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BinarySearch - One Edit Distance


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.


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


Find the first index i that s0[i] is not equal to s1[i]

Based on the length of s0 and s1, compare the rest of the sub string are same or not.

bool solve(string s0, string s1) {
    int m = s0.size(), n = s1.size();
    for (int i = 0; i < min(m, n); i++) {
        if (s0[i] != s1[i]) {
            if (m == n)
                return s0.substr(i + 1) == s1.substr(i + 1);
            else if (m < n)
                return s0.substr(i) == s1.substr(i + 1);
                return s0.substr(i + 1) == s1.substr(i);
    return abs(m - n) <= 1;
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