# 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


$$\begin{equation} LCS(i,j) = \left{ \begin{array}{rcl} & LCS[i-1, j-1] + 1 & ,{s1[i] = s2[j]} \ & max(LCS[i, j-1], LCS[i-1, j]) & ,{s1[i] \neq s2[j]} \end{array} \right. \end{equation}$$

int longestCommonSubsequence(string text1, string text2) {
vector<vector<int>> dp(text1.size() + 1, vector<int>(text2.size() + 1, 0));
for(int i = 1;i < dp.size(); i++) {
for(int j = 1;j < dp.size(); j++) {
if (text1[i-1] == text2[j-1]) {
dp[i][j] = dp[i-1][j-1] + 1;
} else {
dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
}
}
}
return dp[text1.size()][text2.size()];
}


int longestCommonSubsequence(string text1, string text2) {

int n = text1.size() + 1, m = text2.size() + 1;
vector<int> dp1(m, 0);
vector<int> dp2(m);

for(int i = 1;i < n; i++) {
dp1 = 0;
for(int j = 1;j < m; j++) {
if (text1[i-1] == text2[j-1]) {
dp2[j] = dp1[j-1] + 1;
} else {
dp2[j] = max(dp1[j], dp2[j-1]);
}
}
swap(dp1, dp2);
}
return dp1[m-1];
}


int longestCommonSubsequence(string text1, string text2) {
int n = text1.size(), m = text2.size() + 1;
vector<int> dp(m, 0);

for(int i = 0;i < n; i++) {
int prev = 0;
for(int j = 1;j < m; j++) {
int temp = prev;
prev = dp[j];
dp[j] = (text1[i] == text2[j-1]) ? (temp + 1) : (max(dp[j], dp[j-1]));
}
}
return dp[m-1];
}