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 of length n and a 2D array queries, where queries[i] = [li, ri].
For each queries[i]:
[li, ri] in nums.A Zero Array is an array where all elements are equal to 0.
Return true if it is possible to transform nums into a Zero Array after processing all the queries sequentially, otherwise return false.
Example 1:
Input: nums = [1,0,1], queries = [[0,2]]
Output: true
Explanation:
[0, 2] and decrement the values at these indices by 1.[0, 0, 0], which is a Zero Array.Example 2:
Input: nums = [4,3,2,1], queries = [[1,3],[0,2]]
Output: false
Explanation:
[1, 2, 3] and decrement the values at these indices by 1.[4, 2, 1, 0].[0, 1, 2] and decrement the values at these indices by 1.[3, 1, 0, 0], which is not a Zero Array.Constraints:
1 <= nums.length <= 1050 <= nums[i] <= 1051 <= queries.length <= 105queries[i].length == 20 <= li <= ri < nums.lengthProblem summary: You are given an integer array nums of length n and a 2D array queries, where queries[i] = [li, ri]. For each queries[i]: Select a subset of indices within the range [li, ri] in nums. Decrement the values at the selected indices by 1. A Zero Array is an array where all elements are equal to 0. Return true if it is possible to transform nums into a Zero Array after processing all the queries sequentially, otherwise return false.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array
[1,0,1] [[0,2]]
[4,3,2,1] [[1,3],[0,2]]
zero-array-transformation-iv)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3355: Zero Array Transformation I
class Solution {
public boolean isZeroArray(int[] nums, int[][] queries) {
int n = nums.length;
int[] d = new int[n + 1];
for (var q : queries) {
int l = q[0], r = q[1];
++d[l];
--d[r + 1];
}
for (int i = 0, s = 0; i < n; ++i) {
s += d[i];
if (nums[i] > s) {
return false;
}
}
return true;
}
}
// Accepted solution for LeetCode #3355: Zero Array Transformation I
func isZeroArray(nums []int, queries [][]int) bool {
d := make([]int, len(nums)+1)
for _, q := range queries {
l, r := q[0], q[1]
d[l]++
d[r+1]--
}
s := 0
for i, x := range nums {
s += d[i]
if x > s {
return false
}
}
return true
}
# Accepted solution for LeetCode #3355: Zero Array Transformation I
class Solution:
def isZeroArray(self, nums: List[int], queries: List[List[int]]) -> bool:
d = [0] * (len(nums) + 1)
for l, r in queries:
d[l] += 1
d[r + 1] -= 1
s = 0
for x, y in zip(nums, d):
s += y
if x > s:
return False
return True
// Accepted solution for LeetCode #3355: Zero Array Transformation I
/**
* [3355] Zero Array Transformation I
*/
pub struct Solution {}
// submission codes start here
impl Solution {
pub fn is_zero_array(nums: Vec<i32>, queries: Vec<Vec<i32>>) -> bool {
let n = nums.len();
let mut difference_array = vec![0; n + 1];
for query in queries {
let left = query[0] as usize;
let right = query[1] as usize;
difference_array[left] += 1;
difference_array[right + 1] -= 1;
}
let mut prefix_sum = difference_array[0];
for i in 1..=n {
if prefix_sum < nums[i - 1] {
return false;
}
prefix_sum += difference_array[i];
}
true
}
}
// submission codes end
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn test_3355() {
assert!(Solution::is_zero_array(vec![1, 0, 1], vec![vec![0, 2]]));
assert!(!Solution::is_zero_array(
vec![4, 3, 2, 1],
vec![vec![1, 3], vec![0, 2]]
));
}
}
// Accepted solution for LeetCode #3355: Zero Array Transformation I
function isZeroArray(nums: number[], queries: number[][]): boolean {
const n = nums.length;
const d: number[] = Array(n + 1).fill(0);
for (const [l, r] of queries) {
++d[l];
--d[r + 1];
}
for (let i = 0, s = 0; i < n; ++i) {
s += d[i];
if (nums[i] > s) {
return false;
}
}
return true;
}
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.