m = 9, k = 2;
    System.out.println(maxSum(arr, n, m, k));
}
}
 
// This code is contributed by
// Surendra Gangwar

Python3




# Python3 implementation of the approach
def maxSum(arr, n, m, k):
 
    max1 = -1
    max2 = -1
     
    # All the elements in the array are distinct
    # Finding the maximum and the second maximum
    # element from the array
    for i in range(0, n):
        if(arr[i] > max1):
            max2 = max1
            max1 = arr[i]
        elif(arr[i] > max2):
            max2 = arr[i]
     
    # Total times the second maximum element
    # will appear in the generated array
    counter = int(m / (k + 1))
     
    sum = counter * max2 + (m - counter) * max1
     
    # Return the required sum
    return int(sum)
 
# Driver code
arr = [1, 3, 6, 7, 4, 5]
n = len(arr)
m = 9
k = 2
 
print(maxSum(arr, n, m, k))

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the maximum required sum
static int maxSum(int []arr, int n, int m, int k)
{
 
    int max1 = -1, max2 = -1;
 
    // All the elements in the array are distinct
    // Finding the maximum and the second maximum
    // element from the array
    for (int i = 0; i < n; i++)
    {
        if (arr[i] > max1)
        {
            max2 = max1;
            max1 = arr[i];
        }
        else if (arr[i] > max2)
            max2 = arr[i];
    }
 
    // Total times the second maximum element
    // will appear in the generated array
    int counter = m / (k + 1);
 
    int sum = counter * max2 + (m - counter) * max1;
 
    // Return the required sum
    return sum;
}
 
// Driver code
public static void Main(String []args)
{
    int []arr = { 1, 3, 6, 7, 4, 5 };
    int n = arr.Length;
    int m = 9, k = 2;
    Console.WriteLine(maxSum(arr, n, m, k));
}
}
 
/* This code contributed by PrinciRaj1992 */

Javascript




<script>
 
// JavaScript implementation of the approach
 
// Function to return the maximum required sum
function maxSum(arr, n, m, k)
{
 
    var max1 = -1, max2 = -1;
 
    // All the elements in the array are distinct
    // Finding the maximum and the second maximum
    // element from the array
    for (var i = 0; i < n; i++) {
        if (arr[i] > max1) {
            max2 = max1;
            max1 = arr[i];
        }
        else if (arr[i] > max2)
            max2 = arr[i];
    }
 
    // Total times the second maximum element
    // will appear in the generated array
    var counter = m / (k + 1);
 
    var sum = counter * max2 + (m - counter) * max1;
 
    // Return the required sum
    return sum;
}
 
// Driver code
var arr = [1, 3, 6, 7, 4, 5];
var n = arr.length;
var m = 9, k = 2;
document.write( maxSum(arr, n, m, k));
 
 
</script>
Output: 
60

 

Time Complexity: O(n), where n is the size of the given array.
Auxiliary Space: O(1), no extra space used.


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