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.
Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i].
The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.
You must write an algorithm that runs in O(n) time and without using the division operation.
Example 1:
Input: nums = [1,2,3,4] Output: [24,12,8,6]
Example 2:
Input: nums = [-1,1,0,-3,3] Output: [0,0,9,0,0]
Constraints:
2 <= nums.length <= 105-30 <= nums[i] <= 30answer[i] is guaranteed to fit in a 32-bit integer.Follow up: Can you solve the problem in O(1) extra space complexity? (The output array does not count as extra space for space complexity analysis.)
Problem summary: Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i]. The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer. You must write an algorithm that runs in O(n) time and without using the division operation.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array
[1,2,3,4]
[-1,1,0,-3,3]
trapping-rain-water)maximum-product-subarray)paint-house-ii)minimum-difference-in-sums-after-removal-of-elements)construct-product-matrix)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #238: Product of Array Except Self
class Solution {
public int[] productExceptSelf(int[] nums) {
int n = nums.length;
int[] ans = new int[n];
for (int i = 0, left = 1; i < n; ++i) {
ans[i] = left;
left *= nums[i];
}
for (int i = n - 1, right = 1; i >= 0; --i) {
ans[i] *= right;
right *= nums[i];
}
return ans;
}
}
// Accepted solution for LeetCode #238: Product of Array Except Self
func productExceptSelf(nums []int) []int {
n := len(nums)
ans := make([]int, n)
left, right := 1, 1
for i, x := range nums {
ans[i] = left
left *= x
}
for i := n - 1; i >= 0; i-- {
ans[i] *= right
right *= nums[i]
}
return ans
}
# Accepted solution for LeetCode #238: Product of Array Except Self
class Solution:
def productExceptSelf(self, nums: List[int]) -> List[int]:
n = len(nums)
ans = [0] * n
left = right = 1
for i, x in enumerate(nums):
ans[i] = left
left *= x
for i in range(n - 1, -1, -1):
ans[i] *= right
right *= nums[i]
return ans
// Accepted solution for LeetCode #238: Product of Array Except Self
impl Solution {
pub fn product_except_self(nums: Vec<i32>) -> Vec<i32> {
let n = nums.len();
let mut ans = vec![1; n];
for i in 1..n {
ans[i] = ans[i - 1] * nums[i - 1];
}
let mut r = 1;
for i in (0..n).rev() {
ans[i] *= r;
r *= nums[i];
}
ans
}
}
// Accepted solution for LeetCode #238: Product of Array Except Self
function productExceptSelf(nums: number[]): number[] {
const n = nums.length;
const ans: number[] = new Array(n);
for (let i = 0, left = 1; i < n; ++i) {
ans[i] = left;
left *= nums[i];
}
for (let i = n - 1, right = 1; i >= 0; --i) {
ans[i] *= right;
right *= nums[i];
}
return ans;
}
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.