Shortest common supersequence
HardBoth strings, sharing the LCS once
Problem
Return the shortest string that has both a and b as subsequences.
Merge the two strings but write their common subsequence only once: len(a)+len(b)-LCS.
The idea
The shortest string containing both has length m + n - lcs, because the common subsequence need only appear once. Walking the LCS table backwards reconstructs the actual supersequence.
The trick
- Length = m + n - lcs.
- When characters match take one; otherwise take from whichever side the table came from.
·
∅
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 = "abac", b = "cab"
- Output:
- 5
- Explanation:
- Shortest string containing both has length 5.
Example 2
- Input:
- a = "abc", b = "abc"
- Output:
- 3
- Explanation:
- They're equal → length 3.
Example 3
- Input:
- a = "a", b = "b"
- Output:
- 2
- Explanation:
- Need both letters → length 2.
Practice this problem:LeetCode(opens in a new tab)Search GeeksforGeeks(opens in a new tab)
Finished the walkthrough? Add it to your streak.