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# 1143. Longest Common Subsequence

### Description

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.

### Constraints

* `1 <= text1.length, text2.length <= 1000`
* `text1` and `text2` consist of only lowercase English characters.

### Approach

### Links

* Binarysearch
* GeeksforGeeks
* [Leetcode](https://leetcode.com/problems/longest-common-subsequence/)
* ProgramCreek
* YouTube

### **Examples**

{% tabs %}
{% tab title="Example 1" %}
**Input:** text1 = "abc", text2 = "abc"

**Output:** 3

**Explanation:** The longest common subsequence is "abc" and its length is 3.
{% endtab %}

{% tab title="Example 2" %}
**Input:** text1 = "abc", text2 = "def"

**Output:** 0

**Explanation:** There is no such common subsequence, so the result is 0.
{% endtab %}
{% endtabs %}

### **Solutions**

{% tabs %}
{% tab title="Solution 1" %}

```java
/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        if(text1 == null || text2 == null || text1.isEmpty() || text2.isEmpty()) {
            return 0;
        }
        int t1Len = text1.length(), t2Len = text2.length();
        int[][] dp = new int[t1Len+1][t2Len+1];
        
        for(int i = 1; i <= t1Len; i++) {
            for(int j = 1; j <= t2Len; j++) {
                if(text1.charAt(i-1) == text2.charAt(j-1)) {
                    dp[i][j] = 1 + dp[i-1][j-1];
                } else {
                    dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]);
                }
            }
        }
        
        return dp[t1Len][t2Len];
    }
}
```

{% endtab %}
{% endtabs %}

### **Follow up**

*