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Maximum number on 7-segment display using N segments : Recursive
• Last Updated : 23 Jun, 2020

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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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
```0 - 6
1 - 2
2 - 5
4 - 4
6 - 6
9 - 6
```

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
```3 - 5
5 - 5
7 - 3
8 - 7
```

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)NLargest number from N using segmentSegments usedSegments remaining
111738
28126
36124
44122
52120

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 ``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 ``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`
Output:
```71111
```

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)

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