Check if string can be made lexicographically smaller by reversing any substring

Given a string S, the task is to check if we can make the string lexicographically smaller by reversing any sub-string of the given string.

Examples:

Input: S = “striver”
Output: Yes
Reverse “rive” to get “stevirr” which is lexicographically smaller.

Input: S = “rxz”
Output: No

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Iterate in the string and check if for any index s[i] > s[i + 1]. If there exists at least one such index, then it is possible else not.

Below is the implementation of the above approach:

C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function that returns true if s ` `// can be made lexicographically smaller ` `// by reversing a sub-string in s ` `bool` `check(string s) ` `{ ` `    ``int` `n = s.size(); ` ` `  `    ``// Traverse in the string ` `    ``for` `(``int` `i = 0; i < n - 1; i++) { ` ` `  `        ``// Check if s[i+1] < s[i] ` `        ``if` `(s[i] > s[i + 1]) ` `            ``return` `true``; ` `    ``} ` ` `  `    ``// Not possible ` `    ``return` `false``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string s = ``"geeksforgeeks"``; ` ` `  `    ``if` `(check(s)) ` `        ``cout << ``"Yes"``; ` `    ``else` `        ``cout << ``"No"``; ` ` `  `    ``return` `0; ` `} `

Java

 `// Java implementation of the approach  ` `class` `GFG ` `{ ` ` `  `// Function that returns true if s  ` `// can be made lexicographically smaller  ` `// by reversing a sub-string in s  ` `static` `boolean` `check(String s)  ` `{  ` `    ``int` `n = s.length();  ` ` `  `    ``// Traverse in the string  ` `    ``for` `(``int` `i = ``0``; i < n - ``1``; i++) ` `    ``{  ` ` `  `        ``// Check if s[i+1] < s[i]  ` `        ``if` `(s.charAt(i) > s.charAt(i + ``1``))  ` `            ``return` `true``;  ` `    ``}  ` ` `  `    ``// Not possible  ` `    ``return` `false``;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String args[]) ` `{  ` `    ``String s = ``"geeksforgeeks"``;  ` ` `  `    ``if` `(check(s))  ` `        ``System.out.println(``"Yes"``);  ` `    ``else` `        ``System.out.println(``"No"``);  ` ` `  `}  ` `} ` ` `  `// This code is contributed by Arnab Kundu `

Python3

 `# Python 3 implementation of the approach ` ` `  `# Function that returns true if s ` `# can be made lexicographically smaller ` `# by reversing a sub-string in s ` `def` `check(s): ` `    ``n ``=` `len``(s) ` ` `  `    ``# Traverse in the string ` `    ``for` `i ``in` `range``(n``-``1``): ` `         `  `        ``# Check if s[i+1] < s[i] ` `        ``if` `(s[i] > s[i ``+` `1``]): ` `            ``return` `True` ` `  `    ``# Not possible ` `    ``return` `False` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``s ``=` `"geeksforgeeks"` ` `  `    ``if` `(check(s)): ` `        ``print``(``"Yes"``) ` `    ``else``: ` `        ``print``(``"No"``) ` ` `  `# This code is contributed by ` `# Surendra_Gangwar `

C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Function that returns true if s  ` `// can be made lexicographically smaller  ` `// by reversing a sub-string in s  ` `static` `bool` `check(String s)  ` `{  ` `    ``int` `n = s.Length;  ` ` `  `    ``// Traverse in the string  ` `    ``for` `(``int` `i = 0; i < n - 1; i++) ` `    ``{  ` ` `  `        ``// Check if s[i+1] < s[i]  ` `        ``if` `(s[i] > s[i + 1])  ` `            ``return` `true``;  ` `    ``}  ` ` `  `    ``// Not possible  ` `    ``return` `false``;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main(String []args) ` `{  ` `    ``String s = ``"geeksforgeeks"``;  ` ` `  `    ``if` `(check(s))  ` `        ``Console.WriteLine(``"Yes"``);  ` `    ``else` `        ``Console.WriteLine(``"No"``);  ` ` `  `}  ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

PHP

 ` ``\$s``[``\$i` `+ 1]) ` `            ``return` `true; ` `    ``} ` ` `  `    ``// Not possible ` `    ``return` `false; ` `} ` ` `  `    ``// Driver code ` `    ``\$s` `= ``"geeksforgeeks"``; ` ` `  `    ``if` `(check(``\$s``)) ` `    ``echo` `"Yes"``; ` `    ``else` `        ``echo` `"No"``; ` ` `  `// This code is contributed by jit_t ` `?> `

Output:

```Yes
```

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