Given an array arr[] consisting of N positive integers, the task is to make all array elements even by replacing any pair of array elements with their sum.
Examples:
Input: arr[] = {5, 6, 3, 7, 20}
Output: 3
Explanation:
Operation 1: Replace arr[0] and arr[2] by their sum ( = 5 + 3 = 8) modifies arr[] to {8, 6, 8, 7, 20}.
Operation 2: Replace arr[2] and arr[3] by their sum ( = 7 + 8 = 15) modifies arr[] to {8, 6, 15, 15, 20}.
Operation 3: Replace arr[2] and arr[3] by their sum ( = 15 + 15 = 30) modifies arr[] to {8, 6, 30, 30, 20}.Input: arr[] = {2, 4, 16, 8, 7, 9, 3, 1}
Output: 2
Approach: The idea is to keep replacing two odd array elements by their sum until all array elements are even. Follow the steps below to solve the problem:
- Initialize a variable, say moves, to store the minimum number of replacements required.
- Calculate the total number of odd elements present in the given array and store it in a variable, say cnt.
- If the value of cnt is odd, then print (cnt / 2 + 2) as the result. Otherwise, print cnt / 2 as the result.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the minimum number // of replacements required to make // all array elements even void minMoves( int arr[], int N)
{ // Stores the count of odd elements
int odd_element_cnt = 0;
// Traverse the array
for ( int i = 0; i < N; i++) {
// Increase count of odd elements
if (arr[i] % 2 != 0) {
odd_element_cnt++;
}
}
// Store number of replacements required
int moves = (odd_element_cnt) / 2;
// Two extra moves will be required
// to make the last odd element even
if (odd_element_cnt % 2 != 0)
moves += 2;
// Print the minimum replacements
cout << moves;
} // Driver Code int main()
{ int arr[] = { 5, 6, 3, 7, 20 };
int N = sizeof (arr) / sizeof (arr[0]);
// Function call
minMoves(arr, N);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function to find the minimum number // of replacements required to make // all array elements even static void minMoves( int arr[], int N)
{ // Stores the count of odd elements
int odd_element_cnt = 0 ;
// Traverse the array
for ( int i = 0 ; i < N; i++)
{
// Increase count of odd elements
if (arr[i] % 2 != 0 )
{
odd_element_cnt++;
}
}
// Store number of replacements required
int moves = (odd_element_cnt) / 2 ;
// Two extra moves will be required
// to make the last odd element even
if (odd_element_cnt % 2 != 0 )
moves += 2 ;
// Print the minimum replacements
System.out.print(moves);
} // Driver Code public static void main(String[] args)
{ int arr[] = { 5 , 6 , 3 , 7 , 20 };
int N = arr.length;
// Function call
minMoves(arr, N);
} } // This code is contributed by shikhasingrajput |
// C# program for the above approach using System;
public class GFG
{ // Function to find the minimum number
// of replacements required to make
// all array elements even
static void minMoves( int []arr, int N)
{
// Stores the count of odd elements
int odd_element_cnt = 0;
// Traverse the array
for ( int i = 0; i < N; i++)
{
// Increase count of odd elements
if (arr[i] % 2 != 0)
{
odd_element_cnt++;
}
}
// Store number of replacements required
int moves = (odd_element_cnt) / 2;
// Two extra moves will be required
// to make the last odd element even
if (odd_element_cnt % 2 != 0)
moves += 2;
// Print the minimum replacements
Console.Write(moves);
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 5, 6, 3, 7, 20 };
int N = arr.Length;
// Function call
minMoves(arr, N);
}
} // This code is contributed by 29AjayKumar |
# Python program for the above approach # Function to find the minimum number # of replacements required to make # all array elements even def minMoves(arr, N):
# Stores the count of odd elements
odd_element_cnt = 0 ;
# Traverse the array
for i in range (N):
# Increase count of odd elements
if (arr[i] % 2 ! = 0 ):
odd_element_cnt + = 1 ;
# Store number of replacements required
moves = (odd_element_cnt) / / 2 ;
# Two extra moves will be required
# to make the last odd element even
if (odd_element_cnt % 2 ! = 0 ):
moves + = 2 ;
# Print the minimum replacements
print (moves);
# Driver Code if __name__ = = '__main__' :
arr = [ 5 , 6 , 3 , 7 , 20 ];
N = len (arr);
# Function call
minMoves(arr, N);
# This code is contributed by 29AjayKumar
|
<script> // javascript program for the above approach // Function to find the minimum number // of replacements required to make // all array elements even function minMoves(arr, N)
{ // Stores the count of odd elements
var odd_element_cnt = 0;
var i;
// Traverse the array
for (i = 0; i < N; i++) {
// Increase count of odd elements
if (arr[i] % 2 != 0) {
odd_element_cnt++;
}
}
// Store number of replacements required
var moves = Math.floor((odd_element_cnt)/2);
// Two extra moves will be required
// to make the last odd element even
if (odd_element_cnt % 2 != 0)
moves += 2;
// Print the minimum replacements
document.write(moves);
} // Driver Code var arr = [5, 6, 3, 7, 20];
N = arr.length;
// Function call
minMoves(arr, N);
</script> |
3
Time complexity: O(N)
Auxiliary Space: O(1)