Keith Number

21 Jun, 2022
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.
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++
Java
Python3
C#
PHP
Javascript
// 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 programint 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 program to check if a number is Keith or notimport 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 programpublic static void main(String[] args){if (isKeith(14))System.out.println("Yes");elseSystem.out.println("No");if(isKeith(12))System.out.println("Yes");elseSystem.out.println("No");if(isKeith(197))System.out.println("Yes");elseSystem.out.println("No"); }}// this code is contributed by mits
# 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 Codeprint("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 notusing 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 programpublic static void Main(){if (isKeith(14))Console.WriteLine("Yes");elseConsole.WriteLine("No");if(isKeith(12))Console.WriteLine("Yes");elseConsole.WriteLine("No");if(isKeith(197))Console.WriteLine("Yes");elseConsole.WriteLine("No"); }}// this code is contributed by mits
<?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 CodeisKeith(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?>
<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 CodeisKeith(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)This article is contributed by Pratik Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
Mithun Kumar
_saurabh_jaiswal
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