Given two strings text1 and text2, return the length of their longest common subsequence. If there is no common subsequence, return 0.
A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.
For example, "ace" is a subsequence of "abcde".
A common subsequence of two strings is a subsequence that is common to both strings.
Example 1:
Input: text1 = "abcde", text2 = "ace" Output: 3 Explanation: The longest common subsequence is "ace" and its length is 3. Solution:
Bottom up Approach(Memoization) :
Time complexity : O(m*n) where is m and n is the length of two strings a and b
space complexity : o(m*n) for using 2d dp and O(m+n) for auxiliary stack space because in worst case we will make m+n recursive calls.
class Solution { public int longestCommonSubsequence(String text1, String text2) { // we will start with the last index if its a match then we will decrement both the index //else we will decrement text1 index keeping text2 index same and in second method call we will decrement text2 index keeping the text1 index same // by this we will cover all the possibility // and we will be able to get substring with the largest length common in both the Strings // lets optimize with dp int dp[][] = new int[text1.length()][text2.length()]; for(int row[]: dp) Arrays.fill(row,-1); return findLcsLength(text1,text2,text1.length()-1,text2.length()-1,dp); } public int findLcsLength(String a, String b, int i,int j,int dp[][]){ if(i<0 || j<0) return 0; if(dp[i][j]!=-1) return dp[i][j]; if(a.charAt(i) ==b.charAt(j)){ return dp[i][j] = 1 + findLcsLength(a,b,i-1,j-1,dp); } else { return dp[i][j]= 0+Integer.max(findLcsLength(a,b,i-1,j,dp),findLcsLength(a,b,i,j-1,dp)); } } } Tabulation
class Solution { public int longestCommonSubsequence(String str1, String str2) { int dp[][] = new int[str1.length()+1][str2.length()+1]; for(int i=0;i<=str1.length();i++){ dp[i][0] =0; } for( int i =0;i<=str2.length();i++){ dp[0][i] = 0; } for( int i =1;i<=str1.length();i++){ for(int j =1;j<=str2.length();j++){ if(str1.charAt(i-1)==str2.charAt(j-1)){ dp[i][j] = 1 + dp[i-1][j-1]; } else dp[i][j] = Integer.max(dp[i][j-1],dp[i-1][j]); } } return dp[str1.length()][str2.length()]; } } If we have to print the longest common subsequence string, just add the following in the above code.
int p = str1.length(), q = str2.length(); while(p>0 && q>0){ if(str1.charAt(p-1) == str2.charAt(q-1)){ str = str1.charAt(p-1)+ str; p--; q--; } else if(dp[p][q-1] > dp[p-1][q]){ q--; } else p--; } System.out.println(str); 
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