Three-Hundred-One Commit: Add Longest Common Substring problem to Longest Common Substring section

Author: AAdewunmi

src/Dynamic_Programming/Longest_Common_SubString/Problem_1_Longest_Common_Substring.java

@@ -0,0 +1,63 @@
+package Dynamic_Programming.Longest_Common_SubString;
+
+// Problem Statement: Longest Common Substring
+// LeetCode Question:
+
+public class Problem_1_Longest_Common_Substring {
+ // Brute Force Approach
+ public int findLCSLength(String s1, String s2) {
+ return findLCSLengthRecursive(s1, s2, 0, 0, 0);
+ }
+
+ private int findLCSLengthRecursive(String s1, String s2, int i1, int i2, int count) {
+ if(i1 == s1.length() || i2 == s2.length())
+ return count;
+
+ if(s1.charAt(i1) == s2.charAt(i2))
+ count = findLCSLengthRecursive(s1, s2, i1+1, i2+1, count+1);
+
+ int c1 = findLCSLengthRecursive(s1, s2, i1, i2+1, 0);
+ int c2 = findLCSLengthRecursive(s1, s2, i1+1, i2, 0);
+
+ return Math.max(count, Math.max(c1, c2));
+ }
+
+ // Top-down Dynamic Programming with Memoization Approach
+ public int findLCSLength_1(String s1, String s2) {
+ int maxLength = Math.min(s1.length(), s2.length());
+ Integer[][][] dp = new Integer[s1.length()][s2.length()][maxLength];
+ return findLCSLengthRecursive(dp, s1, s2, 0, 0, 0);
+ }
+
+ private int findLCSLengthRecursive(Integer[][][] dp, String s1, String s2, int i1, int i2, int count) {
+ if(i1 == s1.length() || i2 == s2.length())
+ return count;
+
+ if(dp[i1][i2][count] == null) {
+ int c1 = count;
+ if(s1.charAt(i1) == s2.charAt(i2))
+ c1 = findLCSLengthRecursive(dp, s1, s2, i1+1, i2+1, count+1);
+ int c2 = findLCSLengthRecursive(dp, s1, s2, i1, i2+1, 0);
+ int c3 = findLCSLengthRecursive(dp, s1, s2, i1+1, i2, 0);
+ dp[i1][i2][count] = Math.max(c1, Math.max(c2, c3));
+ }
+
+ return dp[i1][i2][count];
+ }
+
+ // Bottom-up Dynamic Programming Approach
+ public int findLCSLength_2(String s1, String s2) {
+ int[][] dp = new int[2][s2.length()+1];
+ int maxLength = 0;
+ for(int i=1; i <= s1.length(); i++) {
+ for(int j=1; j <= s2.length(); j++) {
+ dp[i%2][j] = 0;
+ if(s1.charAt(i-1) == s2.charAt(j-1)) {
+ dp[i%2][j] = 1 + dp[(i-1)%2][j-1];
+ maxLength = Math.max(maxLength, dp[i%2][j]);
+ }
+ }
+ }
+ return maxLength;
+ }
+}