Keith Number

Difficulty Level : Easy
Last Updated : 26 Dec, 2018

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, ...

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 trms of a sequence generated 
    // using digits of x until we either reach x or a 
    // number greate 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 trms of a sequence generated 
    // using digits of x until we either reach x or a 
    // number greate 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 trms of a sequence  
    # generated using digits of x until we  
    # either reach x or a number greate 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#

// 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 trms of a sequence generated 
    // using digits of x until we either reach x or a 
    // number greate 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 trms of a sequence  
    // generated using digits of x until we  
    // either reach x or a number greate 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 
?> 

Output:

Yes
No
Yes

This article is contributed by Pratik Agarwal. 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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Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

PHP