Time Conversion


Problem Statement :


Given a time in 12 -hour AM/PM format, convert it to military (24-hour) time.

Note: - 12:00:00AM on a 12-hour clock is 00:00:00 on a 24-hour clock.
- 12:00:00PM on a 12-hour clock is 12:00:00 on a 24-hour clock.

Example 
s = '12 : 01: 00PM'
   Return '12:01:00'.

s = '12 : 01: 00AM'
  Return '00:01:00'.


Function Description

Complete the timeConversion function in the editor below. It should return a new string representing the input time in 24 hour format.

timeConversion has the following parameter(s):

    string s: a time in 12 hour format 

Input Format

A single string s that represents a time in 12 -hour clock format (i.e.: hh:mm:ss AM or hh:mm:ss PM)

Constraints

    All input times are valid



Solution :



title-img


                            Solution in C :

In C :

char* timeConversion(char* s) {
    char test[2];
    static char final[9];
final[0] = s[0];
final[1] = s[1];
final[2] = ':';
final[5] = ':';
final[3] = s[3];
final[4] = s[4];
final[6] = s[6];
final[7] = s[7];
   
    if(s[8] == 'A')
    {
        if(s[0]=='1' && s[1]=='2')
        {
            final[0] = '0';
            final[1] = '0';
        }
        else{
            final[0] = s[0];
            final[1] = s[1];
        }
    }


    else if(s[8]== 'P')
    {
       if(s[0] == '0' && s[1] < '8')
       {
           final[0] = '1';
           char hr = (int)s[1];
           hr = hr+2;
           final[1]  = (char)hr;
       }
       else if(s[0] == '0' && (s[1] == '8' || s[1] == '9'))
       {
           final[0] = '2';
          if(s[2] == '8') final[1] = '0';
          else final[1] = '1';
          
       }
       else if(s[0] != '0' && (s[1] == '0' || s[1] == '1'))
       {   final[0] = '2';
           if(s[1] == '0') final[1] = '2';
           else if(s[1] == '1')  final[1] = '3';
       }
    
    }//end of PM if

    

return final;
}





In Java :


import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

   	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String s = sc.nextLine(); //07:05:45PM
		DateFormat inFormat = new SimpleDateFormat( "hh:mm:ssaa");
		DateFormat outFormat = new SimpleDateFormat( "HH:mm:ss");

		Date date = null;
		try {
			date = inFormat.parse(s); 
		}catch (ParseException e ){
			e.printStackTrace();
		}
		if( date != null ){
			String myDate = outFormat.format(date);
			System.out.println(myDate);
		}
	}

}





In Python3 :


ins = input().strip()

is_pm = ins[-2:].lower() == 'pm'
time_list = list(map(int, ins[:-2].split(':')))

if is_pm and time_list[0] < 12:
    time_list[0] += 12

if not is_pm and time_list[0] == 12:
    time_list[0] = 0

print(':'.join(map(lambda x: str(x).rjust(2, '0'), time_list)))





In C ++ :


#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
   
    string s;
    string h;
    int hr;
    cin>>s;
    hr = ((s[0]-'0')*10)+(s[1]-'0');
    if(s[8]=='P'&&s[9]=='M'&& hr ==12) cout<<to_string(hr);
    else if(s[8]=='P'&&s[9]=='M') cout<<to_string(hr+12);
    else if(s[8]=='A'&&s[9]=='M'&&hr==12) cout<<"00";
    
    else cout<< s[0]<<s[1];
    
   
    for(int i =2;i<8;i++)
        cout<<s[i];
    cout<<endl;
    return 0;
}
                        








View More Similar Problems

Print the Elements of a Linked List

This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode

View Solution →

Insert a Node at the Tail of a Linked List

You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink

View Solution →

Insert a Node at the head of a Linked List

Given a pointer to the head of a linked list, insert a new node before the head. The next value in the new node should point to head and the data value should be replaced with a given value. Return a reference to the new head of the list. The head pointer given may be null meaning that the initial list is empty. Function Description: Complete the function insertNodeAtHead in the editor below

View Solution →

Insert a node at a specific position in a linked list

Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e

View Solution →

Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

View Solution →

Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

View Solution →