leetcode

Author: ayoubzulfiqar

GreatestCommonDivisorOfStrings/greatest_common_divisor_of_strings.dart

@@ -0,0 +1,93 @@
+/*
+
+
+
+-* 1071. Greatest Common Divisor of Strings *-
+
+
+For two strings s and t, we say "t divides s" if and only if s = t + ... + t (i.e., t is concatenated with itself one 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: ""
+ 
+
+Constraints:
+
+1 <= str1.length, str2.length <= 1000
+str1 and str2 consist of English uppercase letters.
+
+*/
+class A {
+  String gcdOfStrings(String str1, String str2) {
+    return (str1 + str2 == str2 + str1)
+        ? str1.substring(0, str1.length.gcd(str2.length))
+        : "";
+  }
+}
+
+class B {
+  bool cal(String str, String ans) {
+    int n = str.length;
+    int m = ans.length;
+    int idx = 0;
+    for (int i = 0; i < n; i++) {
+      if (str[i] != ans[idx]) return false;
+      idx++;
+      idx %= m;
+    }
+    return true;
+  }
+
+  String gcdOfStrings(String str1, String str2) {
+    int n = str1.length;
+    int m = str2.length;
+    int ss = n.gcd(m);
+    String ans = str1.substring(0, ss);
+    if (cal(str1, ans) && cal(str2, ans)) return ans;
+    return "";
+  }
+}
+
+class C {
+  String gcdOfStrings(String str1, String str2) {
+    String ans = "";
+    for (int i = 0; i < str2.length; i++) {
+      String cur = str2.substring(0, i + 1);
+      List<String> curArray = str2.split(cur);
+      bool flag = true;
+      for (int j = 0; j < curArray.length; j++) {
+        if (curArray[j] != "") {
+          flag = false;
+          break;
+        }
+      }
+      if (flag) {
+        List<String> cur2Array = str1.split(cur);
+        for (int j = 0; j < cur2Array.length; j++) {
+          if (cur2Array[j] != "") {
+            flag = false;
+            break;
+          }
+        }
+        if (flag) {
+          ans = cur;
+        }
+      }
+    }
+    return ans;
+  }
+}

GreatestCommonDivisorOfStrings/greatest_common_divisor_of_strings.go

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+package main
+
+func gcdOfStrings(str1 string, str2 string) string {
+	if str1 == str2 {
+		return str1
+	}
+
+	if len(str2) > len(str1) {
+		str1, str2 = str2, str1
+	}
+
+	if str1[:len(str2)] != str2 {
+		return ""
+	}
+
+	return gcdOfStrings(str1[len(str2):], str2)
+}

GreatestCommonDivisorOfStrings/greatest_common_divisor_of_strings.md

@@ -0,0 +1,83 @@
+# 🔥 3 Approaches 🔥 || Simple Fast and Easy || with Explanation
+
+### Approach
+
+str1+str2 == str2+str1 if and only if str1 and str2 have a gcd .
+E.g. str1=abcabc, str2=abc, then str1+str2 = abcabcabc = str2+str1
+This(str1+str2==str2+str1) is a requirement for the strings to have a gcd. If one of them is NOT a common part then gcd is "".It means we will return empty string
+Proof:-str1 = mGCD, str2 = nGCD, str1 + str2 = (m + n)GCD = str2 + str1
+Both the strings are made of same substring added multiple times.
+Since they are multiples, next step is simply to find the gcd of the lengths of 2 strings e.g. gcd(6,3) = 3, (we can use gcd function to find that)and get the substring of length 3 from either str1 or str2.
+
+### Code - 1
+
+```dart
+class Solution {
+  String gcdOfStrings(String str1, String str2) {
+    return (str1 + str2 == str2 + str1)
+        ? str1.substring(0, str1.length.gcd(str2.length))
+        : "";
+  }
+}
+```
+
+## Code - 2
+
+```dart
+class Solution {
+  bool cal(String str, String ans) {
+    int n = str.length;
+    int m = ans.length;
+    int idx = 0;
+    for (int i = 0; i < n; i++) {
+      if (str[i] != ans[idx]) return false;
+      idx++;
+      idx %= m;
+    }
+    return true;
+  }
+
+  String gcdOfStrings(String str1, String str2) {
+    int n = str1.length;
+    int m = str2.length;
+    int ss = n.gcd(m);
+    String ans = str1.substring(0, ss);
+    if (cal(str1, ans) && cal(str2, ans)) return ans;
+    return "";
+  }
+}
+```
+
+## Code - 3 Brute Force
+
+```dart
+class Solution {
+  String gcdOfStrings(String str1, String str2) {
+    String ans = "";
+    for (int i = 0; i < str2.length; i++) {
+      String cur = str2.substring(0, i + 1);
+      List<String> curArray = str2.split(cur);
+      bool flag = true;
+      for (int j = 0; j < curArray.length; j++) {
+        if (curArray[j] != "") {
+          flag = false;
+          break;
+        }
+      }
+      if (flag) {
+        List<String> cur2Array = str1.split(cur);
+        for (int j = 0; j < cur2Array.length; j++) {
+          if (cur2Array[j] != "") {
+            flag = false;
+            break;
+          }
+        }
+        if (flag) {
+          ans = cur;
+        }
+      }
+    }
+    return ans;
+  }
+}
+```

README.md

@@ -200,6 +200,7 @@ This repo contain leetcode solution using DART and GO programming language. Most
 - [**460.** LFU Cache](LFUCache/lfu_cache.dart)
 - [**1137.** N-th Tribonacci Number](N-thTribonacciNumber/n_th_tribonacci_number.dart)
 - [**1626.** Best Team With No Conflicts](BestTeamWithNoConflicts/best_team_with_no_conflicts.dart)
+- [**1071.** Greatest Common Divisor of Strings](GreatestCommonDivisorOfStrings/greatest_common_divisor_of_strings.dart)
 
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