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Formatted question description: https://leetcode.ca/all/1745.html

# 1745. Palindrome Partitioning IV

Hard

## Description

Given a string s, return true if it is possible to split the string s into three non-empty palindromic substrings. Otherwise, return false.

A string is said to be palindrome if it the same string when reversed.

Example 1:

Input: s = “abcbdd”

Output: true

Explanation: “abcbdd” = “a” + “bcb” + “dd”, and all three substrings are palindromes.

Example 2:

Input: s = “bcbddxy”

Output: false

Explanation: s cannot be split into 3 palindromes.

Constraints:

• 3 <= s.length <= 2000
• s consists only of lowercase English letters.

## Solution

First calculate whether each substring of s is a palindrome. Use a 2D array isPalindrome to store the results so that the results can be obtained efficiently.

Next, loop over i from 0 to s.length() - 3 and loop over j from i + 1 to s.length() - 2. If the substring from index 0 to index i is a palindrome, then check whether the substring from index i + 1 to index j and the substring from index j + 1 to index s.length() - 1 are palindromes. If such i and j exist, return true. Otherwise, return false.

• class Solution {
public boolean checkPartitioning(String s) {
int length = s.length();
boolean[][] isPalindrome = new boolean[length][length];
for (int i = 0; i < length; i++)
isPalindrome[i][i] = true;
for (int i = 1; i < length; i++)
isPalindrome[i - 1][i] = s.charAt(i - 1) == s.charAt(i);
for (int i = length - 3; i >= 0; i--) {
for (int j = i + 2; j < length; j++)
isPalindrome[i][j] = isPalindrome[i + 1][j - 1] && s.charAt(i) == s.charAt(j);
}
int maxFirst = length - 3, maxSecond = length - 2;
for (int i = 0; i <= maxFirst; i++) {
if (isPalindrome[0][i]) {
for (int j = i + 1; j <= maxSecond; j++) {
if (isPalindrome[i + 1][j] && isPalindrome[j + 1][length - 1])
return true;
}
}
}
return false;

/* also working, but a little lower efficiency

for (int i = 0; i <= maxFirst; i++) {
for (int j = i + 1; j <= maxSecond; j++) {
if (isPalindrome[0][i] && isPalindrome[i + 1][j] && isPalindrome[j + 1][length - 1])
return true;
}
}
*/

}
}

############

class Solution {
public boolean checkPartitioning(String s) {
int n = s.length();
boolean[][] g = new boolean[n][n];
for (var e : g) {
Arrays.fill(e, true);
}
for (int i = n - 1; i >= 0; --i) {
for (int j = i + 1; j < n; ++j) {
g[i][j] = s.charAt(i) == s.charAt(j) && (i + 1 == j || g[i + 1][j - 1]);
}
}
for (int i = 0; i < n - 2; ++i) {
for (int j = i + 1; j < n - 1; ++j) {
if (g[0][i] && g[i + 1][j] && g[j + 1][n - 1]) {
return true;
}
}
}
return false;
}
}

• // OJ: https://leetcode.com/problems/palindrome-partitioning-iv/
// Time: O(N^2)
// Space: O(N^2)
class Solution {
public:
bool checkPartitioning(string s) {
unsigned N = s.size(), h[2001][2001] = {}, rh[2001][2001] = {}, d = 16777619;
for (int i = 0; i < N; ++i) {
int hash = 0;
for (int j = i; j < N; ++j) {
hash = hash * d + s[j] - 'a';
h[i][j] = hash;
}
}
reverse(begin(s), end(s));
for (int i = 0; i < N; ++i) {
int hash = 0;
for (int j = i; j < N; ++j) {
hash = hash * d + s[j] - 'a';
rh[N - j - 1][N - i - 1] = hash;
}
}
for (int i = 0; i < N; ++i) { // first part [0,i]
if (h[0][i] != rh[0][i]) continue;
for (int j = i + 1; j < N - 1; ++j) { // second part [i+1,j], last part [j+1,N-1]
if (h[i + 1][j] == rh[i + 1][j] && h[j + 1][N - 1] == rh[j + 1][N - 1]) return true;
}
}
return false;
}
};

• # 1745. Palindrome Partitioning IV
# https://leetcode.com/problems/palindrome-partitioning-iv/

class Solution():
def checkPartitioning(self, s):
n = len(s)

dp = [[False] * n for _ in range(n)]
for i in range(n-1, -1, -1):
for j in range(i, n):
if s[i] == s[j]:
# j-i<=2:
#   case-1: j-i==0, single char itself
#   case-12: j-i==1, e.g. 'aa'
#   case-1: j-i==j, e.g. 'aba'
dp[i][j] = (j - i <= 2) or dp[i+1][j-1]

return any(dp[0][i-1] and dp[i][j-1] and dp[j][n-1] for i in range(1, n-1) for j in range(i+1, n))

# below is 2nd solution
from typing import List

class Solution:
def checkPartitioning(self, s: str) -> bool:
isPalindrome = [ [False] * len(s) for i in range(len(s))]
for i in range(len(s)):
isPalindrome[i][i] = True
for i in range(1, len(s)):
isPalindrome[i - 1][i] = s[i-1] == s[i]
for i in range(len(s) - 3, -1, -1):
for j in range(i + 2, len(s), 1):
isPalindrome[i][j] = (s[i] == s[j] and isPalindrome[i + 1][j - 1])

for i in range(len(s) - 2):
for j in range(len(s) - 1):
if isPalindrome[0][i] and isPalindrome[i + 1][j] and isPalindrome[j + 1][len(s) - 1]:
return True

return False

if __name__ == "__main__":
print(Solution().checkPartitioning("abcbdd"))


• func checkPartitioning(s string) bool {
n := len(s)
g := make([][]bool, n)
for i := range g {
g[i] = make([]bool, n)
for j := range g[i] {
g[i][j] = true
}
}
for i := n - 1; i >= 0; i-- {
for j := i + 1; j < n; j++ {
g[i][j] = s[i] == s[j] && (i+1 == j || g[i+1][j-1])
}
}
for i := 0; i < n-2; i++ {
for j := i + 1; j < n-1; j++ {
if g[0][i] && g[i+1][j] && g[j+1][n-1] {
return true
}
}
}
return false
}