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Minimum elements to be removed such that sum of adjacent elements is always odd
  • Difficulty Level : Easy
  • Last Updated : 06 Dec, 2020

Given an array of N integers. The task is to eliminate the minimum number of elements such that in the resulting array the sum of any two adjacent values is odd.

Examples:

Input: arr[] = {1, 2, 3} 
Output:
Sum of all adjacent elements is already odd.
Input: arr[] = {1, 3, 5, 4, 2} 
Output:
Eliminate 3, 5 and 2 so that in the resulting array the sum of any two adjacent values is odd.

Approach The sum of 2 numbers is odd if one of them is odd and the other is even. This means for every pair of consecutive numbers that have the same parity, eliminate one of them, it doesn’t matter which. So the following greedy algorithm works: 

  • Go through all the elements in order.
  • If the current number has the same parity as the previous one eliminate it, otherwise don’t.

Below is the implementation of the above approach: 
 



C++




// C++ implementation of the above approach
#include<bits/stdc++.h>
using namespace std;
 
// Returns the minimum number of eliminations
int min_elimination(int n, int arr [])
{
    int count = 0;
 
    // Stores the previous element
    int prev_val = arr[0];
 
    // Stores the new value
    for (int i = 1; i < n; i++)
    {
        int curr_val = arr[i];
 
        // Check if the previous and current
        // values are of same parity
        if (curr_val % 2 == prev_val % 2)
            count++;
 
        // Previous value is now the current value
        prev_val = curr_val;
    }
 
    // Return the counter variable
    return count;
}
 
// Driver code
int main()
{
    int arr [] = { 1, 2, 3, 7, 9 };
    int n = sizeof(arr)/sizeof(arr[0]);
 
    cout << min_elimination(n, arr);
     
    return 0;
}
 
// This code is contributed by ihritik


Java




// Java implementation of the above approach
 
class GFG {
 
    // Returns the minimum number of eliminations
    static int min_elimination(int n, int[] arr)
    {
        int count = 0;
 
        // Stores the previous element
        int prev_val = arr[0];
 
        // Stores the new value
        for (int i = 1; i < n; i++) {
            int curr_val = arr[i];
 
            // Check if the previous and current
            // values are of same parity
            if (curr_val % 2 == prev_val % 2)
                count++;
 
            // Previous value is now the current value
            prev_val = curr_val;
        }
 
        // Return the counter variable
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int[] arr = new int[] { 1, 2, 3, 7, 9 };
        int n = arr.length;
 
        System.out.println(min_elimination(n, arr));
    }
}


Python3




# Python3 implementation of the above approach
 
# Returns the minimum number of eliminations
def min_elimination(n, arr):
 
    count = 0
 
    # Stores the previous element
    prev_val = arr[0]
 
    # Stores the new value
    for i in range (1, n):
        curr_val = arr[i];
 
        # Check if the previous and current
        # values are of same parity
        if (curr_val % 2 == prev_val % 2):
            count = count + 1
 
        # Previous value is now the current value
        prev_val = curr_val
     
 
    # Return the counter variable
    return count
 
# Driver code
arr = [ 1, 2, 3, 7, 9 ]
n = len(arr)
 
print(min_elimination(n, arr));
 
# This code is contributed by ihritik


C#




// C# implementation of the above approach
using System;
 
class GFG
{
 
    // Returns the minimum number of eliminations
    static int min_elimination(int n, int[] arr)
    {
        int count = 0;
 
        // Stores the previous element
        int prev_val = arr[0];
 
        // Stores the new value
        for (int i = 1; i < n; i++)
        {
            int curr_val = arr[i];
 
            // Check if the previous and current
            // values are of same parity
            if (curr_val % 2 == prev_val % 2)
                count++;
 
            // Previous value is now the current value
            prev_val = curr_val;
        }
 
        // Return the counter variable
        return count;
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = new int[] { 1, 2, 3, 7, 9 };
        int n = arr.Length;
 
        Console.WriteLine(min_elimination(n, arr));
    }
}
 
// This code is contributed by ihritik


PHP




<?php
// PHP implementation of the above approach
 
// Returns the minimum number of eliminations
function min_elimination($n, $arr)
{
    $count = 0;
 
    // Stores the previous element
    $prev_val = $arr[0];
 
    // Stores the new value
    for ($i = 1; $i < $n; $i++)
    {
        $curr_val = $arr[$i];
 
        // Check if the previous and current
        // values are of same parity
        if ($curr_val % 2 == $prev_val % 2)
            $count++;
 
        // Previous value is now the
        // current value
        $prev_val = $curr_val;
    }
 
    // Return the counter variable
    return $count;
}
 
// Driver code
$arr = array( 1, 2, 3, 7, 9 );
$n = sizeof($arr);
 
echo min_elimination($n, $arr);
 
// This code is contributed by Ryuga
?>


Output: 

2

 

Time Complexity: O(N)

Auxiliary Space: O(1)

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