g2301_2400.s2399_check_distances_between_same_letters.Solution Maven / Gradle / Ivy

Go to download

Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java21 Show documentation

Java-based LeetCode algorithm problem solutions, regularly updated

There is a newer version: 1.38

package g2301_2400.s2399_check_distances_between_same_letters;

// #Easy #Array #String #Hash_Table #2022_09_19_Time_1_ms_(99.88%)_Space_43.8_MB_(22.33%)

/**
 * 2399 - Check Distances Between Same Letters\.
 *
 * Easy
 *
 * You are given a **0-indexed** string `s` consisting of only lowercase English letters, where each letter in `s` appears **exactly** **twice**. You are also given a **0-indexed** integer array `distance` of length `26`.
 *
 * Each letter in the alphabet is numbered from `0` to `25` (i.e. `'a' -> 0`, `'b' -> 1`, `'c' -> 2`, ... , `'z' -> 25`).
 *
 * In a **well-spaced** string, the number of letters between the two occurrences of the i^th letter is `distance[i]`. If the i^th letter does not appear in `s`, then `distance[i]` can be **ignored**.
 *
 * Return `true` _if_ `s` _is a **well-spaced** string, otherwise return_ `false`.
 *
 * **Example 1:**
 *
 * **Input:** s = "abaccb", distance = [1,3,0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
 *
 * **Output:** true
 *
 * **Explanation:**
 *
 * - 'a' appears at indices 0 and 2 so it satisfies distance[0] = 1.
 *
 * - 'b' appears at indices 1 and 5 so it satisfies distance[1] = 3.
 *
 * - 'c' appears at indices 3 and 4 so it satisfies distance[2] = 0.
 *
 * Note that distance[3] = 5, but since 'd' does not appear in s, it can be ignored.
 *
 * Return true because s is a well-spaced string. 
 *
 * **Example 2:**
 *
 * **Input:** s = "aa", distance = [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
 *
 * **Output:** false
 *
 * **Explanation:**
 *
 * - 'a' appears at indices 0 and 1 so there are zero letters between them. Because distance[0] = 1, s is not a well-spaced string. 
 *
 * **Constraints:**
 *
 * *   `2 <= s.length <= 52`
 * *   `s` consists only of lowercase English letters.
 * *   Each letter appears in `s` exactly twice.
 * *   `distance.length == 26`
 * *   `0 <= distance[i] <= 50`
**/
public class Solution {
    public boolean checkDistances(String s, int[] distance) {
        boolean[] seenFirstIndexYet = new boolean[26];
        for (int idxIntoS = 0; idxIntoS < s.length(); ++idxIntoS) {
            char c = s.charAt(idxIntoS);
            if (!seenFirstIndexYet[c - 'a']) {
                seenFirstIndexYet[c - 'a'] = true;
                distance[c - 'a'] += idxIntoS;
            } else {
                // seenFirstIndexYet[c - 'a']
                distance[c - 'a'] -= idxIntoS;
                if (distance[c - 'a'] != -1) {
                    // early return
                    return false;
                }
            }
        }
        return true;
    }
}