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 a 0-indexed integer array nums containing distinct numbers, an integer start, and an integer goal. There is an integer x that is initially set to start, and you want to perform operations on x such that it is converted to goal. You can perform the following operation repeatedly on the number x:
If 0 <= x <= 1000, then for any index i in the array (0 <= i < nums.length), you can set x to any of the following:
x + nums[i]x - nums[i]x ^ nums[i] (bitwise-XOR)Note that you can use each nums[i] any number of times in any order. Operations that set x to be out of the range 0 <= x <= 1000 are valid, but no more operations can be done afterward.
Return the minimum number of operations needed to convert x = start into goal, and -1 if it is not possible.
Example 1:
Input: nums = [2,4,12], start = 2, goal = 12 Output: 2 Explanation: We can go from 2 → 14 → 12 with the following 2 operations. - 2 + 12 = 14 - 14 - 2 = 12
Example 2:
Input: nums = [3,5,7], start = 0, goal = -4 Output: 2 Explanation: We can go from 0 → 3 → -4 with the following 2 operations. - 0 + 3 = 3 - 3 - 7 = -4 Note that the last operation sets x out of the range 0 <= x <= 1000, which is valid.
Example 3:
Input: nums = [2,8,16], start = 0, goal = 1 Output: -1 Explanation: There is no way to convert 0 into 1.
Constraints:
1 <= nums.length <= 1000-109 <= nums[i], goal <= 1090 <= start <= 1000start != goalnums are distinct.Problem summary: You are given a 0-indexed integer array nums containing distinct numbers, an integer start, and an integer goal. There is an integer x that is initially set to start, and you want to perform operations on x such that it is converted to goal. You can perform the following operation repeatedly on the number x: If 0 <= x <= 1000, then for any index i in the array (0 <= i < nums.length), you can set x to any of the following: x + nums[i] x - nums[i] x ^ nums[i] (bitwise-XOR) Note that you can use each nums[i] any number of times in any order. Operations that set x to be out of the range 0 <= x <= 1000 are valid, but no more operations can be done afterward. Return the minimum number of operations needed to convert x = start into goal, and -1 if it is not possible.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array
[2,4,12] 2 12
[3,5,7] 0 -4
[2,8,16] 0 1
minimum-operations-to-reduce-x-to-zero)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2059: Minimum Operations to Convert Number
class Solution {
public int minimumOperations(int[] nums, int start, int goal) {
IntBinaryOperator op1 = (x, y) -> x + y;
IntBinaryOperator op2 = (x, y) -> x - y;
IntBinaryOperator op3 = (x, y) -> x ^ y;
IntBinaryOperator[] ops = {op1, op2, op3};
boolean[] vis = new boolean[1001];
Queue<int[]> queue = new ArrayDeque<>();
queue.offer(new int[] {start, 0});
while (!queue.isEmpty()) {
int[] p = queue.poll();
int x = p[0], step = p[1];
for (int num : nums) {
for (IntBinaryOperator op : ops) {
int nx = op.applyAsInt(x, num);
if (nx == goal) {
return step + 1;
}
if (nx >= 0 && nx <= 1000 && !vis[nx]) {
queue.offer(new int[] {nx, step + 1});
vis[nx] = true;
}
}
}
}
return -1;
}
}
// Accepted solution for LeetCode #2059: Minimum Operations to Convert Number
func minimumOperations(nums []int, start int, goal int) int {
type pair struct {
x int
step int
}
ops := []func(int, int) int{
func(x, y int) int { return x + y },
func(x, y int) int { return x - y },
func(x, y int) int { return x ^ y },
}
vis := make([]bool, 1001)
q := []pair{{start, 0}}
for len(q) > 0 {
x, step := q[0].x, q[0].step
q = q[1:]
for _, num := range nums {
for _, op := range ops {
nx := op(x, num)
if nx == goal {
return step + 1
}
if nx >= 0 && nx <= 1000 && !vis[nx] {
q = append(q, pair{nx, step + 1})
vis[nx] = true
}
}
}
}
return -1
}
# Accepted solution for LeetCode #2059: Minimum Operations to Convert Number
class Solution:
def minimumOperations(self, nums: List[int], start: int, goal: int) -> int:
op1 = lambda x, y: x + y
op2 = lambda x, y: x - y
op3 = lambda x, y: x ^ y
ops = [op1, op2, op3]
vis = [False] * 1001
q = deque([(start, 0)])
while q:
x, step = q.popleft()
for num in nums:
for op in ops:
nx = op(x, num)
if nx == goal:
return step + 1
if 0 <= nx <= 1000 and not vis[nx]:
q.append((nx, step + 1))
vis[nx] = True
return -1
// Accepted solution for LeetCode #2059: Minimum Operations to Convert Number
/**
* [2059] Minimum Operations to Convert Number
*
* You are given a 0-indexed integer array nums containing distinct numbers, an integer start, and an integer goal. There is an integer x that is initially set to start, and you want to perform operations on x such that it is converted to goal. You can perform the following operation repeatedly on the number x:
* If 0 <= x <= 1000, then for any index i in the array (0 <= i < nums.length), you can set x to any of the following:
*
* x + nums[i]
* x - nums[i]
* x ^ nums[i] (bitwise-XOR)
*
* Note that you can use each nums[i] any number of times in any order. Operations that set x to be out of the range 0 <= x <= 1000 are valid, but no more operations can be done afterward.
* Return the minimum number of operations needed to convert x = start into goal, and -1 if it is not possible.
*
* Example 1:
*
* Input: nums = [2,4,12], start = 2, goal = 12
* Output: 2
* Explanation: We can go from 2 → 14 → 12 with the following 2 operations.
* - 2 + 12 = 14
* - 14 - 2 = 12
*
* Example 2:
*
* Input: nums = [3,5,7], start = 0, goal = -4
* Output: 2
* Explanation: We can go from 0 → 3 → -4 with the following 2 operations.
* - 0 + 3 = 3
* - 3 - 7 = -4
* Note that the last operation sets x out of the range 0 <= x <= 1000, which is valid.
*
* Example 3:
*
* Input: nums = [2,8,16], start = 0, goal = 1
* Output: -1
* Explanation: There is no way to convert 0 into 1.
*
*
* Constraints:
*
* 1 <= nums.length <= 1000
* -10^9 <= nums[i], goal <= 10^9
* 0 <= start <= 1000
* start != goal
* All the integers in nums are distinct.
*
*/
pub struct Solution {}
// problem: https://leetcode.com/problems/minimum-operations-to-convert-number/
// discuss: https://leetcode.com/problems/minimum-operations-to-convert-number/discuss/?currentPage=1&orderBy=most_votes&query=
// submission codes start here
impl Solution {
pub fn minimum_operations(nums: Vec<i32>, start: i32, goal: i32) -> i32 {
0
}
}
// submission codes end
#[cfg(test)]
mod tests {
use super::*;
#[test]
#[ignore]
fn test_2059_example_1() {
let nums = vec![2, 4, 12];
let start = 2;
let goal = 12;
let result = 2;
assert_eq!(Solution::minimum_operations(nums, start, goal), result);
}
#[test]
#[ignore]
fn test_2059_example_2() {
let nums = vec![3, 5, 7];
let start = 0;
let goal = -4;
let result = 2;
assert_eq!(Solution::minimum_operations(nums, start, goal), result);
}
#[test]
#[ignore]
fn test_2059_example_3() {
let nums = vec![2, 8, 16];
let start = 0;
let goal = 1;
let result = -1;
assert_eq!(Solution::minimum_operations(nums, start, goal), result);
}
}
// Accepted solution for LeetCode #2059: Minimum Operations to Convert Number
function minimumOperations(nums: number[], start: number, goal: number): number {
const n = nums.length;
const op1 = function (x: number, y: number): number {
return x + y;
};
const op2 = function (x: number, y: number): number {
return x - y;
};
const op3 = function (x: number, y: number): number {
return x ^ y;
};
const ops = [op1, op2, op3];
let vis = new Array(1001).fill(false);
let quenue: Array<Array<number>> = [[start, 0]];
vis[start] = true;
while (quenue.length) {
let [x, step] = quenue.shift();
for (let i = 0; i < n; i++) {
for (let j = 0; j < ops.length; j++) {
const nx = ops[j](x, nums[i]);
if (nx == goal) {
return step + 1;
}
if (nx >= 0 && nx <= 1000 && !vis[nx]) {
vis[nx] = true;
quenue.push([nx, step + 1]);
}
}
}
}
return -1;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
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.