Check if a binary string has two consecutive occurrences of one everywhere
Given string str consisting of only the characters ‘a’ and ‘b’, the task is to check whether the string is valid or not. In a valid string, every group of consecutive b must be of length 2 and must appear after 1 or more occurrences of character ‘a’ i.e. “abba” is a valid sub-string but “abbb” and aba are not. Print 1 if the string is valid, else print -1.
Examples:
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Input: str = “abbaaabbabba”
Output: 1Input: str = “abbaaababb”
Output: -1
Approach: Find every occurrence of ‘b’ in the string and check whether it is a part of the sub-string “abb”. If the condition fails for any sub-string, then print -1 else print 1.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that returns 1 if str is validbool isValidString(string str, int n){ // Index of first appearance of 'b' int index = find(str.begin(), str.end(), 'b') - str.begin(); // If str starts with 'b' if (index == 0) return false; // While 'b' occurs in str while (index <= n - 1) { // If 'b' doesn't appear after an 'a' if (str[index - 1] != 'a') return false; // If 'b' is not succeeded by another 'b' if (index + 1 < n && str[index + 1] != 'b') return false; // If sub-string is of the type "abbb" if (index + 2 < n && str[index + 2] == 'b') return false; // If str ends with a single b if (index == n - 1) return false; index = find(str.begin() + index + 2, str.end(), 'b') - str.begin(); } return true;}// Driver codeint main(){ string str = "abbaaabbabba"; int n = str.length(); isValidString(str, n) ? cout << "true" : cout << "false"; return 0;}// This code is contributed by// sanjeev2552 |
Java
// Java implementation of the approachclass GFG { // Function that returns 1 if str is valid private static boolean isValidString(String str, int n) { // Index of first appearance of 'b' int index = str.indexOf("b"); // If str starts with 'b' if (index == 0) return false; // While 'b' occurs in str while (index != -1) { // If 'b' doesn't appear after an 'a' if (str.charAt(index - 1) != 'a') return false; // If 'b' is not succeeded by another 'b' if (index + 1 < n && str.charAt(index + 1) != 'b') return false; // If sub-string is of the type "abbb" if (index + 2 < n && str.charAt(index + 2) == 'b') return false; // If str ends with a single b if (index == n - 1) return false; index = str.indexOf("b", index + 2); } return true; } // Driver code public static void main(String[] args) { String str = "abbaaabbabba"; int n = str.length(); System.out.println(isValidString(str, n)); }} |
Python 3
# Python 3 implementation of the approach# Function that returns 1 if str is validdef isValidString(str, n): # Index of first appearance of 'b' idx = str.find("b") # If str starts with 'b' if (idx == 0): return False # While 'b' occurs in str while (idx != -1): # If 'b' doesn't appear after an 'a' if (str[idx - 1] != 'a'): return False # If 'b' is not succeeded by another 'b' if (idx + 1 < n and str[idx + 1] != 'b'): return False # If sub-string is of the type "abbb" if (idx + 2 < n and str[idx + 2] == 'b'): return False # If str ends with a single b if (idx == n - 1): return False idx = str.find("b", idx + 2) return True# Driver codeif __name__ == "__main__": str = "abbaaabbabba" n = len(str) print(isValidString(str, n))# This code is contributed by ita_c |
C#
// C# implementation of the approachusing System;class GFG{ // Function that returns 1 if str is valid private static bool isValidString(string str, int n) { // Index of first appearance of 'b' int index = str.IndexOf("b"); // If str starts with 'b' if (index == 0) return false; // While 'b' occurs in str while (index != -1) { // If 'b' doesn't appear after an 'a' if (str[index - 1] != 'a') return false; // If 'b' is not succeeded by another 'b' if (index + 1 < n && str[index + 1] != 'b') return false; // If sub-string is of the type "abbb" if (index + 2 < n && str[index + 2] == 'b') return false; // If str ends with a single b if (index == n - 1) return false; index = str.IndexOf("b", index + 2); } return true; } // Driver code public static void Main() { string str = "abbaaabbabba"; int n = str.Length; Console.WriteLine(isValidString(str, n)); }}// This code is contributed by Ryuga |
Javascript
<script>// Javascript implementation of the approach// Function that returns 1 if str is validfunction isValidString(str,n){ // Index of first appearance of 'b' let index = str.indexOf("b"); // If str starts with 'b' if (index == 0) return false; // While 'b' occurs in str while (index != -1) { // If 'b' doesn't appear after an 'a' if (str[index - 1] != 'a') return false; // If 'b' is not succeeded by another 'b' if (index + 1 < n && str[index + 1] != 'b') return false; // If sub-string is of the type "abbb" if (index + 2 < n && str[index + 2] == 'b') return false; // If str ends with a single b if (index == n - 1) return false; index = str.indexOf("b", index + 2); } return true;}// Driver codelet str = "abbaaabbabba";let n = str.length;document.write(isValidString(str, n));// This code is contributed by rag2127</script> |
true
