LeetCode #3159 — MEDIUM

Find Occurrences of an Element in an Array

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an integer array nums, an integer array queries, and an integer x.

For each queries[i], you need to find the index of the queries[i]^th occurrence of x in the nums array. If there are fewer than queries[i] occurrences of x, the answer should be -1 for that query.

Return an integer array answer containing the answers to all queries.

Example 1:

Input: nums = [1,3,1,7], queries = [1,3,2,4], x = 1

Output: [0,-1,2,-1]

Explanation:

For the 1^st query, the first occurrence of 1 is at index 0.
For the 2^nd query, there are only two occurrences of 1 in nums, so the answer is -1.
For the 3^rd query, the second occurrence of 1 is at index 2.
For the 4^th query, there are only two occurrences of 1 in nums, so the answer is -1.

Example 2:

Input: nums = [1,2,3], queries = [10], x = 5

Output: [-1]

Explanation:

For the 1^st query, 5 doesn't exist in nums, so the answer is -1.

Constraints:

1 <= nums.length, queries.length <= 10⁵
1 <= queries[i] <= 10⁵
1 <= nums[i], x <= 10⁴

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums, an integer array queries, and an integer x. For each queries[i], you need to find the index of the queries[i]th occurrence of x in the nums array. If there are fewer than queries[i] occurrences of x, the answer should be -1 for that query. Return an integer array answer containing the answers to all queries.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[1,3,1,7]
[1,3,2,4]
1

Example 2

[1,2,3]
[10]
5

Step 02

Core Insight

What unlocks the optimal approach

Compress the array <code>nums</code> and save all the occurrences of each element in the separate arrays.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3159: Find Occurrences of an Element in an Array
class Solution {
    public int[] occurrencesOfElement(int[] nums, int[] queries, int x) {
        List<Integer> ids = new ArrayList<>();
        for (int i = 0; i < nums.length; ++i) {
            if (nums[i] == x) {
                ids.add(i);
            }
        }
        int m = queries.length;
        int[] ans = new int[m];
        for (int i = 0; i < m; ++i) {
            int j = queries[i] - 1;
            ans[i] = j < ids.size() ? ids.get(j) : -1;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3159: Find Occurrences of an Element in an Array
func occurrencesOfElement(nums []int, queries []int, x int) (ans []int) {
	ids := []int{}
	for i, v := range nums {
		if v == x {
			ids = append(ids, i)
		}
	}
	for _, i := range queries {
		if i-1 < len(ids) {
			ans = append(ans, ids[i-1])
		} else {
			ans = append(ans, -1)
		}
	}
	return
}

# Accepted solution for LeetCode #3159: Find Occurrences of an Element in an Array
class Solution:
    def occurrencesOfElement(
        self, nums: List[int], queries: List[int], x: int
    ) -> List[int]:
        ids = [i for i, v in enumerate(nums) if v == x]
        return [ids[i - 1] if i - 1 < len(ids) else -1 for i in queries]

// Accepted solution for LeetCode #3159: Find Occurrences of an Element in an Array
/**
 * [3159] Find Occurrences of an Element in an Array
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn occurrences_of_element(nums: Vec<i32>, queries: Vec<i32>, x: i32) -> Vec<i32> {
        let x_pos: Vec<usize> = nums
            .iter()
            .enumerate()
            .filter_map(|(i, v)| if *v == x { Some(i) } else { None })
            .collect();

        let result = queries
            .into_iter()
            .map(|x| {
                let x = x as usize - 1;
                if x >= x_pos.len() {
                    -1
                } else {
                    x_pos[x] as i32
                }
            })
            .collect();

        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3159() {
        assert_eq!(
            vec![0, -1, 2, -1],
            Solution::occurrences_of_element(vec![1, 3, 1, 7], vec![1, 3, 2, 4], 1)
        );
        assert_eq!(
            vec![-1],
            Solution::occurrences_of_element(vec![1, 2, 3], vec![10], 5)
        );
    }
}

// Accepted solution for LeetCode #3159: Find Occurrences of an Element in an Array
function occurrencesOfElement(nums: number[], queries: number[], x: number): number[] {
    const ids: number[] = nums.map((v, i) => (v === x ? i : -1)).filter(v => v !== -1);
    return queries.map(i => ids[i - 1] ?? -1);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n + m)

Space

O(n + m)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

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.

OPTIMIZED

O(n) time

O(1) space

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.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.

Mutating counts without cleanup

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.