Skip to content
Related Articles

Related Articles

Improve Article

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

  • Difficulty Level : Medium
  • Last Updated : 31 May, 2021

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

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :