Minimum Swaps to Group All 1's Together
Given a string s
, return true
if the s
can be palindrome after deleting at most one character from it.
Example 1:
Input: s = "aba"
Output: true
Example 2:
Input: s = "abca"
Output: true
Explanation: You could delete the character 'c'.
Example 3:
Input: s = "abc"
Output: false
Constraints:
1 <= s.length <= 105
s
consists of lowercase English letters.
Solution
The best approach to solve this problem is using two pointers in opposite directions.
Similar to Valid Palindrome, we will initialize two pointers on the two ends of the string and work our way inside.
We will perform the same palindrome validation by matching the characters at the l
and r
pointers.
If their corresponding characters are different, then we know we must delete one character at one of the l
or r
position.
Then, we can check whether the unprocessed substring (excluding the deleted character) is a palindrome.
In Valid Palindrome, we had implemented is_palindrome
with two pointers, here, we are using an iterative approach for two pointers.
The left pointer is i
and the right pointer is slen - i
, where slen
is len(s) - 1
.
Implementation
1def validPalindrome(self, s: str) -> bool:
2 def isPalindrome(s: str):
3 slen = len(s)-1
4 for i in range(slen//2 + 1):
5 if s[i] != s[slen-i]: return False
6 return True
7
8 l, r = 0, len(s)-1
9 while l < r:
10 if s[l] != s[r]:
11 return isPalindrome(s[l+1:r+1]) or isPalindrome(s[l:r])
12 l += 1
13 r -= 1
14 return True
How does quick sort divide the problem into subproblems?
Solution Implementation
Breadth first search can be used to find the shortest path between two nodes in a directed graph.
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