Consider a series of numbers composed of only digits 4 and 7. The first few numbers in the series are 4, 7, 44, 47, 74, 77, 444, .. etc. Given a number constructed by 4, 7 digits only, we need to find the position of this number in this series.

Examples:

Input : 7 Output : pos = 2 Input : 444 Output : pos = 7

It is reverse of the following article :

Find n-th element in a series with only 2 digits (4 and 7) allowed | Set 2 (log(n) method)

"" / \ 1(4) 2(7) / \ / \ 3(44) 4(47) 5(74) 6(77) / \ / \ / \ / \

The idea is based on the fact that all even positioned numbers have 7 as the last digit and all odd positioned numbers have 4 as the last digit.

If the number is 4 then it is the left node of the tree, then it corresponds to (pos*2)+1. Else right child node(7) corresponds to (pos*2)+2.

## C++

`// C++ program to find position of a number ` `// in a series of numbers with 4 and 7 as the ` `// only digits. ` `#include <iostream> ` `#include <algorithm> ` `using` `namespace` `std; ` ` ` `int` `findpos(string n) ` `{ ` ` ` `int` `i = 0, pos = 0; ` ` ` `while` `(n[i] != ` `'\0'` `) { ` ` ` ` ` `// check all digit position ` ` ` `switch` `(n[i]) ` ` ` `{ ` ` ` ` ` `// if number is left then pos*2+1 ` ` ` `case` `'4'` `: ` ` ` `pos = pos * 2 + 1; ` ` ` `break` `; ` ` ` ` ` `// if number is right then pos*2+2 ` ` ` `case` `'7'` `: ` ` ` `pos = pos * 2 + 2; ` ` ` `break` `; ` ` ` `} ` ` ` `i++; ` ` ` `} ` ` ` `return` `pos; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `// given a number which is constructed ` ` ` `// by 4 and 7 digit only ` ` ` `string n = ` `"774"` `; ` ` ` `cout << findpos(n); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Java

`// java program to find position of a number ` `// in a series of numbers with 4 and 7 as the ` `// only digits. ` `import` `java.util.*; ` ` ` `class` `GFG { ` ` ` ` ` `static` `int` `findpos(String n) ` ` ` `{ ` ` ` ` ` `int` `k = ` `0` `, pos = ` `0` `, i = ` `0` `; ` ` ` `while` `(k != n.length()) { ` ` ` ` ` `// check all digit position ` ` ` `switch` `(n.charAt(i)) { ` ` ` ` ` `// if number is left then pos*2+1 ` ` ` `case` `'4'` `: ` ` ` `pos = pos * ` `2` `+ ` `1` `; ` ` ` `break` `; ` ` ` ` ` `// if number is right then pos*2+2 ` ` ` `case` `'7'` `: ` ` ` `pos = pos * ` `2` `+ ` `2` `; ` ` ` `break` `; ` ` ` `} ` ` ` ` ` `i++; ` ` ` `k++; ` ` ` `} ` ` ` ` ` `return` `pos; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` ` ` `// given a number which is constructed ` ` ` `// by 4 and 7 digit only ` ` ` `String n = ` `"774"` `; ` ` ` ` ` `System.out.println(findpos(n)); ` ` ` `} ` `} ` ` ` `// This code is contributed by Sam007. ` |

*chevron_right*

*filter_none*

## Python3

`# python program to find position ` `# of a number in a series of ` `# numbers with 4 and 7 as the ` `# only digits. ` `def` `findpos(n): ` ` ` `i ` `=` `0` ` ` `j ` `=` `len` `(n) ` ` ` `pos ` `=` `0` ` ` `while` `(i<j): ` ` ` ` ` `# check all digit position ` ` ` `# if number is left then ` ` ` `# pos*2+1 ` ` ` `if` `(n[i] ` `=` `=` `'4'` `): ` ` ` `pos ` `=` `pos ` `*` `2` `+` `1` ` ` ` ` `# if number is right then ` ` ` `# pos*2+2 ` ` ` `if` `(n[i] ` `=` `=` `'7'` `): ` ` ` `pos ` `=` `pos ` `*` `2` `+` `2` ` ` ` ` `i` `=` `i` `+` `1` ` ` ` ` `return` `pos ` ` ` ` ` `# Driver code ` `# given a number which is constructed ` `# by 4 and 7 digit only ` `n ` `=` `"774"` `print` `(findpos(n)) ` ` ` `# This code is contributed by Sam007 ` |

*chevron_right*

*filter_none*

## C#

`// C# program to find position of ` `// a number in a series of numbers ` `// with 4 and 7 as the only digits. ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `static` `int` `findpos(String n) ` ` ` `{ ` ` ` ` ` `int` `k = 0, pos = 0, i = 0; ` ` ` `while` `(k != n.Length) { ` ` ` ` ` `// check all digit position ` ` ` `switch` `(n[i]) { ` ` ` ` ` `// if number is left then pos*2+1 ` ` ` `case` `'4'` `: ` ` ` `pos = pos * 2 + 1; ` ` ` `break` `; ` ` ` ` ` `// if number is right then pos*2+2 ` ` ` `case` `'7'` `: ` ` ` `pos = pos * 2 + 2; ` ` ` `break` `; ` ` ` `} ` ` ` ` ` `i++; ` ` ` `k++; ` ` ` `} ` ` ` ` ` `return` `pos; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `static` `void` `Main() ` ` ` `{ ` ` ` ` ` `// given a number which is constructed ` ` ` `// by 4 and 7 digit only ` ` ` `String n = ` `"774"` `; ` ` ` ` ` `Console.Write(findpos(n)); ` ` ` `} ` ` ` `} ` ` ` `// This code is contributed by Sam007 ` |

*chevron_right*

*filter_none*

## PHP

`<?php ` `// PHP program to find position of a number ` `// in a series of numbers with 4 and 7 as the ` `// only digits. ` ` ` `function` `findpos(` `$n` `) ` `{ ` ` ` `$i` `= 0; ` ` ` `$pos` `= 0; ` ` ` `while` `(` `$i` `< ` `strlen` `(` `$n` `)) { ` ` ` ` ` `// check all digit position ` ` ` `switch` `(` `$n` `[` `$i` `]) ` ` ` `{ ` ` ` ` ` `// if number is left then pos*2+1 ` ` ` `case` `'4'` `: ` ` ` `$pos` `= ` `$pos` `* 2 + 1; ` ` ` `break` `; ` ` ` ` ` `// if number is right then pos*2+2 ` ` ` `case` `'7'` `: ` ` ` `$pos` `= ` `$pos` `* 2 + 2; ` ` ` `break` `; ` ` ` `} ` ` ` `$i` `++; ` ` ` `} ` ` ` `return` `$pos` `; ` `} ` ` ` ` ` `// Driver code ` ` ` `// given a number which ` ` ` `// is constructed by 4 ` ` ` `// and 7 digit only ` ` ` `$n` `= ` `"774"` `; ` ` ` `echo` `findpos(` `$n` `); ` ` ` `// This code is contributed by Sam007 ` `?> ` |

*chevron_right*

*filter_none*

Output:

13

This article is contributed by **Devanshu 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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.