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: 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 $m \neq n$, 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 $O(1)$.

Python3JavaC++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

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;
    }
}

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;
    }
};

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
}

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;
}

161. One Edit Distance 🔒

Description

Solutions

Solution 1: Discuss Different Cases

Comments