Make all the elements of array even with given operations - GeeksforGeeks

31 May, 2021
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 operation.
If there is an odd number, then, increment the element and the next adjacent element by 1.
Each increment costs one operation.
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. The 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 elements for the given array arr[].If all the elements in arr[] are even, then print the number of operations.Else print -1.
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 elements for the given array arr[].
If all the elements in arr[] are even, then print the number of operations.
Else print -1.
Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// 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 evenint 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 program to make all array// element evenclass GFG{ // Function to count the total// number of operations needed to make// all array element evenstatic 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 codepublic 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 program to make all array# element even # Function to count the total# number of operations needed to make# all array element evendef 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# program to make all array// element evenusing System; class GFG{ // Function to count the total// number of operations needed to make// all array element evenstatic 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 codepublic 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
<script>// Javascript program to make all array// element even // Function to count the total// number of operations needed to make// all array element evenfunction countOperations(arr, n){    let count = 0;     // Traverse the given array    for (let 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 (let i = 0; i < n; i++) {        if (arr[i] & 1)            return -1;    }     // Returns the count of operations    return count;}  let arr = [ 2, 3, 4, 5, 6 ];let n = arr.length;document.write(countOperations(arr, n)); // This code is contributed by _saurabh_jaiswal</script>
4
 
Time Complexity: O(N) where N is the number of elements in the array. 
29AjayKumar
ankthon
_saurabh_jaiswal
Algorithms
Arrays
Greedy
Arrays
Greedy
Algorithms
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