Longest Common Subsequence
MediumGrid DP · match → diagonal+1, else max(up, left)
Match two strings letter by letter, reusing answers for their shorter pieces in a grid.
The idea
Fill a grid where dp[i][j] is the LCS of the first i and first j characters. If the characters match, extend the diagonal; otherwise take the better of dropping one character from either string.
·
∅
A
C
∅
0
0
0
A
0
0
0
B
0
0
0
C
0
0
0
Step 1 of 8. LCS grid for "ABC" and "AC". Match → diagonal + 1, else the best of up/left.
1/8
Optimal
timeO(n·m)spaceO(n·m)
Match extends the diagonal.
1for (let i = 1; i <= n; i++)2 for (let j = 1; j <= m; j++)3 dp[i][j] = a[i-1] === b[j-1]4 ? dp[i-1][j-1] + 15 : Math.max(dp[i-1][j], dp[i][j-1]);Input
- grid
- 5 × 4
Memory
- at
- —
- cells marked
- 0
Output
- LCS
- —
Check yourself
3 quick questions about this walkthrough. A wrong answer costs nothing.
Examples
Example 1
- Input:
- a = "abcde", b = "ace"
- Output:
- 3
- Explanation:
- "ace" appears in order in both → length 3.
Example 2
- Input:
- a = "abc", b = "abc"
- Output:
- 3
- Explanation:
- Identical strings share everything → 3.
Example 3
- Input:
- a = "abc", b = "def"
- Output:
- 0
- Explanation:
- No common letters → 0.
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