Tidy Number (Digits in non-decreasing Order)

Given a number, the task is to check if it is tidy or not. A tidy number is a number whose digits are in non-decreasing order.

Examples :

Input : 1234
Output : Yes

Input : 1243
Output : No
Explanation: Digits “4” and “3” violate the property.

Asked in Freshokartz

Algorithm:

• 1- One by one find all the digits.
• 2- Compare every digit with its next digit.
• 3- If any is in decreasing order then return false.
• 4- Otherwise return true.

Implementation :

C++

// C++ program to check if a number is Tidy
// or not.
#include<iostream>
using namespace std;

// Returns true if num is Tidy
bool isTidy(int num)
{
    // To store previous digit (Assigning
    // initial value which is more than any
    // digit)
    int prev = 10;

    // Traverse all digits from right to
    // left and check if any digit is
    // smaller than previous.
    while (num)
    {
        int rem = num % 10;
        num /= 10;
        if (rem > prev)
           return false;
        prev = rem;
    }

    return true;
}

// Driver code
int main()
{
    int num = 1556;
    isTidy(num) ? cout << "Yes"
                : cout << "No";
    return 0;
}

Java

// Java program to check if a number
// is Tidy or not.

class Test
{
    // Returns true if num is Tidy
    static boolean isTidy(int num)
    {
        // To store previous digit
        // (Assigning initial value
            // which is more than any
        // digit)
        int prev = 10;

        // Traverse all digits from right to
        // left and check if any digit is
        // smaller than previous.
        while (num!=0)
        {
            int rem = num % 10;
            num /= 10;
            if (rem > prev)
               return false;
            prev = rem;
        }

        return true;
    }

    // Driver method
    public static void main(String[] args)
    {
        int num = 1556;
        System.out.println(isTidy(num) ? "Yes" : "No");
    }
}

Python3

# Python program to check if a number
# is Tidy or not.

# Returns true if num is Tidy
def isTidy(num):

    # To store previous digit (Assigning
    # initial value which is more than any
    # digit)
    prev = 10

    # Traverse all digits from right to
    # left and check if any digit is
    # smaller than previous.
    while (num):
        rem = num % 10
        num /= 10
        if rem > prev:
            return False
        prev = rem
    return True

# Driver code
num = 1556
if isTidy(num):
    print("Yes")
else:
    print("No")

# This code is contributed by Sharad_Bhardwaj.

C#

// C# program to check if a
// number is Tidy or not.
using System;

class GFG
{
    // Returns true if num is Tidy
    static bool isTidy(int num)
    {
        // To store previous digit
        // (Assigning initial value
        // which is more than any
        // digit)
        int prev = 10;

        // Traverse all digits from
        // right to left and check
        // if any digit is smaller
        // than previous.
        while (num != 0)
        {
            int rem = num % 10;
            num /= 10;
            if (rem > prev)
            return false;
            prev = rem;
        }

        return true;
    }

// Driver Code
public static void Main ()
{
    int num = 1556;

    Console.WriteLine(isTidy(num) ?
                            "Yes" :
                             "No");
}
}

// This code is contributed by m_kit

PHP

<?php
// PHP program to check if a
// number is Tidy or not.

// Returns true if num is Tidy
function isTidy($num)
{
    // To store previous digit
    // (Assigning initial value
    // which is more than any
    // digit)
    $prev = 10;

    // Traverse all digits from
    // right to left and check
    // if any digit is smaller
    // than previous.
    while ($num)
    {
        $rem = $num % 10;
        $num = (int)$num / 10;
        if ($rem > $prev)
            return false;
        $prev = $rem;
    }

    return true;
}

// Driver code
$num = 1556;
if(isTidy($num) == true)
echo "Yes";
else
echo "No";

// This code is contributed by aj_36
?>

Javascript

<script>

// JavaScript program for the above approach

    // Returns true if num is Tidy
    function isTidy(num)
    {
        // To store previous digit
        // (Assigning initial value
            // which is more than any
        // digit)
        let prev = 10;

        // Traverse all digits from right to
        // left and check if any digit is
        // smaller than previous.
        while (num!=0)
        {
            let rem = num % 10;
            num /= 10;
            if (rem > prev)
               return false;
            prev = rem;
        }

        return true;
    }

// Driver Code
    let num = 1556;
    document.write(isTidy(num) ? "Yes" : "No");

// This code is contributed by susmitakundugoaldanga.
</script>

Output

Yes

Time Complexity: O(d) where d is the number of digits in the given number.
Auxiliary Space: O(1) since using constant extra space.

Reference : 
https://www.careercup.com/question?id=5136136486780928
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