LeetCode-One Edit Distance

Given two strings S and T, determine if they are both one edit distance apart.

Solution 1:

public class Solution {
    public boolean isOneEditDistance(String s, String t) {
        if ((s.isEmpty() && t.isEmpty()) || Math.abs(s.length() - t.length()) > 1)
            return false;

        if (s.length() == t.length()) {
            return isOneReplace(s, t);
        }

        if (s.length() > t.length()) {
            return isOneInsert(t, s);
        } else {
            return isOneInsert(s, t);
        }
    }

    public boolean isOneReplace(String s, String t) {
        boolean foundDiff = false;
        for (int i = 0; i < s.length(); i++)
            if (s.charAt(i) != t.charAt(i)) {
                if (foundDiff)
                    return false;
                foundDiff = true;
            }
        return foundDiff;

    }

    public boolean isOneInsert(String small, String large) {
        boolean hasSkip = false;
        int p1 = 0, p2 = 0;
        while (p1 < small.length() && p2 < large.length()) {
            if (small.charAt(p1) != large.charAt(p2)) {
                if (hasSkip)
                    return false;
                hasSkip = true;
                p2++;
            } else {
                p1++;
                p2++;
            }
        }
        return true;
    }
}

NOTE: this solution is too much for this question. We can simply fetch substring and compare.

Solution 2:

public 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;
        }

        int len = Math.min(m, n);
        for (int i = 0; i < len; i++) {
            if (s.charAt(i) != t.charAt(i)) {
                boolean result = s.substring(i).equals(t.substring(i + 1));
                result = result || s.substring(i + 1).equals(t.substring(i));
                result = result || s.substring(i + 1).equals(t.substring(i + 1));
                return result;
            }
        }
        return Math.abs(m - n) == 1;
    }
}