Check if number can be displayed using seven segment led

Given a string str representing an integer and an integer led which is the count of LEDs available. The task is to check if it is possible to display the number using the given LEDs. 
Note that a digit will be displayed as it is displayed on a 7 segment LED. If its possible to display the number then print Yes, otherwise print No.

Here's an example of seven segment display: 

Examples: 

Input: str = "999", led = 5 
Output: NO 
9 takes 6 LEDs to be displayed. So 999 will require 18 LEDs 
Since only 5 LEDs are available, it is not possible to display 999

Input: str = "123456789", led = 43 
Output: YES

Input: str = "123456789", led = 20 
Output: NO 

Approach: Pre-compute the number of segments used by digits from 0 to 9 and store it. Now for each element of the string count the number of segments used by it. Now, if count ? led then print YES else print NO.

The number of segment used by digit: 
0 -> 6 
1 -> 2 
2 -> 5 
3 -> 5 
4 -> 4 
5 -> 5 
6 -> 6 
7 -> 3 
8 -> 7 
9 -> 6 

Below is the implementation of the above approach: 

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;

// Pre-computed values of segment used by digit 0 to 9.
const int seg[10] = { 6, 2, 5, 5, 4, 5, 6, 3, 7, 6 };

// Check if it is possible to display the number
string LedRequired(string s, int led)
{
    int count = 0;

    // Finding sum of the segments used by
    // each digit of the number
    for (int i = 0; i < s.length(); ++i) {
        count += seg[int(s[i]) - 48];
    }

    if (count <= led)
        return "YES";
    else
        return "NO";
}

// Driven Program
int main()
{
    string S = "123456789";
    int led = 20;

    // Function call to print required answer
    cout << LedRequired(S, led) << endl;
    return 0;
}

// Java implementation of the above approach 
public class GfG{

    // Check if it is possible to display the number 
    public static String LedRequired(String s, int led) 
    { 
        // Pre-computed values of segment used by digit 0 to 9. 
        int seg[] = { 6, 2, 5, 5, 4, 5, 6, 3, 7, 6 }; 

        int count = 0; 
        // Finding sum of the segments used by 
        // each digit of the number 
        for (int i = 0; i < s.length(); ++i) { 
            count += seg[(int)(s.charAt(i)) - 48]; 
        } 
      
        if (count <= led) 
            return "YES"; 
        else
            return "NO"; 
    } 

    public static void main(String []args){
        
        String S = "123456789"; 
        int led = 20; 
      
        // Function call to print required answer 
        System.out.println(LedRequired(S, led));
    }
}

// This code is contributed by Rituraj Jain

# Python3 implementation of above approach 

# Pre-computed values of segment 
# used by digit 0 to 9. 
seg = [ 6, 2, 5, 5, 4, 
        5, 6, 3, 7, 6 ] 

# Check if it is possible to
# display the number 
def LedRequired(s, led) : 

    count = 0

    # Finding sum of the segments used 
    # by each digit of the number 
    for i in range(len(s)) :
        count += seg[ord(s[i]) - 48]
    
    if (count <= led) :
        return "YES"
    else :
        return "NO"

# Driver Code
if __name__ == "__main__" : 

    S = "123456789"
    led = 20

    # Function call to print
    # required answer 
    print(LedRequired(S, led)) 

# This code is contributed by Ryuga

// C# implementation of the above approach 
using System;
class GFG
{

// Check if it is possible to display the number 
public static String LedRequired(string s, int led) 
{ 
    // Pre-computed values of segment 
    // used by digit 0 to 9. 
    int[] seg = { 6, 2, 5, 5, 4, 5, 6, 3, 7, 6 }; 

    int count = 0; 
    
    // Finding sum of the segments used by 
    // each digit of the number 
    for (int i = 0; i < s.Length; ++i)
    { 
        count += seg[(int)(s[i]) - 48]; 
    } 

    if (count <= led) 
        return "YES"; 
    else
        return "NO"; 
} 

// Driver Code
public static void Main()
{
    
    string S = "123456789"; 
    int led = 20; 

    // Function call to print required answer 
    Console.WriteLine(LedRequired(S, led));
}
}

// This code is contributed by Akanksha Rai

<?php
// PHP implementation of above approach

// Pre-computed values of segment
// used by digit 0 to 9.
$seg = array(6, 2, 5, 5, 4, 5, 6, 3, 7, 6 );

// Check if it is possible to display the number
function LedRequired($s, $led)
{
    $count = 0;
    global $seg;
    
    // Finding sum of the segments used by
    // each digit of the number
    for ($i = 0; $i < strlen($s) ; ++$i) 
    {
        $count += $seg[ord($s[$i]) - 48];
    }

    if ($count <= $led)
        return "YES";
    else
        return "NO";
}

// Driver Code
$S = "123456789";
$led = 20;

// Function call to print required answer
echo LedRequired($S, $led);

// This code is contributed by ihritik 
?>

<script>

// Javascript implementation of above approach

// Pre-computed values of segment used by digit 0 to 9.
const seg = [ 6, 2, 5, 5, 4, 5, 6, 3, 7, 6 ];

// Check if it is possible to display the number
function LedRequired(s, led)
{
    var count = 0;

    // Finding sum of the segments used by
    // each digit of the number
    for (var i = 0; i < s.length; ++i) {
        count += seg[(s[i]) - 48];
    }

    if (count <= led)
        return "YES";
    else
        return "NO";
}

    var S = "123456789";
    var led = 20;

    // Function call to print required answer
    document.write( LedRequired(S, led) + "<br>");

// This code is contributed by SoumikMondal

</script>

NO

Complexity Analysis:

• Time Complexity: O(n), where n is the size of the given string
• Auxiliary Space: O(1), as extra space of size 10 is used to create an array