Given an array position[] consisting of N integers where position[i] denotes the position of the ith element, the task is to find the minimum cost required to move all the elements to the same position by performing either of the following two operations:
- Move from position[i] to position[i] + 2 or position[i] – 2. Cost = 0.
- Move from position[i] to position[i] + 1 or position[i] – 1. Cost = 1.
Examples:
Input: position[] = {1, 2, 3}
Output: 1
Explanation:
Operation 1: Move the element at position 3 to position 1. Cost = 0.
Operation 2: Move the element at position 2 to position 1. Cost = 1.
Therefore, total cost = 1.Input: position[] = {2, 2, 2, 3, 3}
Output: 2
Explanation: Move the two elements at position 3 to position 2. Cost of each operation = 1. Therefore, total cost = 2.
Approach: The idea is to traverse the array and count the number of odd and even elements. For each operation involving increment or decrement by two indices, the cost will always be 0. The cost changes only on moving an element from odd to even position or vice-versa. Therefore, the minimum cost required is the minimum of the count of odd and even elements present in the array position[].
Below is the implementation of the above approach:
// C++ program to implement // the above approach #include <iostream> using namespace std;
// Function to find the minimum // cost required to place all // elements in the same position int minCost( int arr[], int arr_size)
{ // Stores the count of even
// and odd elements
int odd = 0, even = 0;
// Traverse the array arr[]
for ( int i = 0; i < arr_size; i++)
{
// Count even elements
if (arr[i] % 2 == 0)
even++;
// Count odd elements
else
odd++;
}
// Print the minimum count
cout << min(even, odd);
} // Driver Code int main()
{ // Given array
int arr[] = { 1, 2, 3 };
int arr_size = sizeof (arr) / sizeof (arr[0]);
// Function Call
minCost(arr, arr_size);
} // This code is contributed by khushboogoyal499 |
// Java program to implement // the above approach import java.io.*;
class GFG {
// Function to find the minimum
// cost required to place all
// elements in the same position
public void minCost( int [] arr)
{
// Stores the count of even
// and odd elements
int odd = 0 , even = 0 ;
// Traverse the array arr[]
for ( int i = 0 ;
i < arr.length; i++) {
// Count even elements
if (arr[i] % 2 == 0 )
even++;
// Count odd elements
else
odd++;
}
// Print the minimum count
System.out.print(
Math.min(even, odd));
}
// Driver Code
public static void main(String[] args)
{
GFG obj = new GFG();
// Given array
int arr[] = { 1 , 2 , 3 };
// Function Call
obj.minCost(arr);
}
} |
# Python3 program to implement # the above approach # Function to find the minimum # cost required to place all # elements in the same position def minCost(arr):
# Stores the count of even
# and odd elements
odd = 0
even = 0
# Traverse the array arr[]
for i in range ( len (arr)):
# Count even elements
if (arr[i] % 2 = = 0 ):
even + = 1
# Count odd elements
else :
odd + = 1
# Print the minimum count
print ( min (even, odd))
# Driver Code if __name__ = = '__main__' :
# Given array
arr = [ 1 , 2 , 3 ]
# Function Call
minCost(arr)
# This code is contributed by mohit kumar 29 |
// C# program to implement // the above approach using System;
class GFG{
// Function to find the minimum // cost required to place all // elements in the same position public void minCost( int [] arr)
{ // Stores the count of even
// and odd elements
int odd = 0, even = 0;
// Traverse the array arr[]
for ( int i = 0; i < arr.Length; i++)
{
// Count even elements
if (arr[i] % 2 == 0)
even++;
// Count odd elements
else
odd++;
}
// Print the minimum count
Console.Write(Math.Min(even, odd));
} // Driver Code public static void Main()
{ GFG obj = new GFG();
// Given array
int [] arr = { 1, 2, 3 };
// Function Call
obj.minCost(arr);
} } // This code is contributed by sanjoy_62 |
<script> // JavaScript program to implement // the above approach // Function to find the minimum
// cost required to place all
// elements in the same position
function minCost(arr)
{
// Stores the count of even
// and odd elements
let odd = 0, even = 0;
// Traverse the array arr[]
for (let i = 0;
i < arr.length; i++) {
// Count even elements
if (arr[i] % 2 == 0)
even++;
// Count odd elements
else
odd++;
}
// Print the minimum count
document.write(
Math.min(even, odd));
}
// Driver code // Given array
let arr = [ 1, 2, 3 ];
// Function Call
minCost(arr);
// This code is contributed by susmitakundugoaldanga. </script> |
1
Time Complexity: O(N)
Auxiliary Space: O(1)