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|  | 1 | +package com.leetcode.maps.slidingwindow; | 
|  | 2 | + | 
|  | 3 | +import java.util.HashMap; | 
|  | 4 | +import java.util.Map; | 
|  | 5 | + | 
|  | 6 | +import static org.junit.jupiter.api.Assertions.assertEquals; | 
|  | 7 | + | 
|  | 8 | +/** | 
|  | 9 | + * Level: Hard | 
|  | 10 | + * Link: https://leetcode.com/problems/longest-substring-with-at-most-k-distinct-characters/ | 
|  | 11 | + * Description: | 
|  | 12 | + * Given a string, find the length of the longest substring T that contains at most k distinct characters. | 
|  | 13 | + * <p> | 
|  | 14 | + * Example 1: | 
|  | 15 | + * Input: s = "eceba", k = 2 | 
|  | 16 | + * Output: 3 | 
|  | 17 | + * Explanation: T is "ece" which its length is 3. | 
|  | 18 | + * <p> | 
|  | 19 | + * Example 2: | 
|  | 20 | + * Input: s = "aa", k = 1 | 
|  | 21 | + * Output: 2 | 
|  | 22 | + * Explanation: T is "aa" which its length is 2. | 
|  | 23 | + * | 
|  | 24 | + * @author rampatra | 
|  | 25 | + * @since 2019-08-09 | 
|  | 26 | + */ | 
|  | 27 | +public class LongestSubstringWithKDistinctCharacters { | 
|  | 28 | + | 
|  | 29 | +    /** | 
|  | 30 | +     * Time Complexity: O(n) | 
|  | 31 | +     * Space Complexity: O(k), as we keep at most k characters in the hash table | 
|  | 32 | +     * | 
|  | 33 | +     * @param str | 
|  | 34 | +     * @param k | 
|  | 35 | +     * @return | 
|  | 36 | +     */ | 
|  | 37 | +    public static int lengthOfLongestSubstringKDistinct(String str, int k) { | 
|  | 38 | +        int length = 0; | 
|  | 39 | +        Map<Character, Integer> letterCountInWindow = new HashMap<>(); | 
|  | 40 | + | 
|  | 41 | +        int i = 0; // start of window | 
|  | 42 | +        int j = i; // end of window | 
|  | 43 | + | 
|  | 44 | +        while (j < str.length()) { | 
|  | 45 | +            char ch = str.charAt(j); | 
|  | 46 | + | 
|  | 47 | +            letterCountInWindow.putIfAbsent(ch, 0); | 
|  | 48 | +            letterCountInWindow.put(ch, letterCountInWindow.get(ch) + 1); | 
|  | 49 | + | 
|  | 50 | +            // when number of distinct characters in the window exceeds k: | 
|  | 51 | +            // - update length | 
|  | 52 | +            // - remove the first character in the window or reduce its count if the window had more than one of this character | 
|  | 53 | +            // - lastly, move the window forward | 
|  | 54 | +            if (letterCountInWindow.keySet().size() > k) { | 
|  | 55 | +                char firstChar = str.charAt(i); | 
|  | 56 | +                int firstCharCount = letterCountInWindow.get(firstChar); | 
|  | 57 | +                if (firstCharCount > 1) { | 
|  | 58 | +                    letterCountInWindow.put(firstChar, firstCharCount - 1); | 
|  | 59 | +                } else { | 
|  | 60 | +                    letterCountInWindow.remove(firstChar); | 
|  | 61 | +                } | 
|  | 62 | +                length = Math.max(length, j - i); | 
|  | 63 | +                i++; | 
|  | 64 | +            } | 
|  | 65 | +            j++; | 
|  | 66 | +        } | 
|  | 67 | + | 
|  | 68 | +        return length == 0 ? j - i : length; | 
|  | 69 | +    } | 
|  | 70 | + | 
|  | 71 | +    public static void main(String[] args) { | 
|  | 72 | +        assertEquals(3, lengthOfLongestSubstringKDistinct("eceba", 2)); | 
|  | 73 | +        assertEquals(7, lengthOfLongestSubstringKDistinct("eceeeeeba", 2)); | 
|  | 74 | +        assertEquals(2, lengthOfLongestSubstringKDistinct("abcdef", 2)); | 
|  | 75 | +        assertEquals(1, lengthOfLongestSubstringKDistinct("a", 1)); | 
|  | 76 | +        assertEquals(2, lengthOfLongestSubstringKDistinct("aa", 1)); | 
|  | 77 | +        assertEquals(3, lengthOfLongestSubstringKDistinct("aaa", 1)); | 
|  | 78 | +        assertEquals(0, lengthOfLongestSubstringKDistinct("aa", 0)); | 
|  | 79 | +    } | 
|  | 80 | +} | 
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