Minimum Operations to Reduce X to Zero

Minimum Operations to Reduce X to Zero

February 3, 2024
Sliding Window

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 #

  1. Calculate the total sum of the array.
  2. Initialize variables for the current sum, the maximum length of the subarray, and the start index of the subarray.
  3. Use a sliding window to find the longest subarray that meets the criteria.
  4. Iterate through the array, updating the current sum, and adjust the start index of the window as necessary to maintain the sum condition.
  5. After finding the maximum length subarray, subtract its length from the total number of elements to get the minimum operations required.
  6. 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.