161. One Edit Distance

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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:

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

Example 2:

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

Constraints:

  • 0 <= s.length, t.length <= 10^4
  • s and t consist of lowercase letters, uppercase letters, and digits.

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    bool isOneEditDistance(string s, string t) {
        int idx1 = 0, idx2 = 0;
        int n1 = s.size(), n2 = t.size();
        if (abs(n1 - n2) > 1) {
            return false;
        }
        if (n1 > n2) return isOneEditDistance(t, s);
        // so t is always the one shorter
        for (int i = 0; i < n1; i++) {
            if (s[i] != t[i]) {
                if (n1 != n2) {
                    return s.substr(i) == t.substr(i + 1);
                } else {
                    return s.substr(i + 1) == t.substr(i + 1);
                }
            }
        }
        // if we reach here, it means we didn't find unmatched character
        // if n1 == n2, then they are the same
        return n1 != n2;
    }
};