Minimize adding odd and subtracting even numbers to make all array elements equal to K
Given an array, arr[] of size N and an integer K, the task is to find the minimum number of operations required to make all array elements equal to K by performing the following operations any number of times:
- Convert arr[i] to arr[i] + X, where X is an odd number.
- Convert arr[i] to arr[i] – Y, where Y is an even number.
Examples:
Input: arr[] = {8, 7, 2, 1, 3}, K = 5
Output: 8
Explanation: To make all elements of the given array equal to K(= 5), following operations are required:
arr[0] = arr[0] + X, X = 1
arr[0] = arr[0] – Y, Y = 4
arr[1] = arr[1] – Y, Y = 2
arr[2] = arr[2] + X, X = 3
arr[3] = arr[3] + X, X = 3
arr[3] = arr[3] + X, X = 1
arr[4] = arr[4] + X, X = 1
arr[4] = arr[4] + X, X = 1Input: arr[] = {1, 2, 3, 4, 5, 6, 7}, K = 3
Output: 9
Approach: The problem can be solved using the Greedy technique. Following are the observations:
Even + Even = Even
Even + Odd = Odd
Odd + Odd = Even
Odd + Even = Odd
Follow the steps below to solve the problem:
- Traverse the given array and check the following conditions.
- If K > arr[i] and (K – arr[i]) % 2 == 0 then add two odd numbers(X) into arr[i]. Therefore, total 2 operations required.
- If K > arr[i] and (K – arr[i]) % 2 != 0 then add one odd numbers(X) into arr[i]. Therefore, total 1 operations required.
- If K < arr[i] and (arr[i] – arr[i]) % 2 == 0 then subtract one even numbers(Y) into arr[i]. Therefore, total 1 operations required.
- If K < arr[i] and (K – arr[i]) % 2 != 0 then add an odd numbers(X) into arr[i] and subtract an even numbers(Y) from arr[i]. Therefore, total 2 operations required.
- Finally, print the total number of operations required to make all the array elements equal to K.
Below is the implementation of the above approach
C++
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Function to find the minimum operations // required to make array elements equal to K int MinOperation( int arr[], int N, int K) { // Stores minimum count of operations int cntOpe = 0; // Traverse the given array for ( int i = 0; i < N; i++) { // If K is greater than arr[i] if (K > arr[i]) { // If (K - arr[i]) is even if ((K - arr[i]) % 2 == 0) { // Update cntOpe cntOpe += 2; } else { // Update cntOpe cntOpe += 1; } } // If K is less than arr[i] else if (K < arr[i]) { // If (arr[i] - K) is even if ((K - arr[i]) % 2 == 0) { // Update cntOpe cntOpe += 1; } else { // Update cntOpe cntOpe += 2; } } } return cntOpe; } // Driver Code int main() { int arr[] = { 8, 7, 2, 1, 3 }; int K = 5; int N = sizeof (arr) / sizeof (arr[0]); cout << MinOperation(arr, N, K); return 0; } |
Java
// Java program to implement // the above approach class GFG{ // Function to find the minimum // operations required to make // array elements equal to K public static int MinOperation( int arr[], int N, int K) { // Stores minimum count of // operations int cntOpe = 0 ; // Traverse the given array for ( int i = 0 ; i < N; i++) { // If K is greater than // arr[i] if (K > arr[i]) { // If (K - arr[i]) is even if ((K - arr[i]) % 2 == 0 ) { // Update cntOpe cntOpe += 2 ; } else { // Update cntOpe cntOpe += 1 ; } } // If K is less than // arr[i] else if (K < arr[i]) { // If (arr[i] - K) is // even if ((K - arr[i]) % 2 == 0 ) { // Update cntOpe cntOpe += 1 ; } else { // Update cntOpe cntOpe += 2 ; } } } return cntOpe; } // Driver code public static void main(String[] args) { int arr[] = { 8 , 7 , 2 , 1 , 3 }; int K = 5 ; int N = arr.length; System.out.println( MinOperation(arr, N, K)); } } // This code is contributed by divyeshrabadiya07 |
Python3
# Python3 program to implement # the above approach # Function to find the minimum operations # required to make array elements equal to K def MinOperation(arr, N, K): # Stores minimum count of operations cntOpe = 0 # Traverse the given array for i in range (N): # If K is greater than arr[i] if (K > arr[i]): # If (K - arr[i]) is even if ((K - arr[i]) % 2 = = 0 ): # Update cntOpe cntOpe + = 2 else : # Update cntOpe cntOpe + = 1 # If K is less than arr[i] elif (K < arr[i]): # If (arr[i] - K) is even if ((K - arr[i]) % 2 = = 0 ): # Update cntOpe cntOpe + = 1 else : # Update cntOpe cntOpe + = 2 return cntOpe # Driver Code arr = [ 8 , 7 , 2 , 1 , 3 ] K = 5 N = len (arr) print (MinOperation(arr, N, K)) # This code is contributed by sanjoy_62 |
C#
// C# program to implement // the above approach using System; class GFG{ // Function to find the minimum // operations required to make // array elements equal to K public static int MinOperation( int []arr, int N, int K) { // Stores minimum count of // operations int cntOpe = 0; // Traverse the given array for ( int i = 0; i < N; i++) { // If K is greater than // arr[i] if (K > arr[i]) { // If (K - arr[i]) is even if ((K - arr[i]) % 2 == 0) { // Update cntOpe cntOpe += 2; } else { // Update cntOpe cntOpe += 1; } } // If K is less than // arr[i] else if (K < arr[i]) { // If (arr[i] - K) is // even if ((K - arr[i]) % 2 == 0) { // Update cntOpe cntOpe += 1; } else { // Update cntOpe cntOpe += 2; } } } return cntOpe; } // Driver code public static void Main(String[] args) { int []arr = {8, 7, 2, 1, 3}; int K = 5; int N = arr.Length; Console.WriteLine( MinOperation(arr, N, K)); } } // This code is contributed by Amit Katiyar |
Javascript
<script> // Javascript program to implement // the above approach // Function to find the minimum // operations required to make // array elements equal to K function MinOperation(arr, N, K) { // Stores minimum count of // operations let cntOpe = 0; // Traverse the given array for (let i = 0; i < N; i++) { // If K is greater than // arr[i] if (K > arr[i]) { // If (K - arr[i]) is even if ((K - arr[i]) % 2 == 0) { // Update cntOpe cntOpe += 2; } else { // Update cntOpe cntOpe += 1; } } // If K is less than // arr[i] else if (K < arr[i]) { // If (arr[i] - K) is // even if ((K - arr[i]) % 2 == 0) { // Update cntOpe cntOpe += 1; } else { // Update cntOpe cntOpe += 2; } } } return cntOpe; } // Driver Code let arr = [8, 7, 2, 1, 3]; let K = 5; let N = arr.length; document.write( MinOperation(arr, N, K)); </script> |
8
Time Complexity: O(N)
Auxiliary Space: O(1)
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