Minimum increment operations to make K elements equal

Given an array arr[] of N elements and an integer K, the task is to make any K elements of the array equal by performing only increment operations i.e. in one operation, any element can be incremented by 1. Find the minimum number of operations required to make any K elements equal.

Examples:

Input: arr[] = {3, 1, 9, 10}, K = 3
Output: 14
Increment 3 six times and 1 eight times for a total of
14 operations to make 3 elements equal to 9.



Input: arr[] = {5, 3, 10, 5}, K = 2
Output: 0
No operations are required as first and last
elements are already equal.

Naive approach:

  1. Sort the array in increasing order.
  2. Now select K elements and make them equal.
  3. Choose the ith value as the largest value and make all elements just smaller than it equal to the ith element.
  4. Calculate the number of operations needed to make K elements equal to the ith element for all i.
  5. The answer will be the minimum of all the possibilities.

Time Complexity of this approach will be O(n*K + nlogn).

Efficient approach: the naive approach can be modified to calculate the minimum operations needed to make K elements equal to the ith element faster than O(K) using the sliding window technique in constant time given that the operations required for making the 1st K elements equal to the Kth element are known.

Let C be the operations needed or cost for making the elements in the range [l, l + K – 1] equal to the (l + K – 1)th element. Now to find the cost for the range [l + 1, l + K], the solution for the range [l, l + K – 1] can be used.
Let C’ be the cost for the range [l + 1, l + K].

  1. Since we increment lth element to (l + K – 1)th element, C includes the cost element(l + k – 1) – element(l) but C’ does not need to include this cost.
    So, C’ = C – (element(l + k – 1) – element(l))
  2. Now C’ represents the cost of making all the elements in the range [l + 1, l + K – 1] equal to (l + K – 1)th element.
    Since, we need to make all elements equal to the (l + K)th element instead of the (l + K – 1)th element, we can increment these k – 1 elements to the (l + K)th element which makes C’ = C’ + (k – 1) * (element(l + k) – element(l + k -1))

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the minimum number of
// increment operations required to make
// any k elements of the array equal
int minOperations(vector<int> ar, int k)
{
    // Sort the array in increasing order
    sort(ar.begin(), ar.end());
  
    // Calculate the number of operations
    // needed to make 1st k elements equal to
    // the kth element i.e. the 1st window
    int opsNeeded = 0;
    for (int i = 0; i < k; i++) {
        opsNeeded += ar[k - 1] - ar[i];
    }
  
    // Answer will be the minimum of all
    // possible k sized windows
    int ans = opsNeeded;
  
    // Find the operations needed to make
    // k elements equal to ith element
    for (int i = k; i < ar.size(); i++) {
  
        // Slide the window to the right and
        // subtract increments spent on leftmost
        // element of the previous window
        opsNeeded = opsNeeded - (ar[i - 1] - ar[i - k]);
  
        // Add increments needed to make the 1st k-1
        // elements of this window equal to the
        // kth element of the current window
        opsNeeded += (k - 1) * (ar[i] - ar[i - 1]);
        ans = min(ans, opsNeeded);
    }
    return ans;
}
  
// Driver code
int main()
{
    vector<int> arr = { 3, 1, 9, 100 };
    int n = arr.size();
    int k = 3;
  
    cout << minOperations(arr, k);
  
    return 0;
}

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Python3

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# Python3 implementation of the approach
  
# Function to return the minimum number of
# increment operations required to make
# any k elements of the array equal
def minOperations(ar, k):
  
    # Sort the array in increasing order
    arr = sorted(ar)
  
    # Calculate the number of operations
    # needed to make 1st k elements equal to
    # the kth element i.e. the 1st window
    opsNeeded = 0
    for i in range(k):
        opsNeeded += ar[k - 1] - ar[i]
  
    # Answer will be the minimum of all
    # possible k sized windows
    ans = opsNeeded
  
    # Find the operations needed to make
    # k elements equal to ith element
    for i in range(k, len(ar)):
  
        # Slide the window to the right and
        # subtract increments spent on leftmost
        # element of the previous window
        opsNeeded = opsNeeded - (ar[i - 1] - ar[i - k])
  
        # Add increments needed to make the 1st k-1
        # elements of this window equal to the
        # kth element of the current window
        opsNeeded += (k - 1) * (ar[i] - ar[i - 1])
        ans = min(ans, opsNeeded)
  
    return ans
  
# Driver code
arr = [3, 1, 9, 100]
n = len(arr)
k = 3
  
print(minOperations(arr, k))
  
# This code is contributed by Mohit Kumar

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Output:

14

Time Complexity: O(nlogn)



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Final year BTech IT student at DTU, Upcoming Technology Analyst at Morgan Stanley

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Improved By : mohit kumar 29