g0401_0500.s0467_unique_substrings_in_wraparound_string.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0401_0500.s0467_unique_substrings_in_wraparound_string;

// #Medium #String #Dynamic_Programming #2022_07_19_Time_4_ms_(98.79%)_Space_44_MB_(5.26%)

/**
 * 467 - Unique Substrings in Wraparound String\.
 *
 * Medium
 *
 * We define the string `s` to be the infinite wraparound string of `"abcdefghijklmnopqrstuvwxyz"`, so `s` will look like this:
 *
 * *   `"...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd...."`.
 *
 * Given a string `p`, return _the number of **unique non-empty substrings** of_ `p` _are present in_ `s`.
 *
 * **Example 1:**
 *
 * **Input:** p = "a"
 *
 * **Output:** 1 Explanation: Only the substring "a" of p is in s.
 *
 * **Example 2:**
 *
 * **Input:** p = "cac"
 *
 * **Output:** 2
 *
 * **Explanation:** There are two substrings ("a", "c") of p in s.
 *
 * **Example 3:**
 *
 * **Input:** p = "zab"
 *
 * **Output:** 6
 *
 * **Explanation:** There are six substrings ("z", "a", "b", "za", "ab", and "zab") of p in s.
 *
 * **Constraints:**
 *
 * *   1 <= p.length <= 10⁵
 * *   `p` consists of lowercase English letters.
**/
public class Solution {
    public int findSubstringInWraproundString(String p) {
        char[] str = p.toCharArray();
        int n = str.length;
        int[] map = new int[26];
        int len = 0;
        for (int i = 0; i < n; i++) {
            if (i > 0 && ((str[i - 1] + 1 == str[i]) || (str[i - 1] == 'z' && str[i] == 'a'))) {
                len += 1;
            } else {
                len = 1;
            }
            // we are storing the max len of string for each letter and then we will count all these
            // length.
            map[str[i] - 'a'] = Math.max(map[str[i] - 'a'], len);
        }
        int answer = 0;
        for (int num : map) {
            answer += num;
        }
        return answer;
    }
}