Posted on by Kalkicode
Code Number

Check if a given mobile number is fancy

The given problem is to determine whether a given mobile number is "fancy" or not. A fancy number is defined by the following conditions:

  1. It should not contain three adjacent digits that are the same.
  2. It should not contain three adjacent digits that are either strictly increasing or strictly decreasing.
  3. It should not contain any digit occurring more than three times.

To solve this problem, the code provided uses a Java program that checks these conditions for the given mobile number.

Explanation with Example

Let's take two examples to understand the problem and its solution:

  1. Test A: num = "8785917233"

    • Case A: No three adjacent similar digits.
    • Case B: The digits are not strictly increasing or decreasing.
    • Case C: No digit occurs more than three times. Therefore, the output is "8785917233 is not a Fancy Number."
  2. Test B: num = "9795017233"

    • Case A: No three adjacent similar digits.
    • Case B: The digits 9, 7, and 5 are strictly decreasing.
    • Case C: No digit occurs more than three times. Therefore, the output is "9795017233 is a Fancy Number."

Standard Pseudocode

function condition(num, n):
    for i from 0 to n - 2:
        if num[i] == num[i+1] and num[i+1] == num[i+2]:
            return true
    for i from 0 to n - 2:
        if (num[i] > num[i+1] and num[i+1] > num[i+2]) or (num[i] < num[i+1] and num[i+1] < num[i+2]):
            return true
    initialize an array count[10] with zeros
    for i from 0 to 9:
        count[i] = 0
    for i from 0 to 9:
        count[num[i]-'0']++
        if count[num[i]-'0'] >= 4:
            return true
    return false

function isFancyNumber(num):
    n = length of num
    if condition(num, n):
        print num + " is Fancy Number"
    else:
        print num + " is not Fancy Number"

Algorithm Explanation

  1. The function condition(num, n) checks the three cases described earlier:

    • Case A: It loops through the mobile number num and checks for three adjacent similar digits.
    • Case B: It also loops through the number and checks for three adjacent digits that are strictly increasing or decreasing.
    • Case C: It initializes an array count to store the frequency of each digit and then checks if any digit occurs more than three times.
  2. The function isFancyNumber(num) takes the mobile number num as input and calculates its length n. It then calls the condition function to check if the number satisfies the fancy conditions or not and prints the appropriate output accordingly.

Code Solution

Here given code implementation process.

/*
    Java Program for
    Check if a given mobile number is fancy
*/
public class FancyNumber
{
    public boolean condition(String num, int n)
    {
        // Case A
        // Check if three Adjacent Similar
        for (int i = 0; i < n - 2 ; ++i) 
        {
            if(num.charAt(i)==num.charAt(i+1) && 
                num.charAt(i+1) ==num.charAt(i+2))
            {
                return true;
            }
        }
        // Case B
        // Check if three Adjacent digits are increasing or decreasing. 
        for (int i = 0; i < n - 2 ; ++i ) 
        {
            if((num.charAt(i) > num.charAt(i+1) && 
                num.charAt(i+1) > num.charAt(i+2))||
               (num.charAt(i) < num.charAt(i+1) && 
                num.charAt(i+1) < num.charAt(i+2)) )
            {
                return true;
            }
        }

        // Case C
        // Check that if any digits are occur more than 3 times or not.
        int []count = new int[10];
        // Set initial frequency of possible digit
        for (int i = 0; i < 10 ; ++i ) 
        {
            count[i]=0;
        }
        // count frequency of each digit
        for (int i = 0; i < 10 ; ++i ) 
        {
            count[num.charAt(i)- '0']++;

            if(count[num.charAt(i)- '0'] >= 4)
            {
                return true;
            }
        }

        return false;
    }

    public void isFancyNumber(String num)
    {
        int n = num.length();
        // Assume num is valid number
        if(condition(num,n))
        {
            System.out.println(" "+num+" is Fancy Number");
        }
        else
        {
            System.out.println(" "+num+" is not Fancy Number");
        }
    }
   
    public static void main(String[] args)
    {
        FancyNumber task = new FancyNumber();
        // Test A
        task.isFancyNumber("8785917233");
        // Test B
        task.isFancyNumber("9795017233");
    }
}

Output

 8785917233 is not Fancy Number
 9795017233 is Fancy Number
// Include header file
#include <iostream>
#include <string>

using namespace std;
/*
    C++ Program for
    Check if a given mobile number is fancy
*/
class FancyNumber
{
	public: bool condition(string num, int n)
	{
		// Case A
		// Check if three Adjacent Similar
		for (int i = 0; i < n - 2; ++i)
		{
			if (num[i] == num[i + 1] && 
                num[i + 1] == num[i + 2])
			{
				return true;
			}
		}
		// Case B
		// Check if three Adjacent digits are increasing or decreasing. 
		for (int i = 0; i < n - 2; ++i)
		{
			if ((num[i] > num[i + 1] && 
                 num[i + 1] > num[i + 2]) || 
                (num[i] < num[i + 1] && num[i + 1] < num[i + 2]))
			{
				return true;
			}
		}
		// Case C
		// Check that if any digits are occur more than 3 times or not.
		int *count = new int[10];
		// Set initial frequency of possible digit
		for (int i = 0; i < 10; ++i)
		{
			count[i] = 0;
		}
		// count frequency of each digit
		for (int i = 0; i < 10; ++i)
		{
			count[num[i] - '0']++;
			if (count[num[i] - '0'] >= 4)
			{
				return true;
			}
		}
		return false;
	}
	void isFancyNumber(string num)
	{
		int n = num.length();
		// Assume num is valid number
		if (this->condition(num, n))
		{
			cout << " " << num << " is Fancy Number" << endl;
		}
		else
		{
			cout << " " << num << " is not Fancy Number" << endl;
		}
	}
};
int main()
{
	FancyNumber *task = new FancyNumber();
	// Test A
	task->isFancyNumber("8785917233");
	// Test B
	task->isFancyNumber("9795017233");
	return 0;
}

Output

 8785917233 is not Fancy Number
 9795017233 is Fancy Number
// Include namespace system
using System;
/*
    Csharp Program for
    Check if a given mobile number is fancy
*/
public class FancyNumber
{
	public Boolean condition(String num, int n)
	{
		// Case A
		// Check if three Adjacent Similar
		for (int i = 0; i < n - 2; ++i)
		{
			if (num[i] == num[i + 1] && 
                num[i + 1] == num[i + 2])
			{
				return true;
			}
		}
		// Case B
		// Check if three Adjacent digits are increasing or decreasing. 
		for (int i = 0; i < n - 2; ++i)
		{
			if ((num[i] > num[i + 1] && 
                 num[i + 1] > num[i + 2]) || 
                (num[i] < num[i + 1] && num[i + 1] < num[i + 2]))
			{
				return true;
			}
		}
		// Case C
		// Check that if any digits are occur more than 3 times or not.
		int[] count = new int[10];
		// Set initial frequency of possible digit
		for (int i = 0; i < 10; ++i)
		{
			count[i] = 0;
		}
		// count frequency of each digit
		for (int i = 0; i < 10; ++i)
		{
			count[num[i] - '0']++;
			if (count[num[i] - '0'] >= 4)
			{
				return true;
			}
		}
		return false;
	}
	public void isFancyNumber(String num)
	{
		int n = num.Length;
		// Assume num is valid number
		if (this.condition(num, n))
		{
			Console.WriteLine(" " + num + " is Fancy Number");
		}
		else
		{
			Console.WriteLine(" " + num + " is not Fancy Number");
		}
	}
	public static void Main(String[] args)
	{
		FancyNumber task = new FancyNumber();
		// Test A
		task.isFancyNumber("8785917233");
		// Test B
		task.isFancyNumber("9795017233");
	}
}

Output

 8785917233 is not Fancy Number
 9795017233 is Fancy Number
package main

import "fmt"
/*
    Go Program for
    Check if a given mobile number is fancy
*/

func condition(num string, n int) bool {
	// Case A
	// Check if three Adjacent Similar
	for i := 0 ; i < n - 2 ; i++ {
		if num[i] == num[i + 1] && num[i + 1] == num[i + 2] {
			return true
		}
	}
	// Case B
	// Check if three Adjacent digits are increasing or decreasing. 
	for i := 0 ; i < n - 2 ; i++ {
		if (num[i] > num[i + 1] && num[i + 1] > num[i + 2]) || 
		(num[i] < num[i + 1] && num[i + 1] < num[i + 2]) {
			return true
		}
	}
	// Case C
	// Check that if any digits are occur more than 3 times or not.
	// Set initial frequency of possible digit
	var count = make([] int, 10)
	// count frequency of each digit
	for i := 0 ; i < 10 ; i++ {
		count[int(num[i]) - 48]++
		if count[int(num[i]) - 48] >= 4 {
			return true
		}
	}
	return false
}
func isFancyNumber(num string) {
	var n int = len(num)
	// Assume num is valid number
	if condition(num, n) {
		fmt.Println(" ", num, " is Fancy Number")
	} else {
		fmt.Println(" ", num, " is not Fancy Number")
	}
}
func main() {

	// Test A
	isFancyNumber("8785917233")
	// Test B
	isFancyNumber("9795017233")
}

Output

 8785917233 is not Fancy Number
 9795017233 is Fancy Number
<?php
/*
    Php Program for
    Check if a given mobile number is fancy
*/
class FancyNumber
{
	public	function condition($num, $n)
	{
		// Case A
		// Check if three Adjacent Similar
		for ($i = 0; $i < $n - 2; ++$i)
		{
			if ($num[$i] == $num[$i + 1] && 
                $num[$i + 1] == $num[$i + 2])
			{
				return true;
			}
		}
		// Case B
		// Check if three Adjacent digits are increasing or decreasing. 
		for ($i = 0; $i < $n - 2; ++$i)
		{
			if (($num[$i] > $num[$i + 1] && 
                 $num[$i + 1] > $num[$i + 2]) || 
                ($num[$i] < $num[$i + 1] && 
                 $num[$i + 1] < $num[$i + 2]))
			{
				return true;
			}
		}
		// Case C
		// Check that if any digits are occur more than 3 times or not.
		$count = array_fill(0, 10, 0);
		// Set initial frequency of possible digit
		for ($i = 0; $i < 10; ++$i)
		{
			$count[$i] = 0;
		}
		// count frequency of each digit
		for ($i = 0; $i < 10; ++$i)
		{
			$count[ord($num[$i]) - ord('0')]++;
			if ($count[ord($num[$i]) - ord('0')] >= 4)
			{
				return true;
			}
		}
		return false;
	}
	public	function isFancyNumber($num)
	{
		$n = strlen($num);
		// Assume num is valid number
		if ($this->condition($num, $n))
		{
			echo(" ".$num." is Fancy Number\n");
		}
		else
		{
			echo(" ".$num." is not Fancy Number\n");
		}
	}
}

function main()
{
	$task = new FancyNumber();
	// Test A
	$task->isFancyNumber("8785917233");
	// Test B
	$task->isFancyNumber("9795017233");
}
main();

Output

 8785917233 is not Fancy Number
 9795017233 is Fancy Number
/*
    Node JS Program for
    Check if a given mobile number is fancy
*/
class FancyNumber
{
	condition(num, n)
	{
		// Case A
		// Check if three Adjacent Similar
		for (var i = 0; i < n - 2; ++i)
		{
			if (num.charAt(i) == 
                num.charAt(i + 1) 
                && num.charAt(i + 1) == num.charAt(i + 2))
			{
				return true;
			}
		}
		// Case B
		// Check if three Adjacent digits are increasing or decreasing. 
		for (var i = 0; i < n - 2; ++i)
		{
			if ((num.charAt(i) > num.charAt(i + 1) && 
                 num.charAt(i + 1) > num.charAt(i + 2)) || 
                (num.charAt(i) < num.charAt(i + 1) && 
                 num.charAt(i + 1) < num.charAt(i + 2)))
			{
				return true;
			}
		}
		// Case C
		// Check that if any digits are occur more than 3 times or not.
		var count = Array(10).fill(0);
		// Set initial frequency of possible digit
		for (var i = 0; i < 10; ++i)
		{
			count[i] = 0;
		}
		// count frequency of each digit
		for (var i = 0; i < 10; ++i)
		{
			count[num.charCodeAt(i) - 48]++;
			if (count[num.charCodeAt(i)- 48] >= 4)
			{
				return true;
			}
		}
		return false;
	}
	isFancyNumber(num)
	{
		var n = num.length;
		// Assume num is valid number
		if (this.condition(num, n))
		{
			console.log(" " + num + " is Fancy Number");
		}
		else
		{
			console.log(" " + num + " is not Fancy Number");
		}
	}
}

function main()
{
	var task = new FancyNumber();
	// Test A
	task.isFancyNumber("8785917233");
	// Test B
	task.isFancyNumber("9795017233");
}
main();

Output

 8785917233 is not Fancy Number
 9795017233 is Fancy Number
#    Python 3 Program for
#    Check if a given mobile number is fancy
class FancyNumber :
	def condition(self, num, n) :
		i = 0
		#  Case A
		#  Check if three Adjacent Similar
		while (i < n - 2) :
			if (num[i] == num[i + 1] and num[i + 1] == num[i + 2]) :
				return True
			
			i += 1
		
		i = 0
		#  Case B
		#  Check if three Adjacent digits are increasing or decreasing. 
		while (i < n - 2) :
			if ((num[i] > num[i + 1] and 
                 num[i + 1] > num[i + 2]) or
            (num[i] < num[i + 1] and num[i + 1] < num[i + 2])) :
				return True
			
			i += 1
		
		#  Case C
		#  Check that if any digits are occur more than 3 times or not.
		#  Set initial frequency of possible digit
		count = [0] * (10)
		i = 0
		#  count frequency of each digit
		while (i < 10) :
			count[ord(num[i]) - ord('0')] += 1
			if (count[ord(num[i]) - ord('0')] >= 4) :
				return True
			
			i += 1
		
		return False
	
	def isFancyNumber(self, num) :
		n = len(num)
		#  Assume num is valid number
		if (self.condition(num, n)) :
			print(" ", num ," is Fancy Number")
		else :
			print(" ", num ," is not Fancy Number")
		
	

def main() :
	task = FancyNumber()
	#  Test A
	task.isFancyNumber("8785917233")
	#  Test B
	task.isFancyNumber("9795017233")

if __name__ == "__main__": main()

Output

  8785917233  is not Fancy Number
  9795017233  is Fancy Number
#    Ruby Program for
#    Check if a given mobile number is fancy
class FancyNumber 
	def condition(num, n) 
		i = 0
		#  Case A
		#  Check if three Adjacent Similar
		while (i < n - 2) 
			if (num[i] == num[i + 1] && num[i + 1] == num[i + 2]) 
				return true
			end

			i += 1
		end

		i = 0
		#  Case B
		#  Check if three Adjacent digits are increasing or decreasing. 
		while (i < n - 2) 
			if ((num[i] > num[i + 1] && 
                 num[i + 1] > num[i + 2]) || 
                (num[i] < num[i + 1] && num[i + 1] < num[i + 2])) 
				return true
			end

			i += 1
		end

		#  Case C
		#  Check that if any digits are occur more than 3 times or not.
		#  Set initial frequency of possible digit
		count = Array.new(10) {0}
		i = 0
		#  count frequency of each digit
		while (i < 10) 
			count[num[i].ord - '0'.ord] += 1
			if (count[num[i].ord - '0'.ord] >= 4) 
				return true
			end

			i += 1
		end

		return false
	end

	def isFancyNumber(num) 
		n = num.length
		#  Assume num is valid number
		if (self.condition(num, n)) 
			print(" ", num ," is Fancy Number", "\n")
		else
 
			print(" ", num ," is not Fancy Number", "\n")
		end

	end

end

def main() 
	task = FancyNumber.new()
	#  Test A
	task.isFancyNumber("8785917233")
	#  Test B
	task.isFancyNumber("9795017233")
end

main()

Output

 8785917233 is not Fancy Number
 9795017233 is Fancy Number
/*
    Scala Program for
    Check if a given mobile number is fancy
*/
class FancyNumber()
{
	def condition(num: String, n: Int): Boolean = {
		var i: Int = 0;
		// Case A
		// Check if three Adjacent Similar
		while (i < n - 2)
		{
			if (num.charAt(i) == num.charAt(i + 1) && 
                num.charAt(i + 1) == num.charAt(i + 2))
			{
				return true;
			}
			i += 1;
		}
		i = 0;
		// Case B
		// Check if three Adjacent digits are increasing or decreasing. 
		while (i < n - 2)
		{
			if ((num.charAt(i) > num.charAt(i + 1) && 
                 num.charAt(i + 1) > num.charAt(i + 2)) || 
                (num.charAt(i) < num.charAt(i + 1) && 
                 num.charAt(i + 1) < num.charAt(i + 2)))
			{
				return true;
			}
			i += 1;
		}
		// Case C
		// Check that if any digits are occur more than 3 times or not.
		// Set initial frequency of possible digit
		var count: Array[Int] = Array.fill[Int](10)(0);
		i = 0;
		// count frequency of each digit
		while (i < 10)
		{
			count(num.charAt(i).toInt - '0'.toInt) += 1;
			if (count(num.charAt(i).toInt - '0'.toInt) >= 4)
			{
				return true;
			}
			i += 1;
		}
		return false;
	}
	def isFancyNumber(num: String): Unit = {
		var n: Int = num.length();
		// Assume num is valid number
		if (condition(num, n))
		{
			println(" " + num + " is Fancy Number");
		}
		else
		{
			println(" " + num + " is not Fancy Number");
		}
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: FancyNumber = new FancyNumber();
		// Test A
		task.isFancyNumber("8785917233");
		// Test B
		task.isFancyNumber("9795017233");
	}
}

Output

 8785917233 is not Fancy Number
 9795017233 is Fancy Number
import Foundation;
/*
    Swift 4 Program for
    Check if a given mobile number is fancy
*/
class FancyNumber
{
	func condition(_ text: String, _ n: Int) -> Bool
	{
       	var num = Array(text);
		var i: Int = 0;
		// Case A
		// Check if three Adjacent Similar
		while (i < n - 2)
		{
			if (num[i] == num[i + 1] && num[i + 1] == num[i + 2])
			{
				return true;
			}
			i += 1;
		}
		i = 0;
		// Case B
		// Check if three Adjacent digits are increasing or decreasing. 
		while (i < n - 2)
		{
			if ((num[i] > num[i + 1] && 
                 num[i + 1] > num[i + 2]) || 
                (num[i] < num[i + 1] && num[i + 1] < num[i + 2]))
			{
				return true;
			}
			i += 1;
		}
		// Case C
		// Check that if any digits are occur more than 3 times or not.
		// Set initial frequency of possible digit
		var count: [Int] = Array(repeating: 0, count: 10);
		i = 0;
		// count frequency of each digit
		while (i < 10)
		{
			count[Int(UnicodeScalar(String(num[i]))!.value) - 48] += 1;
			if (count[Int(UnicodeScalar(String(num[i]))!.value) - 48] >= 4)
			{
				return true;
			}
			i += 1;
		}
		return false;
	}
	func isFancyNumber(_ num: String)
	{
		let n: Int = num.count;
		// Assume num is valid number
		if (self.condition(num, n))
		{
			print(" ", num ," is Fancy Number");
		}
		else
		{
			print(" ", num ," is not Fancy Number");
		}
	}
}
func main()
{
	let task: FancyNumber = FancyNumber();
	// Test A
	task.isFancyNumber("8785917233");
	// Test B
	task.isFancyNumber("9795017233");
}
main();

Output

  8785917233  is not Fancy Number
  9795017233  is Fancy Number
/*
    Kotlin Program for
    Check if a given mobile number is fancy
*/
class FancyNumber
{
	fun condition(num: String, n: Int): Boolean
	{
		var i: Int = 0;
		// Case A
		// Check if three Adjacent Similar
		while (i < n - 2)
		{
			if (num.get(i) == num.get(i + 1) && 
                num.get(i + 1) == num.get(i + 2))
			{
				return true;
			}
			i += 1;
		}
		i = 0;
		// Case B
		// Check if three Adjacent digits are increasing or decreasing. 
		while (i < n - 2)
		{
			if ((num.get(i) > num.get(i + 1) && 
                 num.get(i + 1) > num.get(i + 2)) || 
                (num.get(i) < num.get(i + 1) && 
                 num.get(i + 1) < num.get(i + 2)))
			{
				return true;
			}
			i += 1;
		}
		// Case C
		// Check that if any digits are occur more than 3 times or not.
		// Set initial frequency of possible digit
		val count: Array < Int > = Array(10)
		{
			0
		};
		i = 0;
		// count frequency of each digit
		while (i < 10)
		{
			count[num.get(i).toInt() - '0'.toInt()] += 1;
			if (count[num.get(i).toInt() - '0'.toInt()] >= 4)
			{
				return true;
			}
			i += 1;
		}
		return false;
	}
	fun isFancyNumber(num: String): Unit
	{
		val n: Int = num.length;
		// Assume num is valid number
		if (this.condition(num, n))
		{
			println(" " + num + " is Fancy Number");
		}
		else
		{
			println(" " + num + " is not Fancy Number");
		}
	}
}
fun main(args: Array < String > ): Unit
{
	val task: FancyNumber = FancyNumber();
	// Test A
	task.isFancyNumber("8785917233");
	// Test B
	task.isFancyNumber("9795017233");
}

Output

 8785917233 is not Fancy Number
 9795017233 is Fancy Number

Time Complexity Analysis

Let's analyze the time complexity of the code:

  • Case A and Case B both loop through the entire mobile number, so each of them takes O(n) time.
  • Case C uses an array count of size 10 and loops through the mobile number, taking O(n) time.
  • The main function isFancyNumber calls the condition function, which has a time complexity of O(n) as explained above.

Overall, the time complexity of the entire code is O(n).

Finally

The provided Java program effectively checks whether a given mobile number is fancy or not based on the specified conditions. The algorithm presented in the code runs in linear time complexity, making it efficient for practical use.

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