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| 1 | +class Main |
| 2 | +{ |
| 3 | + // Function to return an SCS of substrings `X[0…m-1]`, `Y[0…n-1]` |
| 4 | + public static String SCS(String X, String Y, int m, int n, int[][] lookup) |
| 5 | + { |
| 6 | + // if the end of the first string is reached, |
| 7 | + // return the second string |
| 8 | + if (m == 0) { |
| 9 | + return Y.substring(0, n); |
| 10 | + } |
| 11 | + |
| 12 | + // if the end of the second string is reached, |
| 13 | + // return the first string |
| 14 | + if (n == 0) { |
| 15 | + return X.substring(0, m); |
| 16 | + } |
| 17 | + |
| 18 | + // if the last character of `X` and `Y` matches, include it in SSC |
| 19 | + // and recur to find SCS of substring `X[0…m-2]` and `Y[0…n-1]` |
| 20 | + if (X.charAt(m - 1) == Y.charAt(n - 1)) { |
| 21 | + return SCS(X, Y, m - 1, n - 1, lookup) + X.charAt(m - 1); |
| 22 | + } |
| 23 | + // if the last character of `X` and `Y` don't match |
| 24 | + else { |
| 25 | + |
| 26 | + // if the top cell of a current cell has less value than the left |
| 27 | + // cell, then include the current character of string `X` in SCS |
| 28 | + // and find SCS of substring `X[0…m-2]` and `Y[0…n-2]` |
| 29 | + |
| 30 | + if (lookup[m - 1][n] < lookup[m][n - 1]) { |
| 31 | + return SCS(X, Y, m - 1, n, lookup) + X.charAt(m - 1); |
| 32 | + } |
| 33 | + |
| 34 | + // if the left cell of a current cell has less value than the top |
| 35 | + // cell, then include the current character of string `Y` in SCS |
| 36 | + // and find SCS of substring `X[0…m-1]` and `Y[0…n-2]` |
| 37 | + else { |
| 38 | + return SCS(X, Y, m, n - 1, lookup) + Y.charAt(n - 1); |
| 39 | + } |
| 40 | + } |
| 41 | + } |
| 42 | + |
| 43 | + // Function to fill the lookup table by finding the length of SCS of |
| 44 | + // sequences `X[0…m-1]` and `Y[0…n-1]` |
| 45 | + public static void SCSLength(String X, String Y, int m, int n, int[][] lookup) |
| 46 | + { |
| 47 | + // initialize the first column of the lookup table |
| 48 | + for (int i = 0; i <= m; i++) { |
| 49 | + lookup[i][0] = i; |
| 50 | + } |
| 51 | + |
| 52 | + // initialize the first row of the lookup table |
| 53 | + for (int j = 0; j <= n; j++) { |
| 54 | + lookup[0][j] = j; |
| 55 | + } |
| 56 | + |
| 57 | + // fill the lookup table in a bottom-up manner |
| 58 | + for (int i = 1; i <= m; i++) |
| 59 | + { |
| 60 | + for (int j = 1; j <= n; j++) |
| 61 | + { |
| 62 | + // if the current character of `X` and `Y` matches |
| 63 | + if (X.charAt(i - 1) == Y.charAt(j - 1)) { |
| 64 | + lookup[i][j] = lookup[i - 1][j - 1] + 1; |
| 65 | + } |
| 66 | + // otherwise, if the current character of `X` and `Y` don't match |
| 67 | + else { |
| 68 | + lookup[i][j] = Integer.min(lookup[i - 1][j] + 1, lookup[i][j - 1] + 1); |
| 69 | + } |
| 70 | + } |
| 71 | + } |
| 72 | + } |
| 73 | + |
| 74 | + public static void main(String[] args) |
| 75 | + { |
| 76 | + String X = "ABCBDAB", Y = "BDCABA"; |
| 77 | + |
| 78 | + int m = X.length(), n = Y.length(); |
| 79 | + |
| 80 | + // lookup table stores solution to already computed subproblems |
| 81 | + // lookup[i][j] stores the length of SCS of substring `X[0…i-1]` and `Y[0…j-1]` |
| 82 | + int[][] lookup = new int[m + 1][n + 1]; |
| 83 | + |
| 84 | + // fill the lookup table in a bottom-up manner |
| 85 | + SCSLength(X, Y, m, n, lookup); |
| 86 | + |
| 87 | + // find the shortest common supersequence by reading the lookup |
| 88 | + // table in a top-down manner |
| 89 | + System.out.print("The shortest common supersequence of " + X + |
| 90 | + " and " + Y + " is " + SCS(X, Y, m, n, lookup)); |
| 91 | + } |
| 92 | +} |
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