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Posts tagged as “Palindrome”

花花酱 LeetCode 131. Palindrome Partitioning

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.


Input: "aab"

Solution1: DP

dp[i] := ans of str[0:i]
dp[j] = { x + str[i:len] for x in dp[i] }, 0 <= i < len

Time complexity: O(2^n)
Space complexity: O(2^n)


Solution 2: DFS

Time complexity: O(2^n)
Space complexity: O(n)


花花酱 LeetCode 1177. Can Make Palindrome from Substring

Given a string s, we make queries on substrings of s.

For each query queries[i] = [left, right, k], we may rearrange the substring s[left], ..., s[right], and then choose up to k of them to replace with any lowercase English letter. 

If the substring is possible to be a palindrome string after the operations above, the result of the query is true. Otherwise, the result is false.

Return an array answer[], where answer[i] is the result of the i-th query queries[i].

Note that: Each letter is counted individually for replacement so if for example s[left..right] = "aaa", and k = 2, we can only replace two of the letters.  (Also, note that the initial string s is never modified by any query.)

Example :

Input: s = "abcda", queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]]
Output: [true,false,false,true,true]
queries[0] : substring = "d", is palidrome.
queries[1] : substring = "bc", is not palidrome.
queries[2] : substring = "abcd", is not palidrome after replacing only 1 character.
queries[3] : substring = "abcd", could be changed to "abba" which is palidrome. Also this can be changed to "baab" first rearrange it "bacd" then replace "cd" with "ab".
queries[4] : substring = "abcda", could be changed to "abcba" which is palidrome.


  • 1 <= s.length, queries.length <= 10^5
  • 0 <= queries[i][0] <= queries[i][1] < s.length
  • 0 <= queries[i][2] <= s.length
  • s only contains lowercase English letters.

Solution: Prefix frequency

Compute the prefix frequency of each characters, then we can efficiently compute the frequency of each characters in the substring in O(1) time. Count the number odd frequency characters o, we can convert it to a palindrome if o / 2 <= k.

Time complexity:
preprocessing: O(n)
Query: O(1)
Space complexity: O(n)


花花酱 LeetCode 1147. Longest Chunked Palindrome Decomposition

Return the largest possible k such that there exists a_1, a_2, ..., a_k such that:

  • Each a_i is a non-empty string;
  • Their concatenation a_1 + a_2 + ... + a_k is equal to text;
  • For all 1 <= i <= k,  a_i = a_{k+1 - i}.

Example 1:

Input: text = "ghiabcdefhelloadamhelloabcdefghi"
Output: 7
Explanation: We can split the string on "(ghi)(abcdef)(hello)(adam)(hello)(abcdef)(ghi)".

Example 2:

Input: text = "merchant"
Output: 1
Explanation: We can split the string on "(merchant)".

Example 3:

Input: text = "antaprezatepzapreanta"
Output: 11
Explanation: We can split the string on "(a)(nt)(a)(pre)(za)(tpe)(za)(pre)(a)(nt)(a)".

Example 4:

Input: text = "aaa"
Output: 3
Explanation: We can split the string on "(a)(a)(a)".

Solution: Greedy

Break the string when the shortest palindrome is found.
prefer to use string_view

Time complexity: O(n^2)
Space complexity: O(n)


花花酱 LeetCode 9. Palindrome Number


Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.

Example 1:

Input: 121
Output: true

Example 2:

Input: -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Example 3:

Input: 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

Follow up:

Could you solve it without converting the integer to a string?

Solution 1: Convert to string (cheating)

Time complexity: O(log10(x))

Space complexity: O(log10(x))


Solution 2: Digit by Digit

Every time we compare the first and last digits of x, if they are not the same, return false. Otherwise, remove first and last digit and continue this process.

How can we achieve that via int math?

e.g. x = 9999, t = pow((10, int)log10(x)) = 1000

first digit: x / t, last digit: x % 10

then x = (x – x / t * t) / 10 removes first and last digits.

t /= 100 since we removed two digits.

x / t = 9 = 9 = x % 10, 9999 => 99

9 = 9, 99 => “”

Time complexity: O(log10(x) / 2)

Space complexity: O(1)


花花酱 LeetCode 5. Longest Palindromic Substring


Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example 1:

Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.

Example 2:

Input: "cbbd"
Output: "bb"

Solution: Greedy

Try all possible i and find the longest palindromic string whose center is i (odd case) and i / i + 1 (even case).

Time complexity: O(n^2)

Space complexity: O(1)