Given an array arr[] of positive integers, find the minimum number of operations required to make all the array elements even where:
- If there is an odd number then, increment the element and the next adjacent element by 1.
- Each increment costs one operations.
Note: If there is any number in arr[] which is odd after all operations then, print -1.
Examples
Input: arr[] = {2, 3, 4, 5, 6}
Output: 4
Explanation:
Add 1 to 3(at 1st index) and add 1 to its adjacent element 4(2nd index).
Now the array becomes {2, 4, 5, 5, 6}.
Add 1 to 5(at 2nd index) and add 1 to its adjacent element 5(3rd index).
Now the array becomes {2, 4, 6, 6, 6}.
The resultant array has all even numbers.
The total number of operations for 4 increments is 4.Input: arr[] = {5, 6}
Output: -1
Explanation:
Adding 1 to 5(0th index) then we have to increment 1 to its adjacent element 6(1st index).
Now the array becomes {6, 7}.
And we have 1 odd number left after all possible increments. Therefore we can’t make all array elements even.
Approach:
This problem can be solved using Greedy Approach. Following are the steps:
- Traverse the given array arr[].
- If an odd element occurs then increment that element by 1 to make it even and the next adjacent element by 1.
- Repeat the above step for all the odd element for the given array arr[].
- If all the element in arr[] are even then print the number of operation.
- Else print -1.
Below is the implementation of the above approach:
C++
// C++ program to make all array // element even #include "bits/stdc++.h" using namespace std; // Function to count the total // number of operations needed to make // all array element even int countOperations(int arr[], int n) { int count = 0; // Traverse the given array for (int i = 0; i < n - 1; i++) { // If an odd element occurs // then increment that element // and next adjacent element // by 1 if (arr[i] & 1) { arr[i]++; arr[i + 1]++; count += 2; } } // Traverse the array if any odd // element occurs then return -1 for (int i = 0; i < n; i++) { if (arr[i] & 1) return -1; } // Returns the count of operations return count; } int main() { int arr[] = { 2, 3, 4, 5, 6 }; int n = sizeof(arr) / sizeof(int); cout << countOperations(arr, n); return 0; } |
Java
// Java program to make all array // element even class GFG { // Function to count the total // number of operations needed to make // all array element even static int countOperations(int arr[], int n) { int count = 0; // Traverse the given array for (int i = 0; i < n - 1; i++) { // If an odd element occurs // then increment that element // and next adjacent element // by 1 if (arr[i] % 2 == 1) { arr[i]++; arr[i + 1]++; count += 2; } } // Traverse the array if any odd // element occurs then return -1 for (int i = 0; i < n; i++) { if (arr[i] % 2 == 1) return -1; } // Returns the count of operations return count; } // Driver code public static void main(String[] args) { int arr[] = { 2, 3, 4, 5, 6 }; int n = arr.length; System.out.print(countOperations(arr, n)); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 program to make all array # element even # Function to count the total # number of operations needed to make # all array element even def countOperations(arr, n) : count = 0; # Traverse the given array for i in range(n - 1) : # If an odd element occurs # then increment that element # and next adjacent element # by 1 if (arr[i] & 1) : arr[i] += 1; arr[i + 1] += 1; count += 2; # Traverse the array if any odd # element occurs then return -1 for i in range(n) : if (arr[i] & 1) : return -1; # Returns the count of operations return count; if __name__ == "__main__" : arr = [ 2, 3, 4, 5, 6 ]; n = len(arr); print(countOperations(arr, n)); # This code is contributed by AnkitRai01 |
C#
// C# program to make all array // element even using System; class GFG { // Function to count the total // number of operations needed to make // all array element even static int countOperations(int []arr, int n) { int count = 0; // Traverse the given array for (int i = 0; i < n - 1; i++) { // If an odd element occurs // then increment that element // and next adjacent element // by 1 if (arr[i] % 2 == 1) { arr[i]++; arr[i + 1]++; count += 2; } } // Traverse the array if any odd // element occurs then return -1 for (int i = 0; i < n; i++) { if (arr[i] % 2 == 1) return -1; } // Returns the count of operations return count; } // Driver code public static void Main() { int []arr = { 2, 3, 4, 5, 6 }; int n = arr.Length; Console.Write(countOperations(arr, n)); } } // This code is contributed by AnkitRai01 |
4
Time Complexity: O(N) where N is the number of elements in array.
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