161. One Edit Distance

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

Python Code
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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

Java Code
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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;
    }
}

C++ Code
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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;
    }
};

Go Code
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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
}