Leetcode 821 - Shortest Distance to a Character

Note:

  • Key part is how to find the first element in occurence[] that is bigger than index i.
  • Three situations:
    • Before occurence[0]
    • Between two occurence
    • After last occurence.

Question:

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.

Example:

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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) = 2.
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.

Code:

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/**
* @param {string} s
* @param {character} c
* @return {number[]}
*/
var shortestToChar = function(s, c) {
const occurence = [];
for (let i = 0; i < s.length; i++) {
if (s[i] === c) occurence.push(i);
}
let ans = [...new Array(s.length).fill(0)];
const set = new Set(occurence);
for (let j = 0; j < s.length; j++) {
if (set.has(j)) continue;
let i = 0;
while (i < occurence.length && occurence[i] < j) {
i++;
}
if (i === occurence.length) {
ans[j] = j - occurence[i - 1];
} else if (i === 0) {
ans[j] = occurence[0] - j;
} else {
ans[j] = occurence[i] - j < j - occurence[i - 1] ? occurence[i] - j : j - occurence[i - 1];
}
}
return ans;
};