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 <bits/stdc++.h>
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

Javascript

<script>

// Javascript program for the above approach

// Function to check if a string is
// of the form even odd even odd...
function isEvenOddForm(s)
{
    let n = s.length;
    for(let 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 ...
function isOddEvenForm(s)
{
    let n = s.length;
    for(let 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
function isAlternating(n)
{
    let str = n.toString();
    return (isEvenOddForm(str) ||
            isOddEvenForm(str));
}

// Driver code

// Given number N
let N = 129;

// Function call
if (isAlternating(N))
    document.write("Yes");
else
    document.write("No");

// This code is contributed by divyesh072019

</script>

Output:

Yes

Time Complexity: O(log₁₀N)