dp

Author: idf

src/DistinctSubsequences/Solution.java

@@ -0,0 +1,43 @@
+package DistinctSubsequences;
+
+/**
+ * User: Danyang
+ * Date: 1/28/2015
+ * Time: 18:46
+ *
+ * Given a string S and a string T, count the number of distinct subsequences of T in S.
+
+ A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of
+ the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of
+ "ABCDE" while "AEC" is not).
+
+ Here is an example:
+ S = "rabbbit", T = "rabbit"
+
+ Return 3.
+ */
+public class Solution {
+    /**
+     * let f[i][j] represents the num of distinct subsequence of b[0..j] in a[0..i]
+     * f[i][j] = f[i-1][j-1] + f[i-1][j] if a[i]=b[i]
+     *         = f[i-1][j] otherwise
+     * @param S
+     * @param T
+     * @return
+     */
+    public int numDistinct(String S, String T) {
+        int m = S.length();
+        int n = T.length();
+        int[][] f = new int[m+1][n+1];
+        for(int i=0; i<m+1; i++)
+            f[i][0] = 1;
+        for(int i=1; i<m+1; i++) {
+            for(int j=1; j<n+1; j++) {
+                if(S.charAt(i-1)==T.charAt(j-1))
+                    f[i][j] = f[i-1][j-1];
+                f[i][j] += f[i-1][j];
+            }
+        }
+        return f[m][n];
+    }
+}

src/InterleavingString/Solution.java

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+package InterleavingString;
+
+/**
+ * User: Danyang
+ * Date: 1/28/2015
+ * Time: 17:11
+ *
+ * Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.
+
+ For example,
+ Given:
+ s1 = "aabcc",
+ s2 = "dbbca",
+
+ When s3 = "aadbbcbcac", return true.
+ When s3 = "aadbbbaccc", return false.
+ */
+public class Solution {
+    /**
+     * dp
+     * f[i, j] represents s3[0..i+j] is interleaved from s1[0..i] and s2[0..j]
+     * f[i, j] = f[i-1, j] if s1[i] = s3[i+j]
+     *         = f[i, j-1] if s2[j] = s3[i+j]
+     *
+     * Notice:
+     * 1. initial condition
+     * 2. s3[i+j-1] rather than s3[i+j-2]
+     * @param s1
+     * @param s2
+     * @param s3
+     * @return
+     */
+    public boolean isInterleave(String s1, String s2, String s3) {
+        if(s1.length()+s2.length()!=s3.length())
+            return false;
+
+        boolean[][] f = new boolean[s1.length()+1][s2.length()+1];
+        f[0][0] = true;
+        for(int i=1; i<s1.length()+1; i++)
+            f[i][0] = f[i-1][0] && s1.charAt(i-1)==s3.charAt(i-1);
+        for(int j=1; j<s2.length()+1; j++)
+            f[0][j] = f[0][j-1] && s2.charAt(j-1)==s3.charAt(j-1);
+
+        for(int i=1; i<s1.length()+1; i++)
+            for(int j=1; j<s2.length()+1; j++) {
+                if(!f[i][j] && s1.charAt(i-1)==s3.charAt(i+j-1))
+                    f[i][j] = f[i-1][j];
+                if(!f[i][j] && s2.charAt(j-1)==s3.charAt(i+j-1))
+                    f[i][j] = f[i][j-1];
+            }
+
+        return f[s1.length()][s2.length()];
+    }
+
+    public static void main(String[] args) {
+        assert new Solution().isInterleave("aabcc", "dbbca", "aadbbcbcac");
+    }
+}