g0001_0100.s0076_minimum_window_substring.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 g0001_0100.s0076_minimum_window_substring;

// #Hard #Top_100_Liked_Questions #Top_Interview_Questions #String #Hash_Table #Sliding_Window
// #Level_2_Day_14_Sliding_Window/Two_Pointer #Big_O_Time_O(s.length())_Space_O(1)
// #2023_08_11_Time_2_ms_(99.94%)_Space_43.6_MB_(93.87%)

/**
 * 76 - Minimum Window Substring\.
 *
 * Hard
 *
 * Given two strings `s` and `t` of lengths `m` and `n` respectively, return _the **minimum window substring** of_ `s` _such that every character in_ `t` _( **including duplicates** ) is included in the window. If there is no such substring__, return the empty string_ `""`_._
 *
 * The testcases will be generated such that the answer is **unique**.
 *
 * A **substring** is a contiguous sequence of characters within the string.
 *
 * **Example 1:**
 *
 * **Input:** s = "ADOBECODEBANC", t = "ABC"
 *
 * **Output:** "BANC"
 *
 * **Explanation:** The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t. 
 *
 * **Example 2:**
 *
 * **Input:** s = "a", t = "a"
 *
 * **Output:** "a"
 *
 * **Explanation:** The entire string s is the minimum window. 
 *
 * **Example 3:**
 *
 * **Input:** s = "a", t = "aa"
 *
 * **Output:** ""
 *
 * **Explanation:** Both 'a's from t must be included in the window. Since the largest window of s only has one 'a', return empty string. 
 *
 * **Constraints:**
 *
 * *   `m == s.length`
 * *   `n == t.length`
 * *   1 <= m, n <= 10⁵
 * *   `s` and `t` consist of uppercase and lowercase English letters.
 *
 * **Follow up:** Could you find an algorithm that runs in `O(m + n)` time?
**/
public class Solution {
    public String minWindow(String s, String t) {
        int[] map = new int[128];
        for (int i = 0; i < t.length(); i++) {
            map[t.charAt(i) - 'A']++;
        }
        int count = t.length();
        int begin = 0;
        int end = 0;
        int d = Integer.MAX_VALUE;
        int head = 0;
        while (end < s.length()) {
            if (map[s.charAt(end++) - 'A']-- > 0) {
                count--;
            }
            while (count == 0) {
                if (end - begin < d) {
                    d = end - begin;
                    head = begin;
                }
                if (map[s.charAt(begin++) - 'A']++ == 0) {
                    count++;
                }
            }
        }
        return d == Integer.MAX_VALUE ? "" : s.substring(head, head + d);
    }
}