Keith Number

Last Updated : 11 Sep, 2023

A n digit number x is called Keith number if it appears in a special sequence (defined below) generated using its digits. The special sequence has first n terms as digits of x and other terms are recursively evaluated as sum of previous n terms.
The task is to find if a given number is Keith Number or not.
Examples:

Input : x = 197
Output : Yes
197 has 3 digits, so n = 3
The number is Keith because it appears in the special
sequence that has first three terms as 1, 9, 7 and 
remaining terms evaluated using sum of previous 3 terms.
1, 9, 7, 17, 33, 57, 107, 197, .....

Input : x = 12
Output : No
The number is not Keith because it doesn't appear in
the special sequence generated using its digits.
1, 2, 3, 5, 8, 13, 21, .....

Input : x = 14
Output : Yes
14 is a Keith number since it appears in the sequence,
1, 4, 5, 9, 14, ...

Algorithm:

Store the 'n' digits of given number "x" in an array "terms".
Loop for generating next terms of sequence and adding the previous 'n' terms.
Keep storing the next_terms from step 2 in array "terms".
If the next term becomes equal to x, then x is a Keith number. If next term becomes more than x, then x is not a Keith Number.

C++

// C++ program to check if a number is Keith or not

#include<bits/stdc++.h>
using namespace std;

// Returns true if x is Keith, else false.
bool isKeith(int x)
{
    // Store all digits of x in a vector "terms"
    // Also find number of digits and store in "n".
    vector <int> terms;
    int temp = x, n = 0; // n is number of digits in x
    while (temp > 0)
    {
        terms.push_back(temp%10);
        temp = temp/10;
        n++;
    }

    // To get digits in right order (from MSB to
    // LSB)
    reverse(terms.begin(), terms.end());

    // Keep finding next terms of a sequence generated
    // using digits of x until we either reach x or a
    // number greater than x
    int next_term = 0, i = n;
    while (next_term < x)
    {
        next_term = 0;

        // Next term is sum of previous n terms
        for (int j=1; j<=n; j++)
            next_term += terms[i-j];

        terms.push_back(next_term);
        i++;
    }

    /* When the control comes out of the while loop,
       either the next_term is equal to the number
       or greater than it.
       If next_term is equal to x, then x is a
       Keith number, else not */
    return (next_term == x);
}

// Driver program
int main()
{
   isKeith(14)? cout << "Yes\n" : cout << "No\n";
   isKeith(12)? cout << "Yes\n" : cout << "No\n";
   isKeith(197)? cout << "Yes\n" : cout << "No\n";
   return 0;
}

Java

// Java program to check if a number is Keith or not
import java.io.*; 
import java.util.*;

class GFG{
// Returns true if x is Keith, else false.
static boolean isKeith(int x)
{
    // Store all digits of x in a vector "terms"
    // Also find number of digits and store in "n".
    ArrayList<Integer> terms=new ArrayList<Integer>();
    int temp = x, n = 0; // n is number of digits in x
    while (temp > 0)
    {
        terms.add(temp%10);
        temp = temp/10;
        n++;
    }

    // To get digits in right order (from MSB to
    // LSB)
    Collections.reverse(terms);

    // Keep finding next terms of a sequence generated
    // using digits of x until we either reach x or a
    // number greater than x
    int next_term = 0, i = n;
    while (next_term < x)
    {
        next_term = 0;

        // Next term is sum of previous n terms
        for (int j=1; j<=n; j++)
            next_term += terms.get(i-j);

        terms.add(next_term);
        i++;
    }

    /* When the control comes out of the while loop,
    either the next_term is equal to the number
    or greater than it.
    If next_term is equal to x, then x is a
    Keith number, else not */
    return (next_term == x);
}

// Driver program
public static void main(String[] args)
{
if (isKeith(14))
System.out.println("Yes");
else 
System.out.println("No");
if(isKeith(12)) 
System.out.println("Yes");
else
System.out.println("No");
if(isKeith(197)) 
System.out.println("Yes");
else
System.out.println("No");

}
}
// this code is contributed by mits

Python3

# Python3 program to check if a number 
# is Keith or not 

# Returns true if x is Keith, 
# else false. 
def isKeith(x): 
 
    # Store all digits of x in a vector "terms" 
    # Also find number of digits and store in "n". 
    terms = []; 
    temp = x; 
    n = 0; # n is number of digits in x 
    while (temp > 0):
        terms.append(temp % 10); 
        temp = int(temp / 10); 
        n+=1; 

    # To get digits in right order 
    # (from MSB to LSB) 
    terms.reverse(); 

    # Keep finding next terms of a sequence 
    # generated using digits of x until we 
    # either reach x or a number greater than x 
    next_term = 0; 
    i = n; 
    while (next_term < x): 
        next_term = 0; 

        # Next term is sum of previous n terms 
        for j in range(1,n+1): 
            next_term += terms[i - j]; 

        terms.append(next_term); 
        i+=1; 

    # When the control comes out of the 
    # while loop, either the next_term is 
    # equal to the number or greater than it. 
    # If next_term is equal to x, then x is a 
    # Keith number, else not 
    return (next_term == x); 

# Driver Code 
print("Yes") if (isKeith(14)) else print("No");
print("Yes") if (isKeith(12)) else print("No");
print("Yes") if (isKeith(197)) else print("No");

# This code is contributed by mits

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

class GFG{
// Returns true if x is Keith, else false.
static bool isKeith(int x)
{
    // Store all digits of x in a vector "terms"
    // Also find number of digits and store in "n".
    ArrayList terms = new ArrayList();
    int temp = x, n = 0; // n is number of digits in x
    while (temp > 0)
    {
        terms.Add(temp%10);
        temp = temp/10;
        n++;
    }

    // To get digits in right order (from MSB to
    // LSB)
    terms.Reverse();

    // Keep finding next terms of a sequence generated
    // using digits of x until we either reach x or a
    // number greater than x
    int next_term = 0, i = n;
    while (next_term < x)
    {
        next_term = 0;

        // Next term is sum of previous n terms
        for (int j=1; j<=n; j++)
            next_term += (int)terms[i-j];

        terms.Add(next_term);
        i++;
    }

    /* When the control comes out of the while loop,
    either the next_term is equal to the number
    or greater than it.
    If next_term is equal to x, then x is a
    Keith number, else not */
    return (next_term == x);
}

// Driver program
public static void Main()
{
if (isKeith(14))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
if(isKeith(12)) 
Console.WriteLine("Yes");
else
Console.WriteLine("No");
if(isKeith(197)) 
Console.WriteLine("Yes");
else
Console.WriteLine("No");

}
}
// this code is contributed by mits

PHP

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

// Returns true if x is Keith,
// else false.
function isKeith($x)
{
    // Store all digits of x in a vector "terms"
    // Also find number of digits and store in "n".
    $terms = array();
    $temp = $x;
    $n = 0; // n is number of digits in x
    while ($temp > 0)
    {
        array_push($terms, $temp % 10);
        $temp = (int)($temp / 10);
        $n++;
    }

    // To get digits in right order 
    // (from MSB to LSB)
    $terms=array_reverse($terms);

    // Keep finding next terms of a sequence 
    // generated using digits of x until we 
    // either reach x or a number greater than x
    $next_term = 0;
    $i = $n;
    while ($next_term < $x)
    {
        $next_term = 0;

        // Next term is sum of previous n terms
        for ($j = 1; $j <= $n; $j++)
            $next_term += $terms[$i - $j];

        array_push($terms, $next_term);
        $i++;
    }

    /* When the control comes out of the 
    while loop, either the next_term is 
    equal to the number or greater than it.
    If next_term is equal to x, then x is a
    Keith number, else not */
    return ($next_term == $x);
}

// Driver Code
isKeith(14) ? print("Yes\n") : print("No\n");
isKeith(12) ? print("Yes\n") : print("No\n");
isKeith(197) ? print("Yes\n") : print("No\n");

// This code is contributed by mits
?>

JavaScript

<script>

// Javascript program to check if a number
// is Keith or not

// Returns true if x is Keith,
// else false.
function isKeith(x)
{
    // Store all digits of x in a vector "terms"
    // Also find number of digits and store in "n".
    let terms = [];
    let temp = x;
    let n = 0; // n is number of digits in x
    while (temp > 0)
    {
        terms.push(temp % 10);
        temp = parseInt(temp / 10);
        n++;
    }

    // To get digits in right order
    // (from MSB to LSB)
    terms= terms.reverse();

    // Keep finding next terms of a sequence
    // generated using digits of x until we
    // either reach x or a number greater than x
    let next_term = 0;
    let i = n;
    while (next_term < x)
    {
        next_term = 0;

        // Next term is sum of previous n terms
        for (let j = 1; j <= n; j++)
            next_term += terms[i - j];

        terms.push(next_term);
        i++;
    }

    /* When the control comes out of the
    while loop, either the next_term is
    equal to the number or greater than it.
    If next_term is equal to x, then x is a
    Keith number, else not */
    return (next_term == x);
}

// Driver Code
isKeith(14) ? document.write("Yes<br>") : 
document.write("No<br>");
isKeith(12) ? document.write("Yes<br>") : 
document.write("No<br>");
isKeith(197) ? document.write("Yes<br>") : 
document.write("No<br>");

// This code is contributed by _saurabh_jaiswal

</script>

Output:

Yes
No
Yes

Time complexity: O(n^2) where n is no of digits

Auxiliary Space: O(n)

Position of the K-th set bit in a number

Pratik Agarwal

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Keith Number

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