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
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++

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// 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;
}

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Java

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// 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");
    }
}

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Python3

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# 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.

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C#

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// 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

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PHP

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<?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
?>

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Output :

Yes

Time Complexity : O(d) where d is number of digits in given number.

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

This article is contributed by Sahil Chhabra (akku). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : jit_t