 # 1658. Minimum Operations to Reduce X to Zero

Updated: 2023-10-27 LeetCode problem 1658

## Problem Statement

You have an integer array nums and another integer x. In a single operation, you can either remove the leftmost or the rightmost element from the array nums and decrement x by the value of the removed element. The question is: What is the minimum number of operations required to reduce x to exactly zero? If it’s not feasible, the answer should be -1.

## Naive Solution

The naive approach would be to explore every possible combination of removing elements from the start or the end, recursively. While this approach can find the solution, it’s extremely inefficient and would result in a time complexity beyond O(2^n).

## Approach

The problem can be turned around: Instead of trying to determine which numbers to remove from the ends, think of it as finding the longest subarray inside nums such that the sum of its elements is equal to the total sum of nums minus x. Let’s call this approach in mathematics “from the reverse”.

Why does this work? Well, if we find the maximum-length subarray with a sum equal to total - x, then the numbers we need to remove from the array will be the numbers outside this subarray!

This transformed problem can be tackled using a two-pointer or sliding window approach.

## Steps

1. Calculate the total sum of nums.
2. Set a target sum which is total - x.
3. Using a sliding window approach, find the maximum length of the subarray whose sum equals the target.
4. The result will be the total length of nums minus the length of this subarray. If no such subarray exists, return -1.

## Solution

def minOperations(nums, x):
total = sum(nums)
target = total - x

n = len(nums)
curr_sum = 0            # sum of current subarray
max_len = -1            # maximum length of the subarray that sums up to the target

left = 0                # Starting pointer of sliding window

for right in range(n):                      # Starting pointer of sliding window
curr_sum += nums[right]

while curr_sum > target and left < n:   # If current sum goes beyond target
curr_sum -= nums[left]              # shrink the window from the left
left += 1

if curr_sum == target:
max_len = max(max_len, right - left + 1)

if max_len == -1:
return -1
return n - max_len