letcode1143 Longest Common Subsequence
letcode1143 Longest Common Subsequence
Longest Common Subsequence
Given two strings text1 and text2, return the length of their longest common subsequence.
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. (eg, "ace" is a subsequence of "abcde" while "aec" is not). A common subsequence of two strings is a subsequence that is common to both strings.
If there is no common subsequence, return 0.
Example 1:
Input: text1 = "abcde", text2 = "ace"
Output: 3
Explanation: The longest common subsequence is "ace" and its length is 3.
Example 2:
Input: text1 = "abc", text2 = "abc"
Output: 3
Explanation: The longest common subsequence is "abc" and its length is 3.
Example 3:
Input: text1 = "abc", text2 = "def"
Output: 0
Explanation: There is no such common subsequence, so the result is 0.
Constraints:
1 <= text1.length <= 1000
1 <= text2.length <= 1000
The input strings consist of lowercase English characters only.
Hide Hint #1
Try dynamic programming. DP[i][j] represents the longest common subsequence of text1[0 ... i] & text2[0 ... j].
Hide Hint #2
DP[i][j] = DP[i - 1][j - 1] + 1 , if text1[i] == text2[j] DP[i][j] = max(DP[i - 1][j], DP[i][j - 1]) , otherwise
solution
approach1 dp
class Solution {
public:
int longestCommonSubsequence(string text1, string text2) {
int a=text1.length();
int b=text2.length();
int dp[a+1][b+1];
for(int i =0;i<a;i++)
for(int j=0;j<b;j++)
{
if(i==0)
{
if(text2.find(text1[0])<=j)
dp[i][j]=1;
else
dp[i][j]=0;
continue;
}
if(j==0)
{
if(text1.find(text2[0])<=i)
dp[i][j]=1;
else
dp[i][j]=0;
continue;
}
if(text1[i] == text2[j])
dp[i][j] = dp[i - 1][j - 1] + 1;
else
dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
}
return dp[a-1][b-1];
}
};
