One Edit Distance — LeetCode 161 Python Solution

The problem

Given two strings s and t, return true if they are both one edit distance apart, otherwise return false. A string s is said to be one distance apart from a string t if you can: Insert exactly one character into s to get t.

Example

Input

s = "ab", t = "acb"

Output

true

Explanation

We can insert 'c' into s to get t.

Python solution

class Solution:
    def isOneEditDistance(self, s: str, t: str) -> bool:
        if len(s) < len(t):
            return self.isOneEditDistance(t, s)
        m, n = len(s), len(t)
        if m - n > 1:
            return False
        for i, c in enumerate(t):
            if c != s[i]:
                return s[i + 1 :] == t[i + 1 :] if m == n else s[i + 1 :] == t[i:]
        return m == n + 1

Complexity

Pattern: Two Pointers

Use the order already in the input to discard half the search space at every step. LeetCode 161. One Edit Distance is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Two Pointers.

The two pointers guide has the Python template for the pattern and the 201 LeetCode problems that use it.

Related problems