LeetCode #3488 — MEDIUM

Closest Equal Element Queries

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

Solve on LeetCode

The Problem

Problem Statement

You are given a circular array nums and an array queries.

For each query i, you have to find the following:

The minimum distance between the element at index queries[i] and any other index j in the circular array, where nums[j] == nums[queries[i]]. If no such index exists, the answer for that query should be -1.

Return an array answer of the same size as queries, where answer[i] represents the result for query i.

Example 1:

Input: nums = [1,3,1,4,1,3,2], queries = [0,3,5]

Output: [2,-1,3]

Explanation:

Query 0: The element at queries[0] = 0 is nums[0] = 1. The nearest index with the same value is 2, and the distance between them is 2.
Query 1: The element at queries[1] = 3 is nums[3] = 4. No other index contains 4, so the result is -1.
Query 2: The element at queries[2] = 5 is nums[5] = 3. The nearest index with the same value is 1, and the distance between them is 3 (following the circular path: 5 -> 6 -> 0 -> 1).

Example 2:

Input: nums = [1,2,3,4], queries = [0,1,2,3]

Output: [-1,-1,-1,-1]

Explanation:

Each value in nums is unique, so no index shares the same value as the queried element. This results in -1 for all queries.

Constraints:

1 <= queries.length <= nums.length <= 10⁵
1 <= nums[i] <= 10⁶
0 <= queries[i] < nums.length

Step 01

Brute Force Baseline

Problem summary: You are given a circular array nums and an array queries. For each query i, you have to find the following: The minimum distance between the element at index queries[i] and any other index j in the circular array, where nums[j] == nums[queries[i]]. If no such index exists, the answer for that query should be -1. Return an array answer of the same size as queries, where answer[i] represents the result for query i.

Baseline thinking

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

Pattern signal: Array · Hash Map · Binary Search

Example 1

[1,3,1,4,1,3,2]
[0,3,5]

Example 2

[1,2,3,4]
[0,1,2,3]

Step 02

Core Insight

What unlocks the optimal approach

Use a dictionary that maps each unique value in the array to a sorted list of its indices.
For each query, use binary search on the sorted indices list to find the nearest occurrences of the target value.

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 #3488: Closest Equal Element Queries
class Solution {
    public List<Integer> solveQueries(int[] nums, int[] queries) {
        int n = nums.length;
        int m = n * 2;
        int[] d = new int[m];
        Arrays.fill(d, m);

        Map<Integer, Integer> left = new HashMap<>();
        for (int i = 0; i < m; i++) {
            int x = nums[i % n];
            if (left.containsKey(x)) {
                d[i] = Math.min(d[i], i - left.get(x));
            }
            left.put(x, i);
        }

        Map<Integer, Integer> right = new HashMap<>();
        for (int i = m - 1; i >= 0; i--) {
            int x = nums[i % n];
            if (right.containsKey(x)) {
                d[i] = Math.min(d[i], right.get(x) - i);
            }
            right.put(x, i);
        }

        for (int i = 0; i < n; i++) {
            d[i] = Math.min(d[i], d[i + n]);
        }

        List<Integer> ans = new ArrayList<>();
        for (int query : queries) {
            ans.add(d[query] >= n ? -1 : d[query]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3488: Closest Equal Element Queries
func solveQueries(nums []int, queries []int) []int {
	n := len(nums)
	m := n * 2
	d := make([]int, m)
	for i := range d {
		d[i] = m
	}

	left := make(map[int]int)
	for i := 0; i < m; i++ {
		x := nums[i%n]
		if idx, exists := left[x]; exists {
			d[i] = min(d[i], i-idx)
		}
		left[x] = i
	}

	right := make(map[int]int)
	for i := m - 1; i >= 0; i-- {
		x := nums[i%n]
		if idx, exists := right[x]; exists {
			d[i] = min(d[i], idx-i)
		}
		right[x] = i
	}

	for i := 0; i < n; i++ {
		d[i] = min(d[i], d[i+n])
	}

	ans := make([]int, len(queries))
	for i, query := range queries {
		if d[query] >= n {
			ans[i] = -1
		} else {
			ans[i] = d[query]
		}
	}
	return ans
}

# Accepted solution for LeetCode #3488: Closest Equal Element Queries
class Solution:
    def solveQueries(self, nums: List[int], queries: List[int]) -> List[int]:
        n = len(nums)
        m = n << 1
        d = [m] * m
        left = {}
        for i in range(m):
            x = nums[i % n]
            if x in left:
                d[i] = min(d[i], i - left[x])
            left[x] = i
        right = {}
        for i in range(m - 1, -1, -1):
            x = nums[i % n]
            if x in right:
                d[i] = min(d[i], right[x] - i)
            right[x] = i
        for i in range(n):
            d[i] = min(d[i], d[i + n])
        return [-1 if d[i] >= n else d[i] for i in queries]

// Accepted solution for LeetCode #3488: Closest Equal Element Queries
use std::collections::HashMap;

impl Solution {
    pub fn solve_queries(nums: Vec<i32>, queries: Vec<i32>) -> Vec<i32> {
        let n = nums.len();
        let m = n * 2;
        let mut d = vec![m as i32; m];
        let mut left = HashMap::new();

        for i in 0..m {
            let x = nums[i % n];
            if let Some(&l) = left.get(&x) {
                d[i] = d[i].min((i - l) as i32);
            }
            left.insert(x, i);
        }

        let mut right = HashMap::new();

        for i in (0..m).rev() {
            let x = nums[i % n];
            if let Some(&r) = right.get(&x) {
                d[i] = d[i].min((r - i) as i32);
            }
            right.insert(x, i);
        }

        for i in 0..n {
            d[i] = d[i].min(d[i + n]);
        }

        queries
            .iter()
            .map(|&query| {
                if d[query as usize] >= n as i32 {
                    -1
                } else {
                    d[query as usize]
                }
            })
            .collect()
    }
}

// Accepted solution for LeetCode #3488: Closest Equal Element Queries
function solveQueries(nums: number[], queries: number[]): number[] {
    const n = nums.length;
    const m = n * 2;
    const d: number[] = Array(m).fill(m);

    const left = new Map<number, number>();
    for (let i = 0; i < m; i++) {
        const x = nums[i % n];
        if (left.has(x)) {
            d[i] = Math.min(d[i], i - left.get(x)!);
        }
        left.set(x, i);
    }

    const right = new Map<number, number>();
    for (let i = m - 1; i >= 0; i--) {
        const x = nums[i % n];
        if (right.has(x)) {
            d[i] = Math.min(d[i], right.get(x)! - i);
        }
        right.set(x, i);
    }

    for (let i = 0; i < n; i++) {
        d[i] = Math.min(d[i], d[i + n]);
    }

    return queries.map(query => (d[query] >= n ? -1 : d[query]));
}

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)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

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.

BINARY SEARCH

O(log n) time

O(1) space

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).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Boundary update without `+1` / `-1`

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.