1658. Minimum Operations to Reduce X to Zero

Medium

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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.

Example 2:

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

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.

Constraints:

• 1 <= nums.length <= 105

• 1 <= nums[i] <= 104

• 1 <= x <= 109

Solution : This problem is using the same logic as PB 209

class Solution:
    def minOperations(self, nums: List[int], x: int) -> int:
        return self.minSubArrayLen(nums, sum(nums) - x)
    
    # Smiply using the solution of PB 209
    def minSubArrayLen(self, nums, x):
        index, start, size, total = 0, 0, 0, 0
        found = False
        # Now we will try to get max subarray with matches to the target, because remaining numbers
        # will be the numbers that make x to 0
        while index < len(nums):
            total += nums[index]
            while total > x and start <= index:
                total -= nums[start]
                start += 1
            if total == x:
                found = True
                size = max(size, index-start+1)
            index += 1
        return -1 if not found else len(nums) - size