Given an array, arr[] consisting of N integers, the task is to minimize the number of increments of array elements required to make all differences pairs of array elements even.
Examples:
Input: arr[] = {4, 1, 2}
Output: 1
Explanation:
Operation 1: Increment arr[1] by 1. The array arr[] modifies to {4, 2, 2}.
All pairs: (4, 2) → difference = 2
(4, 2) → difference = 2
(2, 2) → difference = 0
Now, the pairwise differences between array elements is even. Hence, the answer is 1.Input: arr[] = {2, 4}
Output: 0
Explanation: Differences between all pairs of array elements is already even. Therefore, the answer is 0.
Approach: The given problem can be solved by observing the fact that, to make the difference between all pairs of array elements even, both the elements must be of the same parity. Therefore, the idea is to convert all the array elements to either even numbers or odd numbers. The minimum number of increments is equal to the minimum of the count of even and odd array elements.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find the minimum increments // required to difference between all // pairs of array elements even void minimumOperations( int arr[], int N) { // Store the count of odd // and even numbers int oddCnt = 0, evenCnt = 0; // Traverse the array for ( int i = 0; i < N; i++) { if (arr[i] % 2 == 0) { // Increment evenCnt by 1 evenCnt++; } else { // Increment eveCnt by 1 oddCnt++; } } // Print the minimum of oddCnt // and eveCnt cout << min(oddCnt, evenCnt); } // Driver Code int main() { int arr[] = { 4, 1, 2 }; int N = sizeof (arr) / sizeof (arr[0]); minimumOperations(arr, N); return 0; } |
Java
// Java program for the above approach public class GFG { // Function to find the minimum increments // required to difference between all // pairs of array elements even static void minimumOperations( int [] arr, int N) { // Store the count of odd // and even numbers int oddCnt = 0 , evenCnt = 0 ; // Traverse the array for ( int i = 0 ; i < N; i++) { if (arr[i] % 2 == 0 ) { // Increment evenCnt by 1 evenCnt++; } else { // Increment oddCnt by 1 oddCnt++; } } // Print the minimum of oddCnt // and eveCnt System.out.print(Math.min(oddCnt, evenCnt)); } // Driver code public static void main(String args[]) { int [] arr = { 4 , 1 , 2 }; int N = arr.length; minimumOperations(arr, N); } } // This code is contributed by AnkThon |
Python3
# Python program for the above approach # Function to find the minimum increments # required to difference between all # pairs of array elements even def minimumOperations(arr, N) : # Store the count of odd # and even numbers oddCnt = 0 evenCnt = 0 # Traverse the array for i in range (N): if (arr[i] % 2 = = 0 ) : # Increment evenCnt by 1 evenCnt + = 1 else : # Increment eveCnt by 1 oddCnt + = 1 # Prthe minimum of oddCnt # and eveCnt print ( min (oddCnt, evenCnt)) # Driver Code arr = [ 4 , 1 , 2 ] N = len (arr) minimumOperations(arr, N) # This code is contributed by sanjoy_62. |
C#
// C# program for the above approach using System; using System.Collections.Generic; class GFG { // Function to find the minimum increments // required to difference between all // pairs of array elements even static void minimumOperations( int [] arr, int N) { // Store the count of odd // and even numbers int oddCnt = 0, evenCnt = 0; // Traverse the array for ( int i = 0; i < N; i++) { if (arr[i] % 2 == 0) { // Increment evenCnt by 1 evenCnt++; } else { // Increment eveCnt by 1 oddCnt++; } } // Print the minimum of oddCnt // and eveCnt Console.Write(Math.Min(oddCnt, evenCnt)); } // Driver code static void Main() { int [] arr = { 4, 1, 2 }; int N = arr.Length; minimumOperations(arr, N); } } // This code is contributed by divyeshrabadiya07. |
1
Time Complexity: O(N)
Auxiliary Space: O(1)
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