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Commit 4aea629

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Longest common substring & Edit distance added
1 parent 4dac109 commit 4aea629

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3 files changed

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‎AlgorithmCode/EditDistance.java‎

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/**
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* An implementation of the edit distance algorithm
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*
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* <p>Time Complexity: O(nm)
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*
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* @author Micah Stairs
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*/
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public class EditDistance {
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// Computes the cost to convert a string 'a' into a string 'b' using dynamic
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// programming given the insertionCost, deletionCost and substitutionCost, O(nm)
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public static int editDistance(
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String a, String b, int insertionCost, int deletionCost, int substitutionCost) {
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final int AL = a.length(), BL = b.length();
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int[][] arr = new int[AL + 1][BL + 1];
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for (int i = 0; i <= AL; i++) {
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for (int j = (i == 0 ? 1 : 0); j <= BL; j++) {
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int min = Integer.MAX_VALUE;
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// Substitution
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if (i > 0 && j > 0)
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min = arr[i - 1][j - 1] + (a.charAt(i - 1) == b.charAt(j - 1) ? 0 : substitutionCost);
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// Deletion
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if (i > 0) min = Math.min(min, arr[i - 1][j] + deletionCost);
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// Insertion
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if (j > 0) min = Math.min(min, arr[i][j - 1] + insertionCost);
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arr[i][j] = min;
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}
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}
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return arr[AL][BL];
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}
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public static void main(String[] args) {
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String a = "abcdefg";
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String b = "abcdefg";
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// The strings are the same so the cost is zero
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System.out.println(editDistance(a, b, 10, 10, 10));
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a = "aaa";
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b = "aaabbb";
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// 10*3 = 30 because of three insertions
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System.out.println(editDistance(a, b, 10, 2, 3));
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a = "1023";
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b = "10101010";
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// Outputs 2*2 + 4*5 = 24 for 2 substitutions and 4 insertions
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System.out.println(editDistance(a, b, 5, 7, 2));
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a = "923456789";
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b = "12345";
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// Outputs 4*4 + 1 = 16 because we need to delete 4
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// characters and perform one substitution
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System.out.println(editDistance(a, b, 2, 4, 1));
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}
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}
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/**
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* This file contains an implementation of finding the Longest Common Substring (LCS) between two
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* strings using dynamic programming.
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*
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* <p>Time Complexity: O(nm)
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*
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* @author William Fiset, william.alexandre.fiset@gmail.com
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*/
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public class LongestCommonSubstring {
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// Returns a non unique Longest Common Substring
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// between the strings str1 and str2 in O(nm)
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public static String lcs(char[] A, char[] B) {
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if (A == null || B == null) return null;
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final int n = A.length;
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final int m = B.length;
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if (n == 0 || m == 0) return null;
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int[][] dp = new int[n + 1][m + 1];
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// Suppose A = a1a2..an-1an and B = b1b2..bn-1bn
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for (int i = 1; i <= n; i++) {
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for (int j = 1; j <= m; j++) {
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// If ends match the LCS(a1a2..an-1an, b1b2..bn-1bn) = LCS(a1a2..an-1, b1b2..bn-1) + 1
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if (A[i - 1] == B[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1;
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// If the ends do not match the LCS of a1a2..an-1an and b1b2..bn-1bn is
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// max( LCS(a1a2..an-1, b1b2..bn-1bn), LCS(a1a2..an-1an, b1b2..bn-1) )
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else dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
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}
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}
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int lcsLen = dp[n][m];
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char[] lcs = new char[lcsLen];
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int index = 0;
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// Backtrack to find a LCS. We search for the cells
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// where we included an element which are those with
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// dp[i][j] != dp[i-1][j] and dp[i][j] != dp[i][j-1])
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int i = n, j = m;
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while (i >= 1 && j >= 1) {
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int v = dp[i][j];
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// The order of these may output different LCSs
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while (i > 1 && dp[i - 1][j] == v) i--;
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while (j > 1 && dp[i][j - 1] == v) j--;
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// Make sure there is a match before adding
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if (v > 0) lcs[lcsLen - index++ - 1] = A[i - 1]; // or B[j-1];
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i--;
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j--;
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}
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return new String(lcs, 0, lcsLen);
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}
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public static void main(String[] args) {
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char[] A = {'A', 'X', 'B', 'C', 'Y'};
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char[] B = {'Z', 'A', 'Y', 'W', 'B', 'C'};
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System.out.println(lcs(A, B)); // ABC
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A = new char[] {'3', '9', '8', '3', '9', '7', '9', '7', '0'};
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B = new char[] {'3', '3', '9', '9', '9', '1', '7', '2', '0', '6'};
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System.out.println(lcs(A, B)); // 339970
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}
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}

‎ProblemSolvingCodeBOJ/CodingQuestion17_11.java‎

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public static void main(String[] args) {
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dot1();
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System.out.println(max);
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//Output:
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// 9
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}
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static class Point {

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