Minimum Operations to Reduce X to Zero
Problem Statement
You are given an integer array nums and an integer x. In one operation, remove either the leftmost or rightmost element from nums and subtract its value from x. Return the minimum number of operations needed to reduce x exactly to 0. If it is impossible, return -1.
Input
An integer array nums of positive values and an integer x to remove from the two ends.
Output
An integer: the fewest end-removal operations that make x exactly 0, or -1 if no such removal sequence exists.
Constraints
- •
1 <= nums.length <= 10^5 - •
1 <= nums[i] <= 10^4 - •
1 <= x <= 10^9
Examples
Example 1
nums = [1,1,4,2,3], x = 5
2Example 2
nums = [5,6,7,8,9], x = 4
-1Example 3
nums = [3,2,20,1,1,3], x = 10
5Learning Objectives
- Reframe end removals as keeping one contiguous middle subarray.
- Convert minimum removals into maximum kept length.
- Use the positivity of **nums** to find the longest subarray with a target sum by sliding window.
- Handle impossible, remove-all, and exact-target cases without special-case confusion.
Intuition
The direct view is awkward because each operation can happen on either end, creating many prefix-suffix combinations. The reframing trick is to look at what remains. After removing some prefix and some suffix, the leftover elements, if any, form one contiguous middle subarray.
If the removed elements must sum to x, then the kept middle subarray must sum to totalSum - x. To minimize removals, maximize how many elements you keep. Since all numbers are positive, the longest subarray with a target sum can be found with a standard variable sliding window: expand to increase the sum, shrink to decrease it.
Common mistakes
- ×Trying to greedily remove the larger end, which can miss the best prefix-suffix combination.
- ×Searching for a subarray that sums to **x** instead of **totalSum - x**.
- ×Returning the longest kept length instead of converting it to **n - longestLength** operations.
- ×Forgetting that **target = 0** means every element must be removed, so the answer is **n**.
Algorithm Explanation
Window setup
Compute target = totalSum - x. The window represents the middle subarray we keep, and its sum should equal target. Because all values are positive, increasing right only increases the window sum, and moving left only decreases it.
Window visualization
For nums = [1,1,4,2,3] and x = 5, the total is 11, so the target kept sum is 6. The window grows as [1], then [1,1], then [1,1,4] with sum 6, giving kept length 3. Later 2 makes the sum too large, so the left edge moves twice and finds [4,2] with sum 6, but that length is only 2. The longest kept window remains length 3, so the answer is 5 - 3 = 2 removals.
Algorithm
- Sum all values and compute target = totalSum - x.
- If target < 0, return -1 because even removing everything is not enough.
- If target == 0, return n because the whole array must be removed.
- Use a positive-number sliding window with left, right, and windowSum.
- Add nums[right], then shrink from the left while windowSum > target.
- Whenever windowSum == target, update the longest kept length.
- Return n - longestLength, or -1 if no target-sum window was found.
Solutions
Solution: Longest middle window with target sum
Instead of simulating removals, preserve the longest middle subarray whose sum is totalSum - x. The array is positive, so a two-pointer window can find that longest target-sum subarray in one pass.
Step-by-step
- Compute the total sum and derive the target kept sum.
- Reject target < 0 and return n when target == 0.
- Expand right and add each value to the current window sum.
- While the sum is too large, subtract nums[left] and move left forward.
- When the sum equals the target, record the longest window length.
- Convert the longest kept length into operations by subtracting it from n.
O(n)
O(1)
Each pointer only moves forward, and the algorithm stores a few integer variables.
Java implementation
Dry Run
Sample input
nums = [1,1,4,2,3], x = 5. The total is 11, so the target kept middle-window sum is 6.
| right | nums[right] | left after shrink | window sum | longest target window | answer if finished |
|---|---|---|---|---|---|
| 0 | 1 | 0 | 1 | none | not found |
| 1 | 1 | 0 | 2 | none | not found |
| 2 | 4 | 0 | 6 | [0..2], length 3 | 2 |
| 3 | 2 | 2 | 6 | [0..2], length 3 | 2 |
| 4 | 3 | 3 | 5 | [0..2], length 3 | 2 |
The longest kept middle subarray has length 3, so the fewest removals is 5 - 3 = 2. Those removals are the suffix values 2 and 3.
Interview Tips
This problem is mostly about reframing. Say that removing from the two ends leaves one contiguous middle segment, so the question becomes longest subarray with sum totalSum - x. Also mention that the sliding-window solution depends on all numbers being positive; if negatives were allowed, you would need a prefix-sum hash map instead.
Likely follow-ups
- How would you solve it if **nums** could contain negative numbers?
- How would you return the actual sequence of left and right removals?
- What if each removal had a different cost and you wanted minimum total cost?
- Can you solve the prefix-suffix version directly with prefix sums and a hash map?
Similar Problems
Key Takeaways
- End removals leave one contiguous middle subarray.
- Minimum operations equals **n - longest kept length**.
- The kept window must sum to **totalSum - x**, not **x**.
- Positive values make target-sum longest-window solvable with two pointers.