# Minimum Operations to Reduce X to Zero

##### February 3, 2024

## Problem Statement #

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`

. The goal is to exactly reduce `x`

to zero using the minimum number of operations, and return this number. If it is not possible, return `-1`

.

## Solution Approach #

The solution involves finding the longest subarray with the sum equal to `totalSum - x`

(where `totalSum`

is the sum of all elements in `nums`

). This approach is based on the idea that removing elements from the ends to reduce `x`

to zero is equivalent to finding the longest contiguous segment of the array whose sum is equal to `totalSum - x`

. The number of operations required is the total length of the array minus the length of this subarray.

## Algorithm Steps #

- Calculate the total sum of the array.
- Initialize variables for the current sum, the maximum length of the subarray, and the start index of the subarray.
- Use a sliding window to find the longest subarray that meets the criteria.
- Iterate through the array, updating the current sum, and adjust the start index of the window as necessary to maintain the sum condition.
- After finding the maximum length subarray, subtract its length from the total number of elements to get the minimum operations required.
- If the maximum length subarray is not found, return
`-1`

.

## Code (Python) #

```
def minOperations(nums, x):
totalSum = sum(nums)
target = totalSum - x
n = len(nums)
maxLength = -1
currentSum = 0
start = 0
for end in range(n):
currentSum += nums[end]
while currentSum > target and start <= end:
currentSum -= nums[start]
start += 1
if currentSum == target:
maxLength = max(maxLength, end - start + 1)
return n - maxLength if maxLength != -1 else -1
# Example usage
print(minOperations([1,1,4,2,3], 5)) # Output: 2
```

## Time Complexity #

O(n), where `n`

is the length of the array. The array is traversed once.

## Space Complexity #

O(1), as the solution uses a constant amount of space.