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 :0Input :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 –

## Python3

`# 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

**Time Complexity:** O(N )

**Auxiliary Space:** O(N)

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