Consider a series of numbers composed of only digits 4 and 7. First few numbers in the series are 4, 7, 44, 47, 74, 77, 444, .. etc. Given a number n, we need to find n-th number in the series.
Examples:
Input : n = 2 Output : 7 Input : n = 3 Output : 44 Input : n = 5 Output : 74 Input : n = 6 Output : 77
We have discussed a O(n) solution in below post.
Find n-th element in a series with only 2 digits (4 and 7) allowed
In this post, a O(log n) solution is discussed which is based on below pattern in numbers. The numbers can be seen
""
/ \
4 7
/ \ / \
44 47 74 77
/ \ / \ / \ / \
The idea is to fill the required number from end. We know can observe that the last digit is 4 if n is odd and last digit is 7 if n is even. After filling last digit, we move to parent node in tree. If n is odd, then parent node corresponds to (n-1/2. Else parent node corresponds to (n-2)/2.
C++
// C++ program to find n-th number containing// only 4 and 7.#include<bits/stdc++.h>using namespace std;string findNthNo(int n){ string res = ""; while (n >= 1) { // If n is odd, append 4 and // move to parent if (n & 1) { res = res + "4"; n = (n-1)/2; } // If n is even, append 7 and // move to parent else { res = res + "7"; n = (n-2)/2; } } // Reverse res and return. reverse(res.begin(), res.end()); return res;}// Driver codeint main(){ int n = 13; cout << findNthNo(n); return 0;} |
Java
// java program to find n-th number // containing only 4 and 7.public class GFG { static String findNthNo(int n) { String res = ""; while (n >= 1) { // If n is odd, append // 4 and move to parent if ((n & 1) == 1) { res = res + "4"; n = (n - 1) / 2; } // If n is even, append // 7 and move to parent else { res = res + "7"; n = (n - 2) / 2; } } // Reverse res and return. StringBuilder sb = new StringBuilder(res); sb.reverse(); return new String(sb); } // Driver code public static void main(String args[]) { int n = 13; System.out.print( findNthNo(n) ); }}// This code is contributed by Sam007 |
Python3
# Python3 program to find# n-th number containing# only 4 and 7.def reverse(s): if len(s) == 0: return s else: return reverse(s[1:]) + s[0] def findNthNo(n): res = ""; while (n >= 1): # If n is odd, append # 4 and move to parent if (n & 1): res = res + "4"; n = (int)((n - 1) / 2); # If n is even, append7 # and move to parent else: res = res + "7"; n = (int)((n - 2) / 2); # Reverse res # and return. return reverse(res);# Driver coden = 13;print(findNthNo(n));# This code is contributed# by mits |
C#
// C# program to find n-th number // containing only 4 and 7.using System;class GFG {static string findNthNo(int n){ string res = ""; while (n >= 1) { // If n is odd, append 4 and // move to parent if ((n & 1) == 1) { res = res + "4"; n = (n - 1) / 2; } // If n is even, append 7 and // move to parent else { res = res + "7"; n = (n - 2) / 2; } } // Reverse res and return. char[] arr = res.ToCharArray(); Array.Reverse(arr); return new string(arr);}// Driver Codepublic static void Main(){ int n = 13; Console.Write( findNthNo(n) );}}// This code is contributed by Sam007 |
PHP
<?php// PHP program to find// n-th number containing// only 4 and 7.function findNthNo($n){ $res = ""; while ($n >= 1) { // If n is odd, append // 4 and move to parent if ($n & 1) { $res = $res . "4"; $n = (int)(($n - 1) / 2); } // If n is even, append // 7 and move to parent else { $res = $res . "7"; $n = (int)(($n - 2) / 2); } } // Reverse res // and return.return strrev($res);}// Driver code$n = 13;echo findNthNo($n);// This code is contributed// by mits?> |
Javascript
<script>// javascript program to find n-th number // containing only 4 and 7. function findNthNo(n){res = "";while (n >= 1){// If n is odd, append // 4 and move to parent if ((n & 1) == 1){res = res + "4";n = (n - 1) / 2; }// If n is even, append // 7 and move to parent else{res = res + "7";n = parseInt((n - 2) / 2); }}// Reverse res and return.return res.split("").reverse().join("");}// Driver codevar n = 13;document.write( findNthNo(n) );// This code is contributed by 29AjayKumar </script> |
Output:
774
Time Complexity: O(logN), where N represents the given integer.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
In this code the total complexity is O(log n). Because while loop run log (n) times.
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