Given an array arr[] of size N, the task is to make all array elements even by replacing a pair of adjacent elements with their sum.
Examples:
Input: arr[] = { 2, 4, 5, 11, 6 }
Output: 1
Explanation:
Replacing a pair (arr[2], arr[3]) with their sum ( = 5 + 11 = 16) modifies arr[] to { 2, 4, 16, 16, 6 }
Since all array elements are even, the required output is 1.Input: arr[] = { 1, 2, 4, 3, 11 }
Output: 3
Explanation:
Replacing the pair (arr[3], arr[4]) and replacing them with their sum ( = 3 + 11 = 14) modifies arr[] to { 1, 2, 4, 14, 14 }
Replacing the pair (arr[0], arr[1]) and replacing them with their sum ( = 1 + 2 = 3) modifies arr[] to { 3, 3, 4, 14, 14 }
Replacing the pair (arr[0], arr[1]) with their sum ( = 3 + 3 = 6) modifies arr[] to { 6, 6, 4, 14, 14 }.
Therefore, the required output is 3.
Approach: The idea is to use the fact that the sum of two odd numbers generates an even number. Follow the steps below to solve the problem:
- Initialize two integers, say res, to count the number of replacements, and odd_continuous_segment, to count the number of continuous odd numbers
-
Traverse the array and check the following conditions for every array element:
- If arr[i] is odd, then increment the count of odd_continuous_segment by 1
- Otherwise, if odd_continuous_segment is odd, then increment res by odd_continuous_segment/2. Otherwise, increment res by odd_continuous_segment / 2 + 2 and assign odd_continuous_segment to 0.
- Check if odd_continuous_segment is odd. If found to be true, then increment res by odd_continuous_segment / 2. Otherwise increment res by (odd_continuous_segment / 2 + 2)
- Finally, print the obtained value of res
Below is the implementation of the above approach:
// C++ program to implement // the above approach #include <iostream> using namespace std;
// Function to find minimum count of operations // required to make all array elements even int make_array_element_even( int arr[], int N)
{ // Stores minimum count of replacements
// to make all array elements even
int res = 0;
// Stores the count of odd
// continuous numbers
int odd_cont_seg = 0;
// Traverse the array
for ( int i = 0; i < N; i++) {
// If arr[i] is an odd number
if (arr[i] % 2 == 1) {
// Update odd_cont_seg
odd_cont_seg++;
}
else {
if (odd_cont_seg > 0) {
// If odd_cont_seg is even
if (odd_cont_seg % 2 == 0) {
// Update res
res += odd_cont_seg / 2;
}
else {
// Update res
res += (odd_cont_seg / 2) + 2;
}
// Reset odd_cont_seg = 0
odd_cont_seg = 0;
}
}
}
// If odd_cont_seg exceeds 0
if (odd_cont_seg > 0) {
// If odd_cont_seg is even
if (odd_cont_seg % 2 == 0) {
// Update res
res += odd_cont_seg / 2;
}
else {
// Update res
res += odd_cont_seg / 2 + 2;
}
}
// Print the result
return res;
} // Drivers Code int main()
{ int arr[] = { 2, 4, 5, 11, 6 };
int N = sizeof (arr) / sizeof (arr[0]);
cout << make_array_element_even(arr, N);
return 0;
} |
// Java program to implement // the above approach import java.util.*;
class GFG{
// Function to find minimum count of operations
// required to make all array elements even
static int make_array_element_even( int arr[], int N)
{
// Stores minimum count of replacements
// to make all array elements even
int res = 0 ;
// Stores the count of odd
// continuous numbers
int odd_cont_seg = 0 ;
// Traverse the array
for ( int i = 0 ; i < N; i++)
{
// If arr[i] is an odd number
if (arr[i] % 2 == 1 )
{
// Update odd_cont_seg
odd_cont_seg++;
}
else
{
if (odd_cont_seg > 0 )
{
// If odd_cont_seg is even
if (odd_cont_seg % 2 == 0 )
{
// Update res
res += odd_cont_seg / 2 ;
}
else
{
// Update res
res += (odd_cont_seg / 2 ) + 2 ;
}
// Reset odd_cont_seg = 0
odd_cont_seg = 0 ;
}
}
}
// If odd_cont_seg exceeds 0
if (odd_cont_seg > 0 )
{
// If odd_cont_seg is even
if (odd_cont_seg % 2 == 0 )
{
// Update res
res += odd_cont_seg / 2 ;
}
else
{
// Update res
res += odd_cont_seg / 2 + 2 ;
}
}
// Print the result
return res;
}
// Drivers Code
public static void main(String[] args)
{
int arr[] = { 2 , 4 , 5 , 11 , 6 };
int N = arr.length;
System.out.print(make_array_element_even(arr, N));
}
} // This code is contributed by shikhasingrajput |
# Python program to implement # the above approach # Function to find minimum count of operations # required to make all array elements even def make_array_element_even(arr, N):
# Stores minimum count of replacements
# to make all array elements even
res = 0
# Stores the count of odd
# continuous numbers
odd_cont_seg = 0
# Traverse the array
for i in range ( 0 , N):
# If arr[i] is an odd number
if (arr[i] % 2 = = 1 ):
# Update odd_cont_seg
odd_cont_seg + = 1
else :
if (odd_cont_seg > 0 ):
# If odd_cont_seg is even
if (odd_cont_seg % 2 = = 0 ):
# Update res
res + = odd_cont_seg / / 2
else :
# Update res
res + = (odd_cont_seg / / 2 ) + 2
# Reset odd_cont_seg = 0
odd_cont_seg = 0
# If odd_cont_seg exceeds 0
if (odd_cont_seg > 0 ):
# If odd_cont_seg is even
if (odd_cont_seg % 2 = = 0 ):
# Update res
res + = odd_cont_seg / / 2
else :
# Update res
res + = odd_cont_seg / / 2 + 2
# Print the result
return res
# Drivers Code arr = [ 2 , 4 , 5 , 11 , 6 ]
N = len (arr)
print (make_array_element_even(arr, N))
# This code is contributed by shubhamsingh10 |
// C# program to implement // the above approach using System;
public class GFG
{ // Function to find minimum count of operations
// required to make all array elements even
static int make_array_element_even( int []arr, int N)
{
// Stores minimum count of replacements
// to make all array elements even
int res = 0;
// Stores the count of odd
// continuous numbers
int odd_cont_seg = 0;
// Traverse the array
for ( int i = 0; i < N; i++)
{
// If arr[i] is an odd number
if (arr[i] % 2 == 1)
{
// Update odd_cont_seg
odd_cont_seg++;
}
else
{
if (odd_cont_seg > 0)
{
// If odd_cont_seg is even
if (odd_cont_seg % 2 == 0)
{
// Update res
res += odd_cont_seg / 2;
}
else
{
// Update res
res += (odd_cont_seg / 2) + 2;
}
// Reset odd_cont_seg = 0
odd_cont_seg = 0;
}
}
}
// If odd_cont_seg exceeds 0
if (odd_cont_seg > 0)
{
// If odd_cont_seg is even
if (odd_cont_seg % 2 == 0)
{
// Update res
res += odd_cont_seg / 2;
}
else
{
// Update res
res += odd_cont_seg / 2 + 2;
}
}
// Print the result
return res;
}
// Drivers Code
public static void Main(String[] args)
{
int []arr = { 2, 4, 5, 11, 6 };
int N = arr.Length;
Console.Write(make_array_element_even(arr, N));
}
} // This code is contributed by 29AjayKumar |
<script> // Javascript program to implement // the above approach // Function to find minimum count of operations // required to make all array elements even function make_array_element_even(arr, N)
{ // Stores minimum count of replacements
// to make all array elements even
let res = 0;
// Stores the count of odd
// continuous numbers
let odd_cont_seg = 0;
// Traverse the array
for (let i = 0; i < N; i++)
{
// If arr[i] is an odd number
if (arr[i] % 2 == 1)
{
// Update odd_cont_seg
odd_cont_seg++;
}
else
{
if (odd_cont_seg > 0)
{
// If odd_cont_seg is even
if (odd_cont_seg % 2 == 0)
{
// Update res
res += odd_cont_seg / 2;
}
else
{
// Update res
res += (odd_cont_seg / 2) + 2;
}
// Reset odd_cont_seg = 0
odd_cont_seg = 0;
}
}
}
// If odd_cont_seg exceeds 0
if (odd_cont_seg > 0)
{
// If odd_cont_seg is even
if (odd_cont_seg % 2 == 0)
{
// Update res
res += odd_cont_seg / 2;
}
else
{
// Update res
res += odd_cont_seg / 2 + 2;
}
}
// Print the result
return res;
} // Driver Code // Given array arr[] let arr = [ 2, 4, 5, 11, 6 ]; let N = arr.length; document.write(make_array_element_even(arr, N)); // This code is contributed by splevel62 </script> |
1
Time complexity: O(N)
Auxiliary space: O(1)