Check If every group of a’s is followed by a group of b’s of same length

Given string str, the task is to check whether every group of consecutive a’s is followed by a group of consecutive b’s of the same length. If the condition is true for every group then print 1 else print 0.

Examples:

Input: str = “ababaabb”
Output:
ab, ab, aabb. All groups are valid

Input: str = “aabbabb”
Output:
aabb, abb (A single ‘a’ followed by 2 ‘b’)

Approach:

• For every a in the string increment the count.
• Starting from the first b, decrement the count for every b.
• If at the end of the above cycle, count != 0 then return false.
• Else repeat the first two steps for the rest of the string.
• Return true if the condition is satisfied for all the cycles else print 0.

Below is the implementation of the above approach:

C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function to match whether there are always n consecutive b's` `// followed by n consecutive a's throughout the string` `int` `matchPattern(string s)` `{` `    ``int` `count = 0;` `    ``int` `n = s.length();`   `    ``// Traverse through the string` `    ``int` `i = 0;` `    ``while` `(i < n) {`   `        ``// Count a's in current segment` `        ``while` `(i < n && s[i] == ``'a'``) {` `            ``count++;` `            ``i++;` `        ``}`   `        ``// Count b's in current segment` `        ``while` `(i < n && s[i] == ``'b'``) {` `            ``count--;` `            ``i++;` `        ``}`   `        ``// If both counts are not same.` `        ``if` `(count != 0)` `            ``return` `false``;` `    ``}`   `    ``return` `true``;` `}`   `// Driver code` `int` `main()` `{` `    ``string s = ``"bb"``;` `    ``if` `(matchPattern(s) == ``true``)` `        ``cout << ``"Yes"``;` `    ``else` `        ``cout << ``"No"``;` `    ``return` `0;` `}`

Java

 `// Java implementation of the above approach`   `public` `class` `GFG{`   `// Function to match whether there are always n consecutive b's` `// followed by n consecutive a's throughout the string` `static` `boolean` `matchPattern(String s)` `{` `    ``int` `count = ``0``;` `    ``int` `n = s.length();`   `    ``// Traverse through the string` `    ``int` `i = ``0``;` `    ``while` `(i < n) {`   `        ``// Count a's in current segment` `        ``while` `(i < n && s.charAt(i) == ``'a'``) {` `            ``count++;` `            ``i++;` `        ``}`   `        ``// Count b's in current segment` `        ``while` `(i < n && s.charAt(i) == ``'b'``) {` `            ``count--;` `            ``i++;` `        ``}`   `        ``// If both counts are not same.` `        ``if` `(count != ``0``)` `            ``return` `false``;` `    ``}`   `    ``return` `true``;` `}`   `// Driver code` `public` `static` `void` `main(String []args)` `{` `    ``String s = ``"bb"``;` `    ``if` `(matchPattern(s) == ``true``)` `        ``System.out.println(``"Yes"``);` `    ``else` `        ``System.out.println(``"No"``);` `}`   `// This code is contributed by Ryuga` `}`

Python3

 `# Python 3 implementation of the approach`   `# Function to match whether there are ` `# always n consecutive b's followed by ` `# n consecutive a's throughout the string` `def` `matchPattern(s):`   `    ``count ``=` `0``;` `    ``n ``=` `len``(s);`   `    ``# Traverse through the string` `    ``i ``=` `0``;` `    ``while` `(i < n) :`   `        ``# Count a's in current segment` `        ``while` `(i < n ``and` `s[i] ``=``=` `'a'``):`   `            ``count ``+``=` `1` `;` `            ``i ``=``+` `1``;`   `        ``# Count b's in current segment` `        ``while` `(i < n ``and` `s[i] ``=``=` `'b'``): ` `            ``count ``-``=` `1` `;` `            ``i ``+``=` `1``;` `    `  `        ``# If both counts are not same.` `        ``if` `(count !``=` `0``):` `            ``return` `False``;`   `    ``return` `True``;`   `# Driver code` `s ``=` `"bb"``;` `if` `(matchPattern(s) ``=``=` `True``):` `    ``print``(``"Yes"``);` `else``:` `    ``print``(``"No"``);`   `# This code is contributed ` `# by Akanksha Rai`

C#

 `// C# implementation of the above approach` ` `  `using` `System;` `public` `class` `GFG{` ` `  `// Function to match whether there are always n consecutive b's` `// followed by n consecutive a's throughout the string` `static` `bool` `matchPattern(``string` `s)` `{` `    ``int` `count = 0;` `    ``int` `n = s.Length;` ` `  `    ``// Traverse through the string` `    ``int` `i = 0;` `    ``while` `(i < n) {` ` `  `        ``// Count a's in current segment` `        ``while` `(i < n && s[i] == ``'a'``) {` `            ``count++;` `            ``i++;` `        ``}` ` `  `        ``// Count b's in current segment` `        ``while` `(i < n && s[i] == ``'b'``) {` `            ``count--;` `            ``i++;` `        ``}` ` `  `        ``// If both counts are not same.` `        ``if` `(count != 0)` `            ``return` `false``;` `    ``}` ` `  `    ``return` `true``;` `}` ` `  `// Driver code` `public` `static` `void` `Main()` `{` `    ``string` `s = ``"bb"``;` `    ``if` `(matchPattern(s) == ``true``)` `        ``Console.Write(``"Yes"``);` `    ``else` `        ``Console.Write(``"No"``);` `}`   `}`

PHP

 ``

Javascript

 ``

Output

`No`

Complexity Analysis:

• Time Complexity : O( | s | ) ,where | s | is length of given string s.
• Space Complexity : O(1) ,as we are not using any extra space

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