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Suppose we have an array called nums and another value k. In one operation, we can select two elements from nums whose sum is equals to k and remove them from the array. We have to find the maximum number of operations we can perform on the array.

So, if the input is like nums = [8,3,6,1,5] k = 9, then the output will be 2 as we can delete [3,6] whose sum is 9, then remove [8,1] whose sum is also 9.

To solve this, we will follow these steps −

- counter := a map holding frequency of each item present in nums
- res := 0
- for each num in counter, do
- if counter[k-num] is non-zero, then
- if num is not same as k - num, then
- res := res + minimum of counter[num] and counter[k-num]
- counter[k-num] := 0
- counter[num] := 0

- otherwise,
- res := res + quotient of (counter[num] / 2)

- if num is not same as k - num, then

- if counter[k-num] is non-zero, then
- return res

Let us see the following implementation to get better understanding −

from collections import Counter def solve(nums, k): counter = Counter(nums) res = 0 for num in counter: if counter.get(k-num, 0): if num != k - num: res += min(counter[num], counter[k-num]) counter[k-num] = 0 counter[num] = 0 else: res += int(counter[num] / 2) return res nums = [8,3,6,1,5] k = 9 print(solve(nums, k))

[8,3,6,1,5], 9

2

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