Closest Distance to Character - Google Top Interview Questions

Problem Statement :

Given a string s and a character c, return a new list of integers of the same length as s where for each index i its value is set the closest distance of s[i] to c. You can assume c exists in s.


n ≤ 100,000 where n is the length of s

Example 1


s = "aabaab"

c = "b"


[2, 1, 0, 1, 1, 0]

Solution :


                        Solution in C++ :

vector<int> solve(string s, string c) {
    vector<int> res(s.size(), s.size());
    set<int> pos;

    for (int i = 0; i < s.size(); i++)
        if (s[i] == c[0]) pos.insert(i);

    for (int i = 0; i < s.size(); i++) {
        auto it = pos.lower_bound(i);
        // check your right
        if (it != pos.end()) res[i] = *it - i;
        // check your left
        if (it != pos.begin()) res[i] = min(res[i], i - *prev(it));
    return res;

                        Solution in Java :

import java.util.*;

class Solution {
    public int[] solve(String s, String c) {
        Integer[] ret = new Integer[s.length()];
        Queue<Integer> q = new LinkedList();
        for (int i = 0; i < s.length(); i++) {
            if (String.valueOf(s.charAt(i)).equals(c)) {
        Set<Integer> vis = new HashSet();
        int dist = 0;
        while (!q.isEmpty()) {
            int size = q.size();
            for (int x = 0; x < size; x++) {
                int temp = q.poll();
                if (temp < 0 || temp >= s.length() || vis.contains(temp) || ret[temp] != null) {
                ret[temp] = dist;
                q.offer(temp + 1);
                q.offer(temp - 1);
        int[] convert = new int[ret.length];
        for (int i = 0; i < ret.length; i++) convert[i] = ret[i];
        return convert;

                        Solution in Python : 
class Solution:
    def solve(self, s, c):
        result = [inf] * len(s)

        last_found = inf

        for pos, char in enumerate(s):
            if char == c:
                result[pos] = 0
                last_found = 0
            elif last_found != inf:
                last_found += 1
                result[pos] = last_found

        for pos in range(len(s) - 2, -1, -1):
            if result[pos + 1] + 1 < result[pos]:
                result[pos] = result[pos + 1] + 1

        return result

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