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Minimum minutes needed to make the time palindromic

  • Difficulty Level : Easy
  • Last Updated : 11 May, 2021
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Given string str which stores the time in the 24 hours format as “HH: MM”. The task is to find the minimum minutes that need to be added to make time palindromic.

Examples:

Input: str = “05:39” 
Output: 11 
Explanation: It takes 11 minutes for minute value to become 50, 05:50 is a palindromic time

Examples: 
Input: str = “13:31” 
Output:
Explanation: Since, 13:31 is already palindromic therefore, 0 minutes is required

Approach: 
The idea is to greedily increment the minute value until the time value becomes palindrome. Run a while loop to increment minute value and simultaneously check if hour value and minute value form a palindrome or not. 
While incrementing minute and hour values make sure to check for the base condition when the minute value is 60 and the hour value is 24.
Below is the implementation of the above approach :



C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to get the required minutes
int get_palindrome_time(string str)
{
    int hh, mm;
 
    // Storing hour and minute value
    // in integral form
    hh
        = (str[0] - 48) * 10
        + (str[1] - 48);
    mm
        = (str[3] - 48) * 10
        + (str[4] - 48);
 
    int requiredTime = 0;
 
    // Keep iterating till first digit
    // hour becomes equal to second
    // digit of minute and second digit
    // of hour becomes equal to first
    // digit of minute
    while (hh % 10 != mm / 10
        || hh / 10 != mm % 10) {
 
        ++mm;
 
        // If mins is 60, increase hour, and
        // reinitilialized to 0
        if (mm == 60) {
            mm = 0;
            ++hh;
        }
 
        // If hours is 60, reinitialized to 0
        if (hh == 24)
            hh = 0;
        ++requiredTime;
    }
 
    // Return the required time
    return requiredTime;
}
 
// Driver Code
int main()
{
    // Given Time as a string
    string str = "05:39";
 
    // Function Call
    cout << get_palindrome_time(str)
        << endl;
}

Java




// Java program for the above approach
class GFG{
     
// Function to get the required minutes
public static int get_palindrome_time(String str)
{
    int hh, mm;
 
    // Storing hour and minute value
    // in integral form
    hh = (str.charAt(0) - 48) * 10 +
        (str.charAt(1) - 48);
    mm = (str.charAt(3) - 48) * 10 +
        (str.charAt(4) - 48);
 
    int requiredTime = 0;
 
    // Keep iterating till first digit
    // hour becomes equal to second
    // digit of minute and second digit
    // of hour becomes equal to first
    // digit of minute
    while (hh % 10 != mm / 10 ||
        hh / 10 != mm % 10)
    {
        ++mm;
 
        // If mins is 60, increase hour, and
        // reinitilialized to 0
        if (mm == 60)
        {
            mm = 0;
            ++hh;
        }
 
        // If hours is 60, reinitialized to 0
        if (hh == 24)
            hh = 0;
        ++requiredTime;
    }
 
    // Return the required time
    return requiredTime;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given Time as a string
    String str = "05:39";
 
    // Function Call
    System.out.println(get_palindrome_time(str));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3




# Python3 program for the above approach
 
# Function to get the required minutes 
def get_palindrome_time(str):
     
    # Storing hour and minute value
    # in integral form
    hh = ((ord(str[0]) - 48) * 10 +
          (ord(str[1]) - 48))
    mm = ((ord(str[3]) - 48) * 10 +
          (ord(str[4]) - 48))
     
    requiredTime = 0
 
    # Keep iterating till first digit
    # hour becomes equal to second
    # digit of minute and second digit
    # of hour becomes equal to first
    # digit of minute
    while (hh % 10 != mm // 10 or
          hh // 10 != mm % 10):
        mm += 1
 
        # If mins is 60, increase hour, and
        # reinitilialized to 0
        if (mm == 60):
            mm = 0
            hh += 1
 
        # If hours is 60, reinitialized to 0
        if (hh == 24):
            hh = 0
             
        requiredTime += 1;
         
    # Return the required time
    return requiredTime
     
if __name__=="__main__":
     
    # Given Time as a string
    str = "05:39";
 
    # Function call
    print(get_palindrome_time(str));
 
# This code is contributed by rutvik_56

C#




// C# program for the above approach
using System;
 
class GFG{
     
// Function to get the required minutes
public static int get_palindrome_time(string str)
{
    int hh, mm;
 
    // Storing hour and minute value
    // in integral form
    hh = (str[0] - 48) * 10 +
        (str[1] - 48);
    mm = (str[3] - 48) * 10 +
        (str[4] - 48);
 
    int requiredTime = 0;
 
    // Keep iterating till first digit
    // hour becomes equal to second
    // digit of minute and second digit
    // of hour becomes equal to first
    // digit of minute
    while (hh % 10 != mm / 10 ||
        hh / 10 != mm % 10)
    {
        ++mm;
 
        // If mins is 60, increase hour,
        // and reinitilialized to 0
        if (mm == 60)
        {
            mm = 0;
            ++hh;
        }
 
        // If hours is 60, reinitialized to 0
        if (hh == 24)
            hh = 0;
        ++requiredTime;
    }
 
    // Return the required time
    return requiredTime;
}
 
// Driver code
public static void Main(string[] args)
{
     
    // Given Time as a string
    string str = "05:39";
 
    // Function Call
    Console.Write(get_palindrome_time(str));
}
}
 
// This code is contributed by rutvik_56

Javascript




<script>
 
// Javascript program for the above approach
  
// Function to get the required minutes
function get_palindrome_time(str)
{
    let hh, mm;
  
    // Storing hour and minute value
    // in letegral form
    hh = (str[0].charCodeAt() - 48) * 10 +
        (str[1].charCodeAt() - 48);
    mm = (str[3].charCodeAt() - 48) * 10 +
        (str[4].charCodeAt() - 48);
  
    let requiredTime = 0;
  
    // Keep iterating till first digit
    // hour becomes equal to second
    // digit of minute and second digit
    // of hour becomes equal to first
    // digit of minute
    while (hh % 10 != Math.floor(mm / 10) ||
        Math.floor(hh / 10) != mm % 10)
    {
        ++mm;
  
        // If mins is 60, increase hour, and
        // reinitilialized to 0
        if (mm == 60)
        {
            mm = 0;
            ++hh;
        }
  
        // If hours is 60, reinitialized to 0
        if (hh == 24)
            hh = 0;
        ++requiredTime;
    }
  
    // Return the required time
    return requiredTime;
}
// Driver Code
 
    // Given Time as a string
    let str = "05:39";
  
    // Function Call
    document.write(get_palindrome_time(str.split('')));
 
</script>
Output: 
11

Time Complexity: O(1) 
Auxiliary Space: O(1)
 

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