LeetCode 161: One Edit Distance Solution

Master LeetCode problem 161 (One Edit Distance), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

161. One Edit Distance

Medium

Two Pointers String

Problem Explanation

Explanation:

The problem asks us to determine if two given strings are one edit distance apart. An edit distance is defined as inserting a character, deleting a character, or replacing a character in one of the strings to make them equal.

To solve this problem, we can iterate through both strings simultaneously and compare the characters. If we find a difference, we need to check if it can be resolved within one edit operation. We can handle this by considering three cases:

If the lengths of the strings differ by more than 1, they can't be one edit distance apart.
If the lengths are equal, we can allow only one character to be different.
If the lengths differ by 1, we can allow one string to have an additional character.

Time complexity: O(n) Space complexity: O(1)

Solution Code

class Solution {
    public boolean isOneEditDistance(String s, String t) {
        int m = s.length();
        int n = t.length();

        if (Math.abs(m - n) > 1) {
            return false;
        }

        for (int i = 0; i < Math.min(m, n); i++) {
            if (s.charAt(i) != t.charAt(i)) {
                if (m == n) {
                    return s.substring(i + 1).equals(t.substring(i + 1));
                } else if (m < n) {
                    return s.substring(i).equals(t.substring(i + 1));
                } else {
                    return s.substring(i + 1).equals(t.substring(i));
                }
            }
        }

        return Math.abs(m - n) == 1;
    }
}

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