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leetcode_java/src/main/java/LeetCodeJava/DynamicProgramming/MinimumWindowSubsequence.java
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| package LeetCodeJava.DynamicProgramming; | ||
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| // https://leetcode.com/problems/minimum-window-subsequence/description/ | ||
| // https://leetcode.ca/all/727.html | ||
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| /** | ||
| * | ||
| * 727. Minimum Window Subsequence | ||
| * Given strings S and T, find the minimum (contiguous) substring W of S, so that T is a subsequence of W. | ||
| * | ||
| * If there is no such window in S that covers all characters in T, return the empty string "". If there are multiple such minimum-length windows, return the one with the left-most starting index. | ||
| * | ||
| * Example 1: | ||
| * | ||
| * Input: | ||
| * S = "abcdebdde", T = "bde" | ||
| * Output: "bcde" | ||
| * Explanation: | ||
| * "bcde" is the answer because it occurs before "bdde" which has the same length. | ||
| * "deb" is not a smaller window because the elements of T in the window must occur in order. | ||
| * | ||
| * | ||
| * Note: | ||
| * | ||
| * All the strings in the input will only contain lowercase letters. | ||
| * The length of S will be in the range [1, 20000]. | ||
| * The length of T will be in the range [1, 100]. | ||
| * | ||
| * | ||
| * Difficulty: | ||
| * Hard | ||
| * Lock: | ||
| * Prime | ||
| * Company: | ||
| * Amazon Bloomberg eBay Google Houzz Microsoft | ||
| * | ||
| */ | ||
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| public class MinimumWindowSubsequence { | ||
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| // V0 | ||
| // TODO : implement | ||
| // public String minWindow(String s1, String s2) { | ||
| // } | ||
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| // V1 | ||
| // https://leetcode.ca/2017-11-26-727-Minimum-Window-Subsequence/ | ||
| // IDEA : DP | ||
| public String minWindow_1(String s1, String s2) { | ||
| int m = s1.length(), n = s2.length(); | ||
| int[][] f = new int[m + 1][n + 1]; | ||
| for (int i = 1; i <= m; ++i) { | ||
| for (int j = 1; j <= n; ++j) { | ||
| if (s1.charAt(i - 1) == s2.charAt(j - 1)) { | ||
| f[i][j] = j == 1 ? i : f[i - 1][j - 1]; | ||
| } else { | ||
| f[i][j] = f[i - 1][j]; | ||
| } | ||
| } | ||
| } | ||
| int p = 0, k = m + 1; | ||
| for (int i = 1; i <= m; ++i) { | ||
| if (s1.charAt(i - 1) == s2.charAt(n - 1) && f[i][n] > 0) { | ||
| int j = f[i][n] - 1; | ||
| if (i - j < k) { | ||
| k = i - j; | ||
| p = j; | ||
| } | ||
| } | ||
| } | ||
| return k > m ? "" : s1.substring(p, p + k); | ||
| } | ||
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| // V2 | ||
| // https://github.com/doocs/leetcode/blob/main/solution/0700-0799/0727.Minimum%20Window%20Subsequence/README.md | ||
| public String minWindow_2(String s1, String s2) { | ||
| int m = s1.length(), n = s2.length(); | ||
| int[][] f = new int[m + 1][n + 1]; | ||
| for (int i = 1; i <= m; ++i) { | ||
| for (int j = 1; j <= n; ++j) { | ||
| if (s1.charAt(i - 1) == s2.charAt(j - 1)) { | ||
| f[i][j] = j == 1 ? i : f[i - 1][j - 1]; | ||
| } else { | ||
| f[i][j] = f[i - 1][j]; | ||
| } | ||
| } | ||
| } | ||
| int p = 0, k = m + 1; | ||
| for (int i = 1; i <= m; ++i) { | ||
| if (s1.charAt(i - 1) == s2.charAt(n - 1) && f[i][n] > 0) { | ||
| int j = f[i][n] - 1; | ||
| if (i - j < k) { | ||
| k = i - j; | ||
| p = j; | ||
| } | ||
| } | ||
| } | ||
| return k > m ? "" : s1.substring(p, p + k); | ||
| } | ||
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| // V3 | ||
| // https://www.cnblogs.com/grandyang/p/8684817.html | ||
| // IDEA : DP (modified by GPT) | ||
| // TODO : validate below | ||
| public String minWindow_3_1(String S, String T) { | ||
| int m = S.length(); | ||
| int n = T.length(); | ||
| int start = -1; | ||
| int minLen = Integer.MAX_VALUE; | ||
| int[][] dp = new int[m + 1][n + 1]; | ||
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| // Initialize dp array with -1 | ||
| for (int[] row : dp) { | ||
| java.util.Arrays.fill(row, -1); | ||
| } | ||
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| // Base case: empty T can be matched starting at any index in S | ||
| for (int i = 0; i <= m; ++i) { | ||
| dp[i][0] = i; | ||
| } | ||
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| // Fill the dp array | ||
| for (int i = 1; i <= m; ++i) { | ||
| for (int j = 1; j <= Math.min(i, n); ++j) { | ||
| dp[i][j] = (S.charAt(i - 1) == T.charAt(j - 1)) ? dp[i - 1][j - 1] : dp[i - 1][j]; | ||
| } | ||
| if (dp[i][n] != -1) { | ||
| int len = i - dp[i][n]; | ||
| if (minLen > len) { | ||
| minLen = len; | ||
| start = dp[i][n]; | ||
| } | ||
| } | ||
| } | ||
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| return (start != -1) ? S.substring(start, start + minLen) : ""; | ||
| } | ||
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| // V3-1 | ||
| // https://www.cnblogs.com/grandyang/p/8684817.html | ||
| // IDEA : 2 POINTERS (modified by GPT) | ||
| // TODO : validate below | ||
| public String minWindow_3_2(String S, String T) { | ||
| int m = S.length(); | ||
| int n = T.length(); | ||
| int start = -1; | ||
| int minLen = Integer.MAX_VALUE; | ||
| int i = 0; | ||
| int j = 0; | ||
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| while (i < m) { | ||
| if (S.charAt(i) == T.charAt(j)) { | ||
| if (++j == n) { | ||
| int end = i + 1; | ||
| while (--j >= 0) { | ||
| while (i >= 0 && S.charAt(i) != T.charAt(j)) { | ||
| i--; | ||
| } | ||
| i--; | ||
| } | ||
| i++; | ||
| j++; | ||
| if (end - i < minLen) { | ||
| minLen = end - i; | ||
| start = i; | ||
| } | ||
| } | ||
| } | ||
| i++; | ||
| } | ||
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| return (start != -1) ? S.substring(start, start + minLen) : ""; | ||
| } | ||
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| } |
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