Python | Number of elements to be removed such that product of adjacent elements is always even

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

Examples:

Input  : arr[] = {1, 2, 3}
Output : 0

Input  : arr[] = {1, 3, 5, 4, 2}
Output : 2
Remove 1 and 3.

Approach: The product of 2 numbers is even if any one of them is even. This means for every pair of consecutive numbers if both are odd then eliminate one of them.
So, to make the product of the adjacent elements even, either all elements should be even or alternative even-odd elements. So the following greedy algorithm works:

  • Go through all the elements in order.
  • If all elements are even then “0” is returned.
  • If all elements are odd then “n” is returned.
  • Otherwise, count the number of consecutive odd pair.
  • Return the minimum count.

Below is the Python implementation –

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# Python 3 implementation of the  
# above approach 
    
# Function to find minimum number of  
# eliminations such that product of all  
# adjacent elements is even 
def min_elimination(n, arr): 
    countOdd = 0
    counteven = 0
    # Stores the new value 
    for i in range(n): 
            
        # Count odd and even  numbers 
        if (arr[i] % 2): 
            countOdd += 1
        else:
            counteven+= 1
    # if all are even
    if counteven == n: 
        return 0
  
    # if all are odd
    if countOdd == n: 
        return n
    else:
        countpair = 0
        for i in range(1, n):
            if (arr[i] % 2 == 1 and arr[i-1] % 2 == 1):
                countpair+= 1
        return countpair 
    
# Driver code 
if __name__ == '__main__'
    arr = [1, 2, 3, 7, 9
    n = len(arr) 
    
    print(min_elimination(n, arr))

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Output:

2

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