Find position of the given number among the numbers made of 4 and 7

08 Jun, 2022
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++
Java
Python3
C#
PHP
Javascript
// 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 codeint main(){    // given a number which is constructed    // by 4 and 7 digit only    string n = "774";    cout << findpos(n);    return 0;}
// 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.
# 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 onlyn = "774"print(findpos(n)) # This code is contributed by Sam007
// 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
<?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?>
<script>    // Javascript program to find position of a number// in a series of numbers with 4 and 7 as the// only digits.   function findpos(n){    let i = 0;    let pos = 0;    while(i < 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++;    }    return pos;}       // Driver code    // given a number which    // is constructed by 4    // and 7 digit only    let n = "774";    document.write(findpos(n));       // This code is contributed by _saurabh_jaiswal</script>
Output:  
13
Time Complexity: O(n), where n represents the size of the given string.Auxiliary Space: O(1), no extra space is required, so it is a constant. 
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