Given two strings T1 and T2 representing two time instances in 24-hour formats, the task is to convert T1 to T2 using a minimum number of time slots among 1, 5, 15, or 60 minutes.
24-hour times are formatted as “HH: MM”, where HH is between 00 and 23, and MM is between 00 and 59. The earliest 24-hour time is 00:00, and the latest is 23:59.
Examples:
Input: T1 = “02:30”, T2 = “04:35”
Output: 3
Explanation: The time difference between them is 2 hours and 5 minutes.
Increase T1 by 60 minutes 2 times and 1 time by 5 minutes.Input: T1 = “11:00”, T2 = “11:01”
Output: 1
Explanation: Increased the time by 1 minutes for once.
Approach: This problem can be solved using the Greedy approach based on the following idea.
To minimize the required slots, use the ones with the highest time first and then go for the smaller ones, i.e. in the order 60, 15, 5, 1. with 60 used first and 1 used and last
Follow the below steps to implement the above idea:
- Convert strings T1 and T2 to their respective time and store them in minutes (say time1_to_min and time2_to_min).
- Initialise a variable to store the number of time slots (say operations).
- Run a until time1_to_min != time2_to_min
- Find the difference between the times (say diff).
- If diff ≥ 60 then decrement 60 from time2_to_min and increment operations by 1.
- If diff ≥ 15 and diff < 60 then decrement 15 from time2_to_min and increment operations by 1.
- If diff ≥ 5 and diff < 15 then decrement 5 from time2_to_min and increment operations by 1.
- Else decrement 1 from time2_to_min and increment operations by 1.
- Return operations as that is the minimum number of time slots required.
Below is the implementation of the above approach.
C++
// C++ code to implement the approach#include <bits/stdc++.h>using namespace std;// Function for calculating the minimum// number of operation required to// convert time1 to time2int convertTime(string T1, string T2){ // Converting strings into integer int hour_time1 = stoi(T1.substr(0, 2)); int min_time1 = stoi(T1.substr(3, 4)); int hour_time2 = stoi(T2.substr(0, 2)); int min_time2 = stoi(T2.substr(3, 4)); // Converting time into minutes int time1_to_min = (hour_time1 * 60) + min_time1; int time2_to_min = (hour_time2 * 60) + min_time2; int operations = 0; while (time1_to_min != time2_to_min) { int diff = time2_to_min - time1_to_min; if (diff >= 60) { time2_to_min -= 60; operations++; } else if (diff >= 15 && diff < 60) { time2_to_min -= 15; operations++; } else if (diff >= 5 && diff < 15) { time2_to_min -= 5; operations++; } else { time2_to_min -= 1; operations++; } } return operations;}// Driver codeint main(){ // Input two strings string T1 = "02:30"; string T2 = "04:35"; // Function call int result = convertTime(T1, T2); cout << " " << result << endl; return 0;} |
Java
// Java code to implement the approachimport java.util.*;public class GFG { // Function for calculating the minimum // number of operation required to // convert time1 to time2 static int convertTime(String T1, String T2) { // Converting strings into integer int hour_time1 = Integer.parseInt(T1.substring(0, 2)); int min_time1 = Integer.parseInt(T1.substring(3, 5)); int hour_time2 = Integer.parseInt(T2.substring(0, 2)); int min_time2 = Integer.parseInt(T2.substring(3, 5)); // Converting time into minutes int time1_to_min = (hour_time1 * 60) + min_time1; int time2_to_min = (hour_time2 * 60) + min_time2; int operations = 0; while (time1_to_min != time2_to_min) { int diff = time2_to_min - time1_to_min; if (diff >= 60) { time2_to_min -= 60; operations++; } else if (diff >= 15 && diff < 60) { time2_to_min -= 15; operations++; } else if (diff >= 5 && diff < 15) { time2_to_min -= 5; operations++; } else { time2_to_min -= 1; operations++; } } return operations; } // Driver code public static void main(String args[]) { // Input two strings String T1 = "02:30"; String T2 = "04:35"; // Function call int result = convertTime(T1, T2); System.out.println(" " + result); }}// This code is contributed by Samim Hossain Mondal. |
Python3
# Python3 code to implement the above approach# Function for calculating the minimum# number of operation required to convert# time1 to time2def convertTime(T1, T2): # Converting strings into integer hour_time1 = int(T1[0 : 2]) min_time1 = int(T1[3 : 3 + 4]) hour_time2 = int(T2[0 : 2]) min_time2 = int(T2[3 : 3 + 4]) # convertin time into minutes time1_to_min = hour_time1 * 60 + min_time1 time2_to_min = hour_time2 * 60 + min_time2 operations = 0 while time1_to_min != time2_to_min: diff = time2_to_min - time1_to_min if diff >= 60: time2_to_min -= 60 operations += 1 elif diff >= 15 and diff < 60: time2_to_min -= 15 operations += 1 elif diff >= 5 and diff < 15: time2_to_min -= 5 operations += 1 else: time2_to_min -= 1 operations += 1 return operations# Driver Code# Input two stringsT1 = "02:30"T2 = "04:35"# Function Callresult = convertTime(T1, T2)print(result) # This code is contributed by phasing17 |
C#
// C# code to implement the approachusing System;public class GFG{ // Function for calculating the minimum // number of operation required to // convert time1 to time2 static int convertTime(String T1, String T2) { // Converting strings into integer int hour_time1 = Int32.Parse(T1.Substring(0, 2)); int min_time1 = Int32.Parse(T1.Substring(3, 2)); int hour_time2 = Int32.Parse(T2.Substring(0, 2)); int min_time2 = Int32.Parse(T2.Substring(3, 2)); // Converting time into minutes int time1_to_min = (hour_time1 * 60) + min_time1; int time2_to_min = (hour_time2 * 60) + min_time2; int operations = 0; while (time1_to_min != time2_to_min) { int diff = time2_to_min - time1_to_min; if (diff >= 60) { time2_to_min -= 60; operations++; } else if (diff >= 15 && diff < 60) { time2_to_min -= 15; operations++; } else if (diff >= 5 && diff < 15) { time2_to_min -= 5; operations++; } else { time2_to_min -= 1; operations++; } } return operations; } //Driver Code public static void Main(string[] args) { // Input two strings string T1 = "02:30"; string T2 = "04:35"; // Function call int result = convertTime(T1, T2); Console.WriteLine(" " + result); }}//this code is contributed by phasing17 |
Javascript
<script> // JavaScript code to implement the approach // Function for calculating the minimum // number of operation required to // convert time1 to time2 const convertTime = (T1, T2) => { // Converting strings into integer let hour_time1 = parseInt(T1.substring(0, 2)); let min_time1 = parseInt(T1.substring(3, 3 + 4)); let hour_time2 = parseInt(T2.substring(0, 2)); let min_time2 = parseInt(T2.substring(3, 3 + 4)); // Converting time into minutes let time1_to_min = (hour_time1 * 60) + min_time1; let time2_to_min = (hour_time2 * 60) + min_time2; let operations = 0; while (time1_to_min != time2_to_min) { let diff = time2_to_min - time1_to_min; if (diff >= 60) { time2_to_min -= 60; operations++; } else if (diff >= 15 && diff < 60) { time2_to_min -= 15; operations++; } else if (diff >= 5 && diff < 15) { time2_to_min -= 5; operations++; } else { time2_to_min -= 1; operations++; } } return operations; } // Driver code // Input two strings let T1 = "02:30"; let T2 = "04:35"; // Function call let result = convertTime(T1, T2); document.write(` ${result}`);// This code is contributed by rakeshsahni</script> |
Output
3
Time Complexity: O(M) where M is the time difference
Auxiliary Space: O(1)
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