401. Closest Equal Element Queries

For each query index q, we need the closest (circular) index with the same value. Group indices by value; for each query binary-search the group list for the neighbors of q.

Approach

1. groups = {value: sorted list of indices}.
2. For each query q:
   • Let arr = groups[nums[q]]. If length 1, answer is -1.
   • bisect_left(arr, q) gives the position; the left and right neighbors (wrapping around) are the two candidates.
   • Minimum of min(|q - i|, n - |q - i|) for each candidate.

import java.util.*;

class Solution {
    public List<Integer> solveQueries(int[] nums, int[] queries) {
        int n = nums.length;
        Map<Integer, List<Integer>> groups = new HashMap<>();
        for (int i = 0; i < n; i++) {
            groups.computeIfAbsent(nums[i], k -> new ArrayList<>()).add(i);
        }
        List<Integer> ans = new ArrayList<>();
        for (int q : queries) {
            List<Integer> arr = groups.get(nums[q]);
            if (arr.size() < 2) {
                ans.add(-1);
                continue;
            }
            int k = Collections.binarySearch(arr, q);
            int leftIdx = ((k - 1) % arr.size() + arr.size()) % arr.size();
            int rightIdx = (k + 1) % arr.size();
            int left = arr.get(leftIdx);
            int right = arr.get(rightIdx);
            int dLeft = Math.abs(q - left);
            dLeft = Math.min(dLeft, n - dLeft);
            int dRight = Math.abs(q - right);
            dRight = Math.min(dRight, n - dRight);
            ans.add(Math.min(dLeft, dRight));
        }
        return ans;
    }
}