Formatted question description: https://leetcode.ca/all/1071.html

1071. Greatest Common Divisor of Strings (Easy)

For two strings s and t, we say "t divides s" if and only if s = t + ... + t  (t concatenated with itself 1 or more times)

Given two strings str1 and str2, return the largest string x such that x divides both str1 and str2.

 

Example 1:

Input: str1 = "ABCABC", str2 = "ABC"
Output: "ABC"

Example 2:

Input: str1 = "ABABAB", str2 = "ABAB"
Output: "AB"

Example 3:

Input: str1 = "LEET", str2 = "CODE"
Output: ""

Example 4:

Input: str1 = "ABCDEF", str2 = "ABC"
Output: ""

 

Constraints:

  • 1 <= str1.length <= 1000
  • 1 <= str2.length <= 1000
  • str1 and str2 consist of English uppercase letters.

Related Topics:
String

Solution 1. Brute Force

// OJ: https://leetcode.com/problems/greatest-common-divisor-of-strings/
// Time: O(S^2)
// Space: O(S)
class Solution {
private:
    bool divisible(string a, string b) {
        int i = 0, j = 0, M = a.size(), N = b.size();
        for (; i < M; ++i) {
            if (a[i] != b[j]) return false;
            if (++j == N) j = 0;
        }
        return true;
    }
public:
    string gcdOfStrings(string str1, string str2) {
        string d = str1.size() < str2.size() ? str1 : str2;
        for (; d.size(); d.pop_back()) {
            if (str1.size() % d.size() || str2.size() % d.size()) continue;
            if (divisible(str1, d) && divisible(str2, d)) return d;
        }
        return d;
    }
};

Solution 2. GCD

// OJ: https://leetcode.com/problems/greatest-common-divisor-of-strings/
// Time: O(SH) where S is string length and H is depth of recursion
// Space: O(SH)
class Solution {
public:
    string gcdOfStrings(string str1, string str2) {
        if (str1.size() < str2.size()) return gcdOfStrings(str2, str1);
        if (str2.empty()) return str1;
        auto d = str1.substr(0, str2.size());
        return d == str2 ? gcdOfStrings(str1.substr(str2.size()), str2) : "";
    }
};

Java

  • class Solution {
        public String gcdOfStrings(String str1, String str2) {
            if (str1 == null || str2 == null)
                return null;
            while (str1.length() > 0 && str2.length() > 0) {
                if (str1.length() > str2.length()) {
                    String temp = str1;
                    str1 = str2;
                    str2 = temp;
                }
                if (str2.indexOf(str1) < 0)
                    return "";
                str2 = str2.substring(str1.length());
            }
            return str1.length() == 0 ? str2 : str1;
        }
    }
    
  • // OJ: https://leetcode.com/problems/greatest-common-divisor-of-strings/
    // Time: O(S^2)
    // Space: O(S)
    class Solution {
    private:
        bool divisible(string a, string b) {
            int i = 0, j = 0, M = a.size(), N = b.size();
            for (; i < M; ++i) {
                if (a[i] != b[j]) return false;
                if (++j == N) j = 0;
            }
            return true;
        }
    public:
        string gcdOfStrings(string str1, string str2) {
            string d = str1.size() < str2.size() ? str1 : str2;
            for (; d.size(); d.pop_back()) {
                if (str1.size() % d.size() || str2.size() % d.size()) continue;
                if (divisible(str1, d) && divisible(str2, d)) return d;
            }
            return d;
        }
    };
    
  • class Solution(object):
        def gcdOfStrings(self, str1, str2):
            """
            :type str1: str
            :type str2: str
            :rtype: str
            """
            l1, l2 = len(str1), len(str2)
            shorter = min(l1, l2)
            res = ""
            for i in range(shorter, 0, -1):
                if l1 % i or l2 % i:
                    continue
                t1, t2 = l1 // i, l2 // i
                gcd = str1[:i]
                rebuild1 = gcd * t1
                rebuild2 = gcd * t2
                if rebuild1 == str1 and rebuild2 == str2:
                    res = gcd
                    break
            return res
    

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