Suppose we have a list of numbers called nums and another value k. Now let us consider an operation where we can subtract 1 from any element in the list. We can perform this operation k times. We have to find the minimum possible maximum value in the list after k such operations.
So, if the input is like nums = [3, 4, 6, 5] k = 6, then the output will be 3, as we can decrease 4 once, 6 thrice and 5 twice to get [3,3,3,3].
To solve this, we will follow these steps:
Let us see the following implementation to get better understanding:
class Solution: def solve(self, nums, k): nums.sort(reverse=True) i = 0 curr = nums while k > 0: while i < len(nums) and nums[i] == curr: i += 1 if k >= i: k -= i curr -= 1 else: return curr return curr ob = Solution() nums = [3, 4, 6, 5] k = 6 print(ob.solve(nums, k))
[3, 4, 6, 5], 6