 Open in App
Not now

# Maximum number that can be display on Seven Segment Display using N segments

• Difficulty Level : Medium
• Last Updated : 16 Dec, 2022

Given a positive integer N. The task is to find the maximum number that can be displayed on seven segment display using N segments.
Seven Segment Display: A seven-segment display (SSD), or seven-segment indicator, is a form of an electronic display device for displaying decimal numerals that is an alternative to the more complex dot matrix displays. The individual segments of a seven-segment display
Image Source: Wikipedia

Examples:

```Input : N = 5
Output : 71
On 7-segment display, 71 will look like:
_
| |
| |

Input : N = 4
Output : 11```

Observe, the number having a greater number of digits than other numbers will be greater in value. So, we will try to make a number with maximum possible length (number of digits) using given ‘N’ segments.
Also observe, to increase the length of the number we will try to use less segment on each digit as possible. So, number ‘1’ use only 2 segments to represent a digit. No other digit use less than 2 segments.
So, in case N is even, the answer would be 1s N/2 number of time.
In case N is odd, we cannot use all segments if we make 1s N/2 number of time. Also, if we use 3 segments to make a digit of 7 and (N-3)/2 number of 1s, then the number formed will be greater in value than the number formed by N/2 number of 1s.
Below is the implementation of this approach:

## C++

 `#include ``using` `namespace` `std;` `// Function to print maximum number that can be formed``// using N segments``void` `printMaxNumber(``int` `n)``{``    ``// If n is odd``    ``if` `(n & 1) {``        ``// use 3 three segment to print 7``        ``cout << ``"7"``;` `        ``// remaining to print 1``        ``for` `(``int` `i = 0; i < (n - 3) / 2; i++)``            ``cout << ``"1"``;``    ``}` `    ``// If n is even``    ``else` `{``        ``// print n/2 1s.``        ``for` `(``int` `i = 0; i < n / 2; i++)``            ``cout << ``"1"``;``    ``}``}` `// Driver's Code``int` `main()``{``    ``int` `n = 5;` `    ``printMaxNumber(n);` `    ``return` `0;``}`

## Java

 `// Java implementation of above approach``import` `java.io.*;``class` `GFG {` `    ``// Function to print maximum number that``    ``// can be formed using N segments``    ``public` `static` `void` `printMaxNumber(``int` `n)``    ``{``        ``// If n is odd``        ``if` `(n % ``2` `!= ``0``) {``            ``// use 3 three segment to print 7``            ``System.out.print(``"7"``);` `            ``// remaining to print 1``            ``for` `(``int` `i = ``0``; i < (n - ``3``) / ``2``; i++)``                ``System.out.print(``"1"``);``        ``}` `        ``// If n is even``        ``else` `{` `            ``// print n/2 1s.``            ``for` `(``int` `i = ``0``; i < n / ``2``; i++)``                ``System.out.print(``"1"``);``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `n = ``5``;``        ``printMaxNumber(n);``    ``}``}` `// This code is contributed by princiraj1992`

## Python3

 `# Function to print maximum number that can be formed``# using N segments``def` `printMaxNumber(n):``    ` `    ``# If n is odd``    ``if` `(n ``%` `2` `=``=` `1``):``        ` `        ``# use 3 three segment to print 7``        ``print``(``"7"``,end``=``"");` `        ``# remaining to print 1``        ``for` `i ``in` `range``(``int``((n ``-` `3``) ``/` `2``)):``            ``print``(``"1"``,end``=``"");` `    ``# If n is even``    ``else``:``        ` `        ``# print n/2 1s.``        ``for` `i ``in` `range``(n``/``2``):``            ``print``(``"1"``,end``=``"");` `# Driver's Code``n ``=` `5``;``printMaxNumber(n);` `# This code contributed by Rajput-Ji`

## C#

 `// C# implementation of above approach``using` `System;` `class` `GFG``{` `    ``// Function to print maximum number that``    ``// can be formed using N segments``    ``public` `static` `void` `printMaxNumber(``int` `n)``    ``{``        ``// If n is odd``        ``if` `(n % 2 != 0)``        ``{``            ``// use 3 three segment to print 7``            ``Console.Write(``"7"``);` `            ``// remaining to print 1``            ``for` `(``int` `i = 0; i < (n - 3) / 2; i++)``                ``Console.Write(``"1"``);``        ``}` `        ``// If n is even``        ``else``        ``{` `            ``// print n/2 1s.``            ``for` `(``int` `i = 0; i < n / 2; i++)``                ``Console.Write(``"1"``);``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int` `n = 5;``        ``printMaxNumber(n);``    ``}``}` `// This code has been contributed by 29AjayKumar`

## PHP

 ``

## Javascript

 ``

Output:

`71`

Time Complexity: O(n), as there runs a loop.
Auxiliary Space: O(1), as no extra space has been taken.

My Personal Notes arrow_drop_up