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, 100}, 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:
- Sort the array in increasing order.
- Now select K elements and make them equal.
- Choose the ith value as the largest value and make all elements just smaller than it equal to the ith element.
- Calculate the number of operations needed to make K elements equal to the ith element for all i.
- The answer will be the minimum of all the possibilities.
C++14
#include <bits/stdc++.h> using namespace std; int minOperations(vector< int > ar, int & n, int & k) { // Sort the array in increasing order sort(ar.begin(), ar.end()); int opsneeded, ans = INT_MAX; for ( int i = k; i < n; i++) { opsneeded = 0; for ( int j = i - k; j < i; j++) opsneeded += ar[i - 1] - ar[j]; ans = min(ans, opsneeded); } return ans; } int main() { vector< int > arr = { 3, 1, 9, 100 }; int n = arr.size(); int k = 3; cout << minOperations(arr, n, k); return 0; } // this code is contributed by prophet1999 |
Java
// JAVA code for the above approach import java.util.*; class GFG { public static int minOperations(ArrayList<Integer> ar, int n, int k) { // Sort the array in increasing order Collections.sort(ar); int opsneeded, ans = Integer.MAX_VALUE; for ( int i = k; i < n; i++) { opsneeded = 0 ; for ( int j = i - k; j < i; j++) opsneeded += ar.get(i - 1 ) - ar.get(j); ans = Math.min(ans, opsneeded); } return ans; } public static void main(String[] args) { ArrayList<Integer> arr = new ArrayList<Integer>( Arrays.asList( 3 , 1 , 9 , 100 )); int n = arr.size(); int k = 3 ; System.out.print(minOperations(arr, n, k)); } } // This code is contributed by Taranpreet |
Python3
import math def minOperations(ar, n, k): # Sort the array in increasing order ar.sort() opsneeded, ans = math.inf, math.inf for i in range (k, n): opsneeded = 0 for j in range (i - k, i): opsneeded + = ar[i - 1 ] - ar[j] ans = min (ans, opsneeded) return ans # Driver code arr = [ 3 , 1 , 9 , 100 ] n = len (arr) k = 3 print (minOperations(arr, n, k)) |
C#
// C# code for the above approach using System; using System.Collections.Generic; using System.Linq; class GFG { public static int minOperations(List< int > ar, int n, int k) { // Sort the array in increasing order ar.Sort(); int opsneeded, ans = int .MaxValue; for ( int i = k; i < n; i++) { opsneeded = 0; for ( int j = i - k; j < i; j++) opsneeded += ar[i - 1] - ar[j]; ans = Math.Min(ans, opsneeded); } return ans; } // Driver code public static void Main( string [] args) { List< int > arr = new List< int >( new int [] {3, 1, 9, 100}); int n = arr.Count; int k = 3; Console.Write(minOperations(arr, n, k)); } } // This code is contributed by phasing17 |
Javascript
// javascript approach for the same function minOperations(ar,n,k){ // sort the array in increasing order ar.sort((a,b) => a-b); let opsneeded; let ans=Infinity; // loop to find the minimum operations for (let i = k; i < n; i++) { opsneeded = 0 for (let j = i - k; j < i; j++) { opsneeded += ar[i - 1] - ar[j] } ans = Math.min(ans, opsneeded) } return ans; } // Driver code let arr = [3, 1, 9, 100] let k = 3 console.log(minOperations(arr,arr.length,k)); |
14
Time Complexity: Depends on sorting, it will be either O(n^2+n*K) or O(n log n+n*K)
Auxiliary Space: Depends on sorting, it will be either O(n) or O(1)
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 β 1] equal to the (l + K β 1)th element. Now to find the cost for the range [l + 1, l + K], the solution for the range [l, l + K β 1] can be used.
Let Cβ be the cost for the range [l + 1, l + K].
- Since we increment lth element to (l + K β 1)th element, C includes the cost element(l + k β 1) β element(l) but Cβ does not need to include this cost.
So, Cβ = C β (element(l + k β 1) β element(l))
- Now Cβ represents the cost of making all the elements in the range [l + 1, l + K β 1] equal to (l + K β 1)th element.
Since, we need to make all elements equal to the (l + K)th element instead of the (l + K β 1)th element, we can increment these k β 1 elements to the (l + K)th element which makes Cβ = Cβ + (k β 1) * (element(l + k) β element(l + k -1))
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 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; } |
Java
// Java implementation of the approach import java.util.Arrays; class w3wiki { // Function to return the minimum number of // increment operations required to make // any k elements of the array equal static int minOperations( int ar[], int k) { // Sort the array in increasing order Arrays.sort(ar); // 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.length; 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 = Math.min(ans, opsNeeded); } return ans; } // Driver code public static void main(String[] args) { int [] arr = { 3 , 1 , 9 , 100 }; int n = arr.length; int k = 3 ; System.out.printf( "%d" ,minOperations(arr, k)); } } // This code is contributed by Atul_kumar_Shrivastava |
Python3
# 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 ar = 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 |
C#
// C# implementation of the approach using System; class w3wiki { // Function to return the minimum number of // increment operations required to make // any k elements of the array equal static int minOperations( int [] ar, int k) { // Sort the array in increasing order Array.Sort(ar); // 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.Length; 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 = Math.Min(ans, opsNeeded); } return ans; } // Driver code public static void Main() { int [] arr = { 3, 1, 9, 100 }; int n = arr.Length; int k = 3; Console.Write(minOperations(arr, k)); } } // This code is contributed by AbhiThakur |
Javascript
<script> // JavaScript implementation of the approach // Function to return the minimum number of // increment operations required to make // any k elements of the array equal function minOperations(ar,k) { // Sort the array in increasing order ar.sort( function (a,b){ return a-b}); // Calculate the number of operations // needed to make 1st k elements equal to // the kth element i.e. the 1st window let opsNeeded = 0; for (let i = 0; i < k; i++) { opsNeeded += ar[k - 1] - ar[i]; } // Answer will be the minimum of all // possible k sized windows let ans = opsNeeded; // Find the operations needed to make // k elements equal to ith element for (let i = k; i < ar.length; 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 = Math.min(ans, opsNeeded); } return ans; } // Driver code let arr=[3, 1, 9, 100 ]; let n = arr.length; let k = 3; document.write(minOperations(arr, k)); // This code is contributed by patel2127 </script> |
14
Time Complexity: Depends on sorting, it will be either O(n^2) or O(n log n)
Auxiliary Space: Depends on sorting, it will be either O(n) or O(1)