1658. Minimum Operations to Reduce X to Zero
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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
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).
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
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.
- Calculate the total sum of
- Set a target sum which is
total - x.
- Using a sliding window approach, find the maximum length of the subarray whose sum equals the target.
- The result will be the total length of
numsminus the length of this subarray. If no such subarray exists, return
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