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
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) | 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 Codeint 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:
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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