One Edit Distance - Amazon Top Interview Questions


Problem Statement :


Given two strings s0 and s1 determine whether they are one or zero edit distance away. An edit can be described as deleting a character, adding a character, or replacing a character with another character.

Constraints

n ≤ 100,000 where n is the length of s0.
m ≤ 100,000 where m is the length of s1.

Example 1

Input

s0 = "quicksort"

s1 = "quicksort"

Output

True

Explanation

This has the edit distance of 0, since they are the same string.

Example 2

Input

s0 = "mergesort"

s1 = "mergesorts"

Output

True

Explanation

This has the edit distance 1, since s was added to the second string.
Example 3

Input

s0 = "mergeport"

s1 = "mergesorts"

Output

False

Explanation

This has edit distance of 2.


Solution :



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                        Solution in C++ :

bool equal(string& s0, int i, string& s1, int j) {
    while (i < s0.size() && j < s1.size() && s0[i] == s1[j]) {
        i++, j++;
    }
    return i == s0.size() && j == s1.size();
}

bool solve(string s0, string s1) {
    int m = s0.size(), n = s1.size();
    if (m > 1 + n || n > 1 + m) return false;
    for (int i = 0; i < min(m, n); i++) {
        if (s0[i] != s1[i])
            return equal(s0, i + 1, s1, i) || equal(s0, i, s1, i + 1) ||
                   equal(s0, i + 1, s1, i + 1);
    }
    return true;
}
                    


                        Solution in Java :

class Solution {
    public boolean solve(String s0, String s1) {
        int m = s0.length(), n = s1.length();
        if (Math.abs(m - n) > 1) {
            return false;
        }

        int i = 0, j = 0, edit = 0;
        while (i < m && j < n) {
            if (s0.charAt(i) != s1.charAt(j)) {
                edit += 1;
                if (m < n) {
                    i--;
                } else if (m > n) {
                    j--;
                }
            }
            i++;
            j++;
        }
        return (edit < 2);
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, s0, s1):
        # Get dimensions
        n = len(s0)
        m = len(s1)

        # Ensure n < m
        if m < n:
            return self.solve(s1, s0)

        # Check if length diff is > 1
        if abs(n - m) > 1:
            return False

        # index i -> s0, index j -> s1
        i = 0
        j = 0
        edits = 0
        while i < n:
            # chars match, increment i, j :D
            if s0[i] == s1[j]:
                i += 1
                j += 1
            # mismatch, count edit >:(
            else:
                edits += 1

                # same length, replace char in s0, move i, j
                if n == m:
                    i += 1
                    j += 1
                # not same length, delete char from s1 by moving j
                else:
                    j += 1

            # too many edits, terminate
            if edits > 1:
                break

        # 1 edit or less?
        return edits <= 1


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