Given three strings str, A and B. The task is to check whether str = A + B or str = B + A where + denotes concatenation.
Examples:
Input: str = “neveropen”, A = “Geeksfo”, B = “rGeeks”
Output: Yes
str = A + B = “Geeksfo” + “rGeeks” = “neveropen”
Input: str = “Delhicapitals”, B = “Delmi”, C = “capitals”
Output: No
Approach:
- If len(str) != len(A) + len(B) then it is not possible to generate str by concatenating a + b or b + a.
- Else check whether str starts with a and ends with b or it starts with b and ends with a. Print Yes if any of these is true else print No
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that return true// if pre is a prefix of strbool startsWith(string str, string pre){ int strLen = str.length(); int preLen = pre.length(); int i = 0, j = 0; // While there are characters to match while (i < strLen && j < preLen) { // If characters differ at any position if (str[i] != pre[j]) return false; i++; j++; } // str starts with pre return true;}// Function that return true// if suff is a suffix of strbool endsWith(string str, string suff){ int i = str.length() - 0; int j = suff.length() - 0; // While there are characters to match while (i >= 0 && j >= 0) { // If characters differ at any position if (str[i] != suff[j]) return false; i--; j--; } // str ends with suff return true;}// Function that returns true// if str = a + b or str = b + abool checkString(string str, string a, string b){ // str cannot be generated // by concatenating a and b if (str.length() != a.length() + b.length()) return false; // If str starts with a // i.e. a is a prefix of str if (startsWith(str, a)) { // Check if the rest of the characters // are equal to b i.e. b is a suffix of str if (endsWith(str, b)) return true; } // If str starts with b // i.e. b is a prefix of str if (startsWith(str, b)) { // Check if the rest of the characters // are equal to a i.e. a is a suffix of str if (endsWith(str, a)) return true; } return false;}// Driver codeint main(){ string str = "neveropen"; string a = "Geeksfo"; string b = "rGeeks"; if (checkString(str, a, b)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG{ // Function that return true// if pre is a prefix of strstatic boolean startsWith(String str, String pre){ int strLen = str.length(); int preLen = pre.length(); int i = 0, j = 0; // While there are characters to match while (i < strLen && j < preLen) { // If characters differ at any position if (str.charAt(i) != pre.charAt(j)) return false; i++; j++; } // str starts with pre return true;}// Function that return true// if suff is a suffix of strstatic boolean endsWith(String str, String suff){ int i = str.length() - 1; int j = suff.length() - 1; // While there are characters to match while (i >= 0 && j >= 0) { // If characters differ at any position if (str.charAt(i) != suff.charAt(j)) return false; i--; j--; } // str ends with suff return true;}// Function that returns true// if str = a + b or str = b + astatic boolean checkString(String str, String a, String b){ // str cannot be generated // by concatenating a and b if (str.length() != a.length() + b.length()) return false; // If str starts with a // i.e. a is a prefix of str if (startsWith(str, a)) { // Check if the rest of the characters // are equal to b i.e. b is a suffix of str if (endsWith(str, b)) return true; } // If str starts with b // i.e. b is a prefix of str if (startsWith(str, b)) { // Check if the rest of the characters // are equal to a i.e. a is a suffix of str if (endsWith(str, a)) return true; } return false;}// Driver codepublic static void main(String args[]){ String str = "neveropen"; String a = "Geeksfo"; String b = "rGeeks"; if (checkString(str, a, b)) System.out.println("Yes"); else System.out.println("No");}}// This code is contributed by Arnab Kundu |
Python3
# Python3 implementation of the approach # Function that return true # if pre is a prefix of str def startsWith(str, pre): strLen = len(str) preLen = len(pre) i = 0 j = 0 # While there are characters to match while (i < strLen and j < preLen): # If characters differ at any position if (str[i] != pre[j]) : return False i += 1 j += 1 # str starts with pre return True # Function that return true # if suff is a suffix of str def endsWith(str, suff): i = len(str) - 1 j = len(suff) - 1 # While there are characters to match while (i >= 0 and j >= 0): # If characters differ at any position if (str[i] != suff[j]): return False i -= 1 j -= 1 # str ends with suff return True# Function that returns true # if str = a + b or str = b + a def checkString(str, a, b): # str cannot be generated # by concatenating a and b if (len(str) != len(a) + len(b)): return False # If str starts with a # i.e. a is a prefix of str if (startsWith(str, a)): # Check if the rest of the characters # are equal to b i.e. b is a suffix of str if (endsWith(str, b)): return True # If str starts with b # i.e. b is a prefix of str if (startsWith(str, b)): # Check if the rest of the characters # are equal to a i.e. a is a suffix of str if (endsWith(str, a)): return True return False # Driver code str = "neveropen"a = "Geeksfo"b = "rGeeks"if (checkString(str, a, b)): print("Yes")else: print("No")# This code is contributed by SHUBHAMSINGH10 |
C#
// C# implementation of the approachusing System;class GFG{ // Function that return true// if pre is a prefix of strstatic Boolean startsWith(String str, String pre){ int strLen = str.Length; int preLen = pre.Length; int i = 0, j = 0; // While there are characters to match while (i < strLen && j < preLen) { // If characters differ at any position if (str[i] != pre[j]) return false; i++; j++; } // str starts with pre return true;}// Function that return true// if suff is a suffix of strstatic Boolean endsWith(String str, String suff){ int i = str.Length - 1; int j = suff.Length - 1; // While there are characters to match while (i >= 0 && j >= 0) { // If characters differ at any position if (str[i] != suff[j]) return false; i--; j--; } // str ends with suff return true;}// Function that returns true// if str = a + b or str = b + astatic Boolean checkString(String str, String a, String b){ // str cannot be generated // by concatenating a and b if (str.Length != a.Length + b.Length) return false; // If str starts with a // i.e. a is a prefix of str if (startsWith(str, a)) { // Check if the rest of the characters // are equal to b i.e. b is a suffix of str if (endsWith(str, b)) return true; } // If str starts with b // i.e. b is a prefix of str if (startsWith(str, b)) { // Check if the rest of the characters // are equal to a i.e. a is a suffix of str if (endsWith(str, a)) return true; } return false;}// Driver codepublic static void Main(String []args){ String str = "neveropen"; String a = "Geeksfo"; String b = "rGeeks"; if (checkString(str, a, b)) Console.WriteLine("Yes"); else Console.WriteLine("No");}}// This code is contributed by 29AjayKumar |
PHP
<?php// PHP implementation of the approach // Function that return true // if pre is a prefix of str function startsWith($str, $pre) { $strLen = strlen($str); $preLen = strlen($pre); $i = 0; $j = 0; // While there are characters to match while ($i < $strLen && $j < $preLen) { // If characters differ at any position if ($str[$i] != $pre[$j]) return false; $i++; $j++; } // str starts with pre return true; } // Function that return true // if suff is a suffix of str function endsWith($str, $suff) { $i = strlen($str)- 0; $j = strlen($suff)- 0; // While there are characters to match while ($i >= 0 && $j >= 0) { // If characters differ at any position if ($str[$i] != $suff[$j]) return false; $i--; $j--; } // str ends with suff return true; } // Function that returns true // if str = a + b or str = b + a function checkString($str, $a, $b) { // str cannot be generated // by concatenating a and b if (strlen($str) != strlen($a) + strlen($b)) return false; // If str starts with a // i.e. a is a prefix of str if (startsWith($str, $a)) { // Check if the rest of the characters // are equal to b i.e. b is a suffix of str if (endsWith($str, $b)) return true; } // If str starts with b // i.e. b is a prefix of str if (startsWith($str, $b)) { // Check if the rest of the characters // are equal to a i.e. a is a suffix of str if (endsWith($str, $a)) return true; } return false; } // Driver code $str = "neveropen"; $a = "Geeksfo"; $b = "rGeeks"; if (checkString($str, $a, $b)) echo "Yes"; else echo "No"; // This code is contributed by AnkitRai01?> |
Javascript
<script> // JavaScript implementation of the approach // Function that return true // if pre is a prefix of str function startsWith(str, pre) { let strLen = str.length; let preLen = pre.length; let i = 0, j = 0; // While there are characters to match while (i < strLen && j < preLen) { // If characters differ at any position if (str[i] != pre[j]) return false; i++; j++; } // str starts with pre return true; } // Function that return true // if suff is a suffix of str function endsWith(str, suff) { let i = str.length - 1; let j = suff.length - 1; // While there are characters to match while (i >= 0 && j >= 0) { // If characters differ at any position if (str[i] != suff[j]) return false; i--; j--; } // str ends with suff return true; } // Function that returns true // if str = a + b or str = b + a function checkString(str, a, b) { // str cannot be generated // by concatenating a and b if (str.length != a.length + b.length) return false; // If str starts with a // i.e. a is a prefix of str if (startsWith(str, a)) { // Check if the rest of the characters // are equal to b i.e. b is a suffix of str if (endsWith(str, b)) return true; } // If str starts with b // i.e. b is a prefix of str if (startsWith(str, b)) { // Check if the rest of the characters // are equal to a i.e. a is a suffix of str if (endsWith(str, a)) return true; } return false; } let str = "neveropen"; let a = "Geeksfo"; let b = "rGeeks"; if (checkString(str, a, b)) document.write("Yes"); else document.write("No"); </script> |
Output:
Yes
Time Complexity: O(max(a,b,c)), as we are using a loop to traverse a, b, and c times. Where a, b, and c are the length of the strings.
Auxiliary Space: O(1), as we are not using any extra space.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
