# Alternating Numbers

Given a number N, the task is to check if N is an Alternating Number or not. If N is an Alternating Number then print “Yes” else print “No”.

Alternating Number is a positive integer for which, in base-10, the parity of its digits are alternates i.e., the digits in number N is followed by odd, even, odd, … or even, odd, even, … order.

Examples:

Input: N = 129
Output: Yes
Explanation:
129 has digits which is in alternate odd even odd form.
Input: N = 28
Output: No
Explanation:
28 has digits which is not in alternate odd even odd form.

Approach: The idea is to convert the number into a string and check if any digit is followed by the digit of the same parity then N is not an Alternating Numbers, else N is an Alternating Numbers.
Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to check if a string is ` `// of the form even odd even odd... ` `bool` `isEvenOddForm(string s) ` `{ ` `    ``int` `n = s.length(); ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(i % 2 == 0 && s[i] % 2 != 0) ` `            ``return` `false``; ` ` `  `        ``if` `(i % 2 == 1 && s[i] % 2 != 1) ` `            ``return` `false``; ` `    ``} ` `    ``return` `true``; ` `} ` ` `  `// Function to check if a string is ` `// of the form odd even odd even  ... ` `bool` `isOddEvenForm(string s) ` `{ ` `    ``int` `n = s.length(); ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(i % 2 == 0 && s[i] % 2 != 1) ` `            ``return` `false``; ` ` `  `        ``if` `(i % 2 == 1 && s[i] % 2 != 0) ` `            ``return` `false``; ` `    ``} ` `    ``return` `true``; ` `} ` ` `  `// Function to check if n is an ` `// alternating number ` `bool` `isAlternating(``int` `n) ` `{ ` `    ``string str = to_string(n); ` `    ``return` `(isEvenOddForm(str) ` `            ``|| isOddEvenForm(str)); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Given Number N ` `    ``int` `N = 129; ` ` `  `    ``// Function Call ` `    ``if` `(isAlternating(N)) ` `        ``cout << ``"Yes"``; ` `    ``else` `        ``cout << ``"No"``; ` `    ``return` `0; ` `} `

## Java

 `// Java program for the above approach ` `class` `GFG{  ` `     `  `// Function to check if a string is ` `// of the form even odd even odd... ` `static` `boolean` `isEvenOddForm(String s) ` `{ ` `    ``int` `n = s.length(); ` `    ``for``(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `       ``if` `(i % ``2` `== ``0` `&& s.charAt(i) % ``2` `!= ``0``) ` `           ``return` `false``; ` `       ``if` `(i % ``2` `== ``1` `&& s.charAt(i) % ``2` `!= ``1``) ` `           ``return` `false``; ` `    ``} ` `    ``return` `true``; ` `} ` ` `  `// Function to check if a string is ` `// of the form odd even odd even ... ` `static` `boolean` `isOddEvenForm(String s) ` `{ ` `    ``int` `n = s.length(); ` `    ``for``(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `       ``if` `(i % ``2` `== ``0` `&& s.charAt(i) % ``2` `!= ``1``) ` `           ``return` `false``; ` `       ``if` `(i % ``2` `== ``1` `&& s.charAt(i) % ``2` `!= ``0``) ` `           ``return` `false``; ` `    ``} ` `    ``return` `true``; ` `} ` ` `  `// Function to check if n is an ` `// alternating number ` `static` `boolean` `isAlternating(``int` `n) ` `{ ` `    ``String str = Integer.toString(n); ` `    ``return` `(isEvenOddForm(str) || ` `            ``isOddEvenForm(str)); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args)  ` `{ ` `     `  `    ``// Given number N ` `    ``int` `N = ``129``; ` ` `  `    ``// Function call ` `    ``if` `(isAlternating(N)) ` `        ``System.out.println(``"Yes"``); ` `    ``else` `        ``System.out.println(``"No"``); ` `} ` `} ` ` `  `// This Code is contributed by rock_cool `

## Python3

 `# Python3 program for the above approach  ` ` `  `# Function to check if a string is  ` `# of the form even odd even odd...  ` `def` `isEvenOddForm(s): ` `    ``n ``=` `len``(s) ` `    ``for` `i ``in` `range``(n): ` `        ``if` `(i ``%` `2` `=``=` `0` `and`  `            ``int``(s[i]) ``%` `2` `!``=` `0``): ` `            ``return` `False` `        ``if` `(i ``%` `2` `=``=` `1` `and`  `            ``int``(s[i]) ``%` `2` `!``=` `1``): ` `            ``return` `False` `    ``return` `True` ` `  `# Function to check if a string is  ` `# of the form odd even odd even ...  ` `def` `isOddEvenForm(s): ` `    ``n ``=` `len``(s) ` `    ``for` `i ``in` `range``(n): ` `        ``if` `(i ``%` `2` `=``=` `0` `and`  `            ``int``(s[i]) ``%` `2` `!``=` `1``): ` `            ``return` `False` `        ``if` `(i ``%` `2` `=``=` `1` `and`  `            ``int``(s[i]) ``%` `2` `!``=` `0``): ` `            ``return` `False` `    ``return` `True` ` `  `# Function to check if n is an  ` `# alternating number  ` `def` `isAlternating(n): ` `    ``s ``=` `str``(n) ` `    ``return` `(isEvenOddForm(s) ``or` `            ``isOddEvenForm(s)) ` ` `  `# Driver Code ` ` `  `# Given Number N  ` `N ``=` `129` ` `  `# Function Call  ` `if` `(isAlternating(N)):  ` `    ``print``(``"Yes"``)  ` `else``: ` `    ``print``(``"No"``) ` `     `  `# This code is contributed by Vishal Maurya`

## C#

 `// C# program for the above approach ` `using` `System; ` `class` `GFG{  ` `     `  `// Function to check if a string is ` `// of the form even odd even odd... ` `static` `bool` `isEvenOddForm(String s) ` `{ ` `    ``int` `n = s.Length; ` `    ``for``(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``if` `(i % 2 == 0 && s[i] % 2 != 0) ` `            ``return` `false``; ` `        ``if` `(i % 2 == 1 && s[i] % 2 != 1) ` `            ``return` `false``; ` `    ``} ` `    ``return` `true``; ` `} ` ` `  `// Function to check if a string is ` `// of the form odd even odd even ... ` `static` `bool` `isOddEvenForm(String s) ` `{ ` `    ``int` `n = s.Length; ` `    ``for``(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``if` `(i % 2 == 0 && s[i] % 2 != 1) ` `            ``return` `false``; ` `        ``if` `(i % 2 == 1 && s[i] % 2 != 0) ` `            ``return` `false``; ` `    ``} ` `    ``return` `true``; ` `} ` ` `  `// Function to check if n is an ` `// alternating number ` `static` `bool` `isAlternating(``int` `n) ` `{ ` `    ``String str = n.ToString(); ` `    ``return` `(isEvenOddForm(str) || ` `            ``isOddEvenForm(str)); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args)  ` `{ ` `     `  `    ``// Given number N ` `    ``int` `N = 129; ` ` `  `    ``// Function call ` `    ``if` `(isAlternating(N)) ` `        ``Console.WriteLine(``"Yes"``); ` `    ``else` `        ``Console.WriteLine(``"No"``); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

Output:

`Yes`

Time Complexity: O(log10N)

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