Given an array of N integers. The task is to eliminate the minimum number of elements such that in the resulting array the sum of any two adjacent values is odd.
Examples:
Input: arr[] = {1, 2, 3}
Output: 0
Sum of all adjacent elements is already odd.Input: arr[] = {1, 3, 5, 4, 2}
Output: 3
Eliminate 3, 5 and 2 so that in the resulting array the sum of any two adjacent values is odd.
Approach The sum of 2 numbers is odd if one of them is odd and the other is even. This means for every pair of consecutive numbers that have the same parity, eliminate one of them, it doesn’t matter which. So the following greedy algorithm works:
- Go through all the elements in order.
- If the current number has the same parity as the previous one eliminate it, otherwise don’t.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include<bits/stdc++.h>using namespace std;// Returns the minimum number of eliminationsint min_elimination(int n, int arr []){ int count = 0; // Stores the previous element int prev_val = arr[0]; // Stores the new value for (int i = 1; i < n; i++) { int curr_val = arr[i]; // Check if the previous and current // values are of same parity if (curr_val % 2 == prev_val % 2) count++; // Previous value is now the current value prev_val = curr_val; } // Return the counter variable return count;}// Driver codeint main(){ int arr [] = { 1, 2, 3, 7, 9 }; int n = sizeof(arr)/sizeof(arr[0]); cout << min_elimination(n, arr); return 0;}// This code is contributed by ihritik |
Java
// Java implementation of the above approachclass GFG { // Returns the minimum number of eliminations static int min_elimination(int n, int[] arr) { int count = 0; // Stores the previous element int prev_val = arr[0]; // Stores the new value for (int i = 1; i < n; i++) { int curr_val = arr[i]; // Check if the previous and current // values are of same parity if (curr_val % 2 == prev_val % 2) count++; // Previous value is now the current value prev_val = curr_val; } // Return the counter variable return count; } // Driver code public static void main(String[] args) { int[] arr = new int[] { 1, 2, 3, 7, 9 }; int n = arr.length; System.out.println(min_elimination(n, arr)); }} |
Python3
# Python3 implementation of the above approach# Returns the minimum number of eliminationsdef min_elimination(n, arr): count = 0 # Stores the previous element prev_val = arr[0] # Stores the new value for i in range (1, n): curr_val = arr[i]; # Check if the previous and current # values are of same parity if (curr_val % 2 == prev_val % 2): count = count + 1 # Previous value is now the current value prev_val = curr_val # Return the counter variable return count# Driver codearr = [ 1, 2, 3, 7, 9 ]n = len(arr)print(min_elimination(n, arr));# This code is contributed by ihritik |
C#
// C# implementation of the above approachusing System;class GFG{ // Returns the minimum number of eliminations static int min_elimination(int n, int[] arr) { int count = 0; // Stores the previous element int prev_val = arr[0]; // Stores the new value for (int i = 1; i < n; i++) { int curr_val = arr[i]; // Check if the previous and current // values are of same parity if (curr_val % 2 == prev_val % 2) count++; // Previous value is now the current value prev_val = curr_val; } // Return the counter variable return count; } // Driver code public static void Main() { int[] arr = new int[] { 1, 2, 3, 7, 9 }; int n = arr.Length; Console.WriteLine(min_elimination(n, arr)); }}// This code is contributed by ihritik |
PHP
<?php// PHP implementation of the above approach // Returns the minimum number of eliminations function min_elimination($n, $arr) { $count = 0; // Stores the previous element $prev_val = $arr[0]; // Stores the new value for ($i = 1; $i < $n; $i++) { $curr_val = $arr[$i]; // Check if the previous and current // values are of same parity if ($curr_val % 2 == $prev_val % 2) $count++; // Previous value is now the // current value $prev_val = $curr_val; } // Return the counter variable return $count; } // Driver code $arr = array( 1, 2, 3, 7, 9 ); $n = sizeof($arr); echo min_elimination($n, $arr); // This code is contributed by Ryuga?> |
Javascript
<script>// JavaScript implementation of the above approach// Returns the minimum number of eliminationsfunction min_elimination(n,arr){ let count = 0; // Stores the previous element let prev_val = arr[0]; // Stores the new value for (let i = 1; i < n; i++) { let curr_val = arr[i]; // Check if the previous and current // values are of same parity if (curr_val % 2 == prev_val % 2) count++; // Previous value is now the current value prev_val = curr_val; } // Return the counter variable return count;}// Driver codelet arr = [1, 2, 3, 7, 9];let n = arr.length;document.write(min_elimination(n, arr));</script> |
Output
2
Complexity Analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(1)
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