821. Shortest Distance to a Character

Description

Given a string s and a character c that occurs in s, return an array of integers answer where answer.length == s.length and answer[i] is the distance from index i to the closest occurrence of character c in s.

The distance between two indices i and j is abs(i - j), where abs is the absolute value function.

Constraints

• 1 <= s.length <= 104

• s[i] and c are lowercase English letters.

• It is guaranteed that c occurs at least once in s.

Approach

Links

• Binarysearch

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

Input: s = "loveleetcode", c = "e"

Output: [3, 2, 1, 0, 1, 0, 0, 1, 2, 2, 1, 0]

Explanation: The character 'e' appears at indices 3, 5, 6, and 11 (0-indexed).

The closest occurrence of 'e' for index 0 is at index 3, so the distance is abs(0 - 3) = 3.

The closest occurrence of 'e' for index 1 is at index 3, so the distance is abs(1 - 3) = 3.

For index 4, there is a tie between the 'e' at index 3 and the 'e' at index 5, but the distance is still the same: abs(4 - 3) == abs(4 - 5) = 1.

The closest occurrence of 'e' for index 8 is at index 6, so the distance is abs(8 - 6) = 2.

Example 2

Input: s = "aaab", c = "b"

Output: [3,2,1,0]

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public int[] shortestToChar(String s, char c) {
        if(s == null || s.length() == 0) {
            return new int[0];
        }
        int n = s.length();
        int[] result = new int[n];
        
        for(int i = 0, index = Integer.MAX_VALUE; i < n; i++) {
            if(s.charAt(i) == c) {
                index = i;
            }
            result[i] = Math.abs(i-index);
        }
        
        for(int i = n-1, index = Integer.MAX_VALUE; i >= 0; i--) {
            if(s.charAt(i) == c) {
                index = i;
            }
            result[i] = Math.min(result[i], Math.abs(i-index));
        }
        
        return result;
    }
}

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