Possible timings

Given a one/two digit timing, calculate the possibilities of occurrence of other timings(including the glowing one) with relevant to glowing segments, assuming some segments may not be glowing.
Displaying the numbers is done using seven segment display. It is guaranteed that the sticks currently displaying are working fine..
Segment display
Examples:

Input : 78
Output :5

Input :05
Output :8

Explanation:
Example 1: 7 can be replaced by 5 different numbers 9, 3, 8, 0 and 7(if none of the segment is broken) and 8 can be replaced by only 1 number i.e 8 itself(if none of the segment is broken), therefore the answer is 5*1=5.
Example 2: 0 can be replaced by 2 numbers, 8 and 0, while 5 can be replaced by 4 different numbers. So, answer is 4*2=8.

Approach :
Calculate for every number from 09 what all digits are possible by adding or
removing exactly one rod from the display. Store this in an array and the answer will be product of the array value of both the digits of the input.

Below is the implementation of above approach:

C++

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// CPP program to calculate possible
// number of timings
#include <bits/stdc++.h>
using namespace std;
  
// Array storing different numbers
// of digits a particular digit 
// can be replaced with
int num[10] = { 2, 7, 2, 3, 3, 
                4, 2, 5, 1, 2 };
  
// Function performing calculations
void possibleTimings(string n)
{
    cout << num[n[0] - '0'] * 
            num[n[1] - '0'] << endl;
}
  
// Driver function
int main()
{
    string n = "05";
  
    // Calling function
    possibleTimings(n);
  
    return 0;
}

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Java

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// Java program to calculate
// possible timings.
import java.io.*;
  
class Calci {
  
    // Array storing different
    // numbers of digits a particular
    // digit can be replaced with
    static int num[] = { 2, 7, 2, 3, 3,
                         4, 2, 5, 1, 2 };
  
    // Function performing calculations
    public static void possibleTimings(String n)
    {
        System.out.println(num[(n.charAt(0) - '0')]
                          * num[n.charAt(1) - '0']);
    }
  
    // Driver function
    public static void main(String args[])
    {
        String n = "05";
  
        // Calling function
        possibleTimings(n);
    }
}

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Python3

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# python3 program to calculate possible
# number of timings
  
# Array storing different numbers
# of digits a particular digit 
# can be replaced with
num = [ 2,7,2,3,3,4,2,5,1,2 ]
  
# Function performing calculations
def possibleTimings(n):
  
    print(num[int(n[0]) - int('0')] * num[int(n[1]) - int('0')] )
  
  
# Driver function
n = "05"
  
# Calling function
possibleTimings(n)
  
# This code is contributed
# by Smitha Dinesh Semwal

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C#

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// C# program to calculate
// possible timings.
using System;
  
class Calci {
  
    // Array storing different
    // numbers of digits a particular
    // digit can be replaced with
    static int []num = { 2, 7, 2, 3, 3,
                        4, 2, 5, 1, 2 };
  
    // Function performing calculations
    public static void possibleTimings(string n)
    {
        Console.WriteLine(num[(n[0] - '0')]
                        * num[n[1] - '0']);
    }
  
    // Driver function
    public static void Main()
    {
        string n = "05";
  
        // Calling function
        possibleTimings(n);
    }
}
  
// This code is contributed by vt_m.

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PHP

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<?php
// PHP program to calculate 
// possible number of timings
  
// Array storing different 
// numbers of digits a
// particular digit can be
// replaced with
$num = array(2, 7, 2, 3, 3, 
             4, 2, 5, 1, 2);
  
// Function performing
// calculations
function possibleTimings($n)
{
    global $num;
    echo ($num[$n[0] - '0'] * 
          $num[$n[1] - '0']. "\n");
}
  
// Driver Code
$n = "05";
  
// Calling function
possibleTimings($n);
  
// This code is contributed by 
// Manish Shaw(manishshaw1)
?>

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Output:

8


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