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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年08月09日 |
| 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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