LeetCode 0161. One Edit Distance Solution in Java, Python, C++, JavaScript, Go & Rust | Explanation + Code

Description

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.
• Delete exactly one character from s to get t.
• Replace exactly one character of s with a different character to get t.

 

Example 1:

Input: s = "ab", t = "acb"
Output: true
Explanation: We can insert 'c' into s to get t.

Example 2:

Input: s = "", t = ""
Output: false
Explanation: We cannot get t from s by only one step.

 

Constraints:

• 0 <= s.length, t.length <= 104
• s and t consist of lowercase letters, uppercase letters, and digits.

Solutions

Solution 1: Discuss Different Cases

Let m represent the length of string s, and n represent the length of string t. We can assume that m is always greater than or equal to n.

If m-n > 1, return false directly;

Otherwise, iterate through s and t, if s[i] is not equal to t[i]:

• If m ≠ n, compare s[i+1:] with t[i:], return true if they are equal, otherwise return false;
• If m = n, compare s[i:] with t[i:], return true if they are equal, otherwise return false.

If the iteration ends, it means that all the characters of s and t that have been iterated are equal, at this time it needs to satisfy m=n+1.

The time complexity is O(m), where m is the length of string s. The space complexity is O(1).

PythonJavaC++GoTypeScript

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(code-box)

class Solution { public boolean isOneEditDistance(String s, String t) { int m = s.length(), n = t.length(); if (m < n) { return isOneEditDistance(t, s); } if (m - n > 1) { return false; } for (int i = 0; i < n; ++i) { if (s.charAt(i) != t.charAt(i)) { if (m == n) { return s.substring(i + 1).equals(t.substring(i + 1)); } return s.substring(i + 1).equals(t.substring(i)); } } return m == n + 1; } }(code-box)

class Solution { public: bool isOneEditDistance(string s, string t) { int m = s.size(), n = t.size(); if (m < n) return isOneEditDistance(t, s); if (m - n > 1) return false; for (int i = 0; i < n; ++i) { if (s[i] != t[i]) { if (m == n) return s.substr(i + 1) == t.substr(i + 1); return s.substr(i + 1) == t.substr(i); } } return m == n + 1; } };(code-box)

func isOneEditDistance(s string, t string) bool { m, n := len(s), len(t) if m < n { return isOneEditDistance(t, s) } if m-n > 1 { return false } for i := range t { if s[i] != t[i] { if m == n { return s[i+1:] == t[i+1:] } return s[i+1:] == t[i:] } } return m == n+1 }(code-box)

function isOneEditDistance(s: string, t: string): boolean { const [m, n] = [s.length, t.length]; if (m < n) { return isOneEditDistance(t, s); } if (m - n > 1) { return false; } for (let i = 0; i < n; ++i) { if (s[i] !== t[i]) { return s.slice(i + 1) === t.slice(i + (m === n ? 1 : 0)); } } return m === n + 1; }(code-box)