leetcode - 1658. Minimum Operations to Reduce X to Zero

Description

You are given an integer array nums and an integer x. In one operation, you can either remove the leftmost or the rightmost element from the array nums and subtract its value from x. Note that this modifies the array for future operations.

Return the minimum number of operations to reduce x to exactly 0 if it is possible, otherwise, return -1.

Example 1:

Input: nums = [1,1,4,2,3], x = 5
Output: 2
Explanation: The optimal solution is to remove the last two elements to reduce x to zero.
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Example 2:

Input: nums = [5,6,7,8,9], x = 4
Output: -1
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Example 3:

Input: nums = [3,2,20,1,1,3], x = 10
Output: 5
Explanation: The optimal solution is to remove the last three elements and the first two elements (5 operations in total) to reduce x to zero.
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Constraints:

1 <= nums.length <= 10^5
1 <= nums[i] <= 10^4
1 <= x <= 10^9
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Solution

Recursive (TLE)

Every time try with leftest and rightest items.

Time complexity: $o(2^n)$
Space complexity: $o(n)$

Sliding window

Solved after help.

Think in reverse; instead of finding the minimum prefix + suffix, find the maximum subarray.

So the problem becomes: to find the longest subarray with target sum.

Time complexity: $o(n)$
Space complexity: $o(1)$

Code

Recursive (TLE)

class Solution:
    def minOperations(self, nums: List[int], x: int) -> int:
        def helper(left: int, right: int, x: int) -> int:
            if x == 0:
                return 0
            if left > right:
                return -1
            candidates = []
            if x - nums[left] >= 0:
                next_cnt = helper(left + 1, right, x - nums[left])
                if next_cnt != -1:
                    candidates.append(next_cnt + 1)
            if x - nums[right] >= 0:
                next_cnt = helper(left, right - 1, x - nums[right])
                if next_cnt != -1:
                    candidates.append(1 + next_cnt)
            return min(candidates) if candidates else -1
        return helper(0, len(nums) - 1, x)
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Sliding window

class Solution:
    def minOperations(self, nums: List[int], x: int) -> int:
        array_sum = sum(nums) - x
        if array_sum < 0:
            return -1
        res = -1
        left = 0
        cur_sum = 0
        for right in range(len(nums)):
            cur_sum += nums[right]
            while cur_sum > array_sum:
                cur_sum -= nums[left]
                left += 1
            if cur_sum == array_sum:
                res = max(res, right - left + 1)
        return -1 if res == -1 else len(nums) - res
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