Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using array strategy.
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. Note that this modifies the array for future operations.
Return the minimum number of operations to reduce x to exactly 0 if it is possible, otherwise, return -1.
Example 1:
Input: nums = [1,1,4,2,3], x = 5 Output: 2 Explanation: The optimal solution is to remove the last two elements to reduce x to zero.
Example 2:
Input: nums = [5,6,7,8,9], x = 4 Output: -1
Example 3:
Input: nums = [3,2,20,1,1,3], x = 10 Output: 5 Explanation: The optimal solution is to remove the last three elements and the first two elements (5 operations in total) to reduce x to zero.
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 1041 <= x <= 109Problem summary: 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. Note that this modifies the array for future operations. Return the minimum number of operations to reduce x to exactly 0 if it is possible, otherwise, return -1.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Hash Map · Binary Search · Sliding Window
[1,1,4,2,3] 5
[5,6,7,8,9] 4
[3,2,20,1,1,3] 10
minimum-size-subarray-sum)subarray-sum-equals-k)minimum-operations-to-convert-number)removing-minimum-number-of-magic-beans)minimum-operations-to-make-the-integer-zero)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1658: Minimum Operations to Reduce X to Zero
class Solution {
public int minOperations(int[] nums, int x) {
int s = -x;
for (int v : nums) {
s += v;
}
Map<Integer, Integer> vis = new HashMap<>();
vis.put(0, -1);
int mx = -1, t = 0;
int n = nums.length;
for (int i = 0; i < n; ++i) {
t += nums[i];
vis.putIfAbsent(t, i);
if (vis.containsKey(t - s)) {
mx = Math.max(mx, i - vis.get(t - s));
}
}
return mx == -1 ? -1 : n - mx;
}
}
// Accepted solution for LeetCode #1658: Minimum Operations to Reduce X to Zero
func minOperations(nums []int, x int) int {
s := -x
for _, v := range nums {
s += v
}
vis := map[int]int{0: -1}
mx, t := -1, 0
for i, v := range nums {
t += v
if _, ok := vis[t]; !ok {
vis[t] = i
}
if j, ok := vis[t-s]; ok {
mx = max(mx, i-j)
}
}
if mx == -1 {
return -1
}
return len(nums) - mx
}
# Accepted solution for LeetCode #1658: Minimum Operations to Reduce X to Zero
class Solution:
def minOperations(self, nums: List[int], x: int) -> int:
s = sum(nums) - x
vis = {0: -1}
mx, t = -1, 0
for i, v in enumerate(nums):
t += v
if t not in vis:
vis[t] = i
if t - s in vis:
mx = max(mx, i - vis[t - s])
return -1 if mx == -1 else len(nums) - mx
// Accepted solution for LeetCode #1658: Minimum Operations to Reduce X to Zero
use std::collections::HashMap;
impl Solution {
pub fn min_operations(nums: Vec<i32>, x: i32) -> i32 {
let s = nums.iter().sum::<i32>() - x;
let mut vis: HashMap<i32, i32> = HashMap::new();
vis.insert(0, -1);
let mut mx = -1;
let mut t = 0;
for (i, v) in nums.iter().enumerate() {
t += v;
if !vis.contains_key(&t) {
vis.insert(t, i as i32);
}
if let Some(&j) = vis.get(&(t - s)) {
mx = mx.max((i as i32) - j);
}
}
if mx == -1 {
-1
} else {
(nums.len() as i32) - mx
}
}
}
// Accepted solution for LeetCode #1658: Minimum Operations to Reduce X to Zero
function minOperations(nums: number[], x: number): number {
const s = nums.reduce((acc, cur) => acc + cur, -x);
const vis: Map<number, number> = new Map([[0, -1]]);
let [mx, t] = [-1, 0];
const n = nums.length;
for (let i = 0; i < n; ++i) {
t += nums[i];
if (!vis.has(t)) {
vis.set(t, i);
}
if (vis.has(t - s)) {
mx = Math.max(mx, i - vis.get(t - s)!);
}
}
return ~mx ? n - mx : -1;
}
Use this to step through a reusable interview workflow for this problem.
Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.
Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: Zero-count keys stay in map and break distinct/count constraints.
Usually fails on: Window/map size checks are consistently off by one.
Fix: Delete keys when count reaches zero.
Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.
Usually fails on: Two-element ranges never converge.
Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.
Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.
Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.
Fix: Shrink in a `while` loop until the invariant is valid again.