Minimum insertions to make string palindrome
HardKeep the longest palindromic core
Problem
Return the fewest characters to insert to make a string a palindrome.
Insertions needed = length minus the longest palindromic subsequence already inside.
The idea
The characters already forming the longest palindromic subsequence can stay; everything else needs a matching insertion. So the answer is n minus that subsequence's length.
The trick
- Answer = n - longestPalindromicSubsequence(s).
- Which itself is n - LCS(s, reverse(s)).
·
∅
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:
- s = "abcaa"
- Output:
- 2
- Explanation:
- Add 2 letters to mirror it.
Example 2
- Input:
- s = "aa"
- Output:
- 0
- Explanation:
- Already a palindrome → 0.
Example 3
- Input:
- s = "abc"
- Output:
- 2
- Explanation:
- Need 2 insertions.
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