Maximum number on 7-segment display using N segments : Recursive

Last Updated : 16 Dec, 2022

Given an integer N, the task is to find the largest number that can be shown with the help of N segments using any number of 7 segment displays.
Examples:

Input: N = 4
Output: 11
Explanation:
Largest number that can be displayed with the help of 4 segments using 2 seven segment displays turned is 11.

Input: N = 7
Output: 711
Explanation:
Largest number that can be displayed by turning on seven segments is 711 with the help of 3 segments display set.

Approach:
The key observation in seven segment display is to turn on any number from 0 to 9 takes certain amounts of segments, which is described below:

If the problem is observed carefully, then the number N can be of two types that is even or odd and each of them should be solved separately as follows:

For Even: As in the above image, There are 6 numbers that can be displayed using even number of segments which is

As it is observed number 1 uses the minimum count of segments to display a digit. Then, even the number of segments can be displayed using 1 with 2 counts of segments in each digit.
For Odd: As in the above image, there are 5 numbers that can be displayed using an odd number of segments which is

As it is observed number 7 uses the minimum number of odd segments to display a digit. Then an odd number of segments can be displayed using 7 with 3 counts of segments in each digit.

Algorithm:

If the given number N is 0 or 1, then any number cannot be displayed with this much of bits.
If the given number N is odd then the most significant digit will be 7 and the rest of the digits can be displayed with the help of the (N – 3) segments because to display 7 it takes 3 segments.
If given number N is even then the most significant digit will be 1 and the rest of the digits can be displayed with the help of the (N – 2) segments because to display 1, it takes 2 segments only.
The number N is processed digit by digit recursively.

Explanation with Example:
Given number N be – 11

Digit (from MSB to LSB)	N	Largest number from N using segment	Segments used	Segments remaining
1	11	7	3	8
2	8	1	2	6
3	6	1	2	4
4	4	1	2	2
5	2	1	2	0

Then, the largest number will be 71111.
Below is the implementation of the above approach:

C++

// C++ implementation to find the
// maximum number that can be 
// using the N segments in 
// N segments display
 
#include <iostream>
using namespace std;
 
// Function to find the maximum
// number that can be displayed
// using the N segments 
void segments(int n)
{
    // Condition to check base case
    if (n == 1 || n == 0) {
        return;
    }
    // Condition to check if the 
    // number is even
    if (n % 2 == 0) {
        cout << "1";
        segments(n - 2);
    }
     
    // Condition to check if the
    // number is odd
    else if (n % 2 == 1) {
 
        cout << "7";
        segments(n - 3);
    }
}
 
// Driver Code
int main()
{
    int n;
    n = 11;
    segments(n);
    return 0;
}

Java

// Java implementation to find the 
// maximum number that can be 
// using the N segments in 
// N segments display 
import java.io.*;
class GFG {
     
    // Function to find the maximum 
    // number that can be displayed 
    // using the N segments 
    static void segments(int n) 
    { 
        // Condition to check base case 
        if (n == 1 || n == 0) { 
            return; 
        } 
        // Condition to check if the 
        // number is even 
        if (n % 2 == 0) { 
            System.out.print("1"); 
            segments(n - 2); 
        } 
         
        // Condition to check if the 
        // number is odd 
        else if (n % 2 == 1) { 
     
            System.out.print("7"); 
            segments(n - 3); 
        } 
    } 
     
    // Driver Code 
    public static void main (String[] args)
    { 
        int n; 
        n = 11; 
        segments(n); 
    } 
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation to find the 
# maximum number that can be 
# using the N segments in 
# N segments display 
 
# Function to find the maximum 
# number that can be displayed 
# using the N segments 
def segments(n) :
 
    # Condition to check base case 
    if (n == 1 or n == 0) :
        return; 
     
    # Condition to check if the 
    # number is even 
    if (n % 2 == 0) :
        print("1",end=""); 
        segments(n - 2); 
     
    # Condition to check if the 
    # number is odd 
    elif (n % 2 == 1) :
 
        print("7",end=""); 
        segments(n - 3); 
 
# Driver Code 
if __name__ == "__main__" : 
 
    n = 11; 
    segments(n); 
 
# This code is contributed by AnkitRai01

C#

// C# implementation to find the 
// maximum number that can be 
// using the N segments in 
// N segments display 
using System;
 
class GFG {
     
    // Function to find the maximum 
    // number that can be displayed 
    // using the N segments 
    static void segments(int n) 
    { 
        // Condition to check base case 
        if (n == 1 || n == 0) { 
            return; 
        } 
        // Condition to check if the 
        // number is even 
        if (n % 2 == 0) { 
            Console.Write("1"); 
            segments(n - 2); 
        } 
         
        // Condition to check if the 
        // number is odd 
        else if (n % 2 == 1) { 
     
            Console.Write("7"); 
            segments(n - 3); 
        } 
    } 
     
    // Driver Code 
    public static void Main()
    { 
        int n; 
        n = 11; 
        segments(n); 
    } 
}
 
// This code is contributed by AnkitRai01

Javascript

<script>
    // Javascript implementation to find the 
    // maximum number that can be 
    // using the N segments in 
    // N segments display
     
    // Function to find the maximum 
    // number that can be displayed 
    // using the N segments 
    function segments(n) 
    { 
        // Condition to check base case 
        if (n == 1 || n == 0) { 
            return; 
        } 
        // Condition to check if the 
        // number is even 
        if (n % 2 == 0) { 
            document.write("1"); 
            segments(n - 2); 
        } 
           
        // Condition to check if the 
        // number is odd 
        else if (n % 2 == 1) { 
       
            document.write("7"); 
            segments(n - 3); 
        } 
    } 
     
    let n; 
    n = 11; 
    segments(n);
 
// This code is contributed by divyesh072019.
</script>

Output:

Performance Analysis:

Time Complexity: As in the above approach, there is recursive call which takes O(N) time in worst case, Hence the Time Complexity will be O(N).
Auxiliary Space Complexity: As in the above approach, taking consideration of the stack space used in recursive call then the auxiliary space complexity will be O(N)

Suggest improvement

Queries to check if traversal is possible from (1,1) to (2, N)

Construct a matrix such that union of ith row and ith column contains every element from 1 to 2N-1

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Maximum number on 7-segment display using N segments : Recursive

C++

Java

Python3

C#

Javascript

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