Given strings S1, S2 and S, the task is to check if every substring of S which is same as S1 has another substring of S same as S2 before it. It is given that S1 is always present as a substring in the string S.
Examples:
Input: S1 = “code”, S2 = “geek”, S = “sxyneveropengcodetecode”
Output: True
Explanation: Substring S2 is present before both the occurrence of S1.
“sxygeeksgcodetecode“Input: S1 = “code”, S2 = “my”, “sxycodesforneveropenvhgh”
Output: False
Approach: The approach is to check which substring occurs first. If substring S2 occurs first return true. If S1 occurs first return false.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check if S2 is present// before all S1 in string Sbool chekfavstring(string& S, string& S1, string& S2){ bool cod = false; int n = S.size(); int n1 = S1.size(), n2 = S2.size(); for (int i = 0; i <= n - n2; i++) { string str; for (int k = i; k < i + n2; k++) { str.push_back(S[k]); } if (str == S2) { return true; } if (str == S1) { return false; } } return true;}// Driver codeint main(){ string S = "sxyneveropengcodetecode"; string S1 = "code", S2 = "geek"; chekfavstring(S, S1, S2) ? cout << "True" : cout << "False"; return 0;} |
Java
// Java program to implement// the above approachclass GFG { // Function to check if S2 is present // before all S1 in string S static boolean chekfavstring(String S, String S1, String S2) { int n = S.length(); int n1 = S1.length(), n2 = S2.length(); for (int i = 0; i <= n - n2; i++) { String str = ""; for (int k = i; k < i + n2; k++) { str += S.charAt(k); } if (str == S2) { return true; } if (str == S1) { return false; } } return true; } // Driver Code public static void main(String[] args) { String S = "sxyneveropengcodetecode"; String S1 = "code", S2 = "geek"; if (chekfavstring(S, S1, S2)) { System.out.print("True"); } else { System.out.print("False"); } }}// This code is contributed by ukasp. |
Python3
# python3 program for the above approach# Function to check if S2 is present# before all S1 in string Sdef chekfavstring(S, S1, S2): cod = False n = len(S) n1 = len(S1) n2 = len(S2) for i in range(0, n - n2 + 1): str = "" for k in range(i, i + n2): str += S[k] if (str == S2): return True if (str == S1): return False return True# Driver codeif __name__ == "__main__": S = "sxyneveropengcodetecode" S1 = "code" S2 = "geek" print("True") if chekfavstring(S, S1, S2) else print("False") # This code is contributed by rakeshsahni |
C#
// C# program to implement// the above approachusing System;class GFG{// Function to check if S2 is present// before all S1 in string Sstatic bool chekfavstring(string S, string S1, string S2){ int n = S.Length; int n1 = S1.Length, n2 = S2.Length; for (int i = 0; i <= n - n2; i++) { string str = ""; for (int k = i; k < i + n2; k++) { str += S[k]; } if (str == S2) { return true; } if (str == S1) { return false; } } return true;}// Driver Codepublic static void Main(){ string S = "sxyneveropengcodetecode"; string S1 = "code", S2 = "geek"; if(chekfavstring(S, S1, S2)) { Console.Write("True"); } else { Console.Write("False"); }}}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script> // JavaScript code for the above approach // Function to check if S2 is present // before all S1 in string S function chekfavstring(S, S1, S2) { let cod = false; let n = S.length; let n1 = S1.length, n2 = S2.length; for (let i = 0; i <= n - n2; i++) { let str = ""; for (let k = i; k < i + n2; k++) { str += (S[k]); } if (str == S2) { return true; } if (str == S1) { return false; } } return true; } // Driver code let S = "sxyneveropengcodetecode"; let S1 = "code", S2 = "geek"; chekfavstring(S, S1, S2) ? document.write("True") : document.write("False") // This code is contributed by Potta Lokesh </script> |
Output
True
Time Complexity: O(N * K)
Auxiliary Space: O(1)
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