Find the time which is palindromic and comes after the given time

Given a string str which stores the time in the 24 hours format as HH:MM such that 0 ≤ HH ≤ 23 and 0 ≤ MM ≤ 59. The task is to find the next closest time which is a palindrome when read as a string. If no such string exists then print -1.

Examples:

Input: str = “21:12”
Output: 22:22
The only palindromic time possible in the given hour is 21:12
but it is not greater than the given time so the output will be
the palindromic time in the next hour i.e. 22:22



Input: str = “23:32”
Output: -1

Approach: There are three possible cases:

  1. If MM < reverse(HH) then the output will be HH as hours and reverse(HH) as minutes.
  2. If HH = 23 and MM ≥ 32 then output will be -1.
  3. Else output will be HH + 1 as hours and reverse(HH + 1) as minutes.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
  
// Function to return the required time
string getTime(string s, int n)
{
    // To store the resultant time
    string res;
  
    // Hours are stored in h as integer
    int h = stoi(s.substr(0, 2));
  
    // Minutes are stored in m as integer
    int m = stoi(s.substr(3, 2));
  
    // Reverse of h
    int rev_h = (h % 10) * 10 + ((h % 100) - (h % 10)) / 10;
  
    // Reverse of h as a string
    string rev_hs = to_string(rev_h);
  
    if (h == 23 && m >= 32) {
        res = "-1";
    }
  
    // If MM < reverse of (HH)
    else if (m < rev_h) {
        string temp;
  
        // 0 is added if HH < 10
        if (h < 10)
            temp = "0";
        temp = temp + to_string(h);
  
        // 0 is added if rev_h < 10
        if (rev_h < 10)
            res = res + temp + ":0" + rev_hs;
        else
            res = res + temp + ":" + rev_hs;
    }
    else {
  
        // Increment hours
        h++;
  
        // Reverse of the hour after incrementing 1
        rev_h = (h % 10) * 10 + ((h % 100) - (h % 10)) / 10;
        rev_hs = to_string(rev_h);
  
        string temp;
  
        // 0 is added if HH < 10
        if (h < 10)
            temp = "0";
        temp = temp + to_string(h);
  
        // 0 is added if rev_h < 10
        if (rev_h < 10)
            res = res + temp + ":0" + rev_hs;
        else
            res = res + temp + ":" + rev_hs;
    }
    return res;
}
  
// Driver code
int main()
{
    string s = "21:12";
    int n = s.length();
  
    cout << getTime(s, n);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG 
{
  
    // Function to return the required time
    static String getTime(String s, int n) 
    {
  
        // To store the resultant time
        String res = "";
  
        // Hours are stored in h as integer
        int h = Integer.parseInt(s.substring(0, 0 + 2));
  
        // Minutes are stored in m as integer
        int m = Integer.parseInt(s.substring(3, 3 + 2));
  
        // Reverse of h
        int rev_h = (h % 10) * 10
                   ((h % 100) - (h % 10)) / 10;
  
        // Reverse of h as a string
        String rev_hs = Integer.toString(rev_h);
        if (h == 23 && m >= 32)
        {
            res = "-1";
        }
          
        // If MM < reverse of (HH)
        else if (m < rev_h)
        {
            String temp = "";
  
            // 0 is added if HH < 10
            if (h < 10)
                temp = "0";
            temp = temp + Integer.toString(h);
  
            // 0 is added if rev_h < 10
            if (rev_h < 10)
                res = res + temp + ":0" + rev_hs;
            else
                res = res + temp + ":" + rev_hs;
        
        else
        {
  
            // Increment hours
            h++;
  
            // Reverse of the hour after incrementing 1
            rev_h = (h % 10) * 10 + ((h % 100) - 
                    (h % 10)) / 10;
            rev_hs = Integer.toString(rev_h);
  
            String temp = "";
  
            // 0 is added if HH < 10
            if (h < 10)
                temp = "0";
            temp = temp + Integer.toString(h);
  
            // 0 is added if rev_h < 10
            if (rev_h < 10)
                res = res + temp + ":0" + rev_hs;
            else
                res = res + temp + ":" + rev_hs;
        }
        return res;
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        String s = "21:12";
        int n = s.length();
        System.out.println(getTime(s, n));
    }
}
  
// This code is contributed by
// sanjeev2552

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Python3

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# Python3 implementation of the approach 
  
# Function to return the required time 
def getTime(s, n) : 
  
    # Hours are stored in h as integer 
    h = int(s[0 : 2]); 
  
    # Minutes are stored in m as integer 
    m = int(s[3 : 5]); 
  
    # Reverse of h 
    rev_h = (h % 10) * 10 + ((h % 100) - (h % 10)) // 10
  
    # Reverse of h as a string 
    rev_hs = str(rev_h)
      
    temp = ""
    res  = ""
      
    if (h == 23 and m >= 32) :
        res = "-1"
      
  
    # If MM < reverse of (HH) 
    elif (m < rev_h) : 
  
        # 0 is added if HH < 10 
        if (h < 10) :
            temp = "0"
              
        temp = temp + str(h); 
  
        # 0 is added if rev_h < 10 
        if (rev_h < 10) :
            res = res + temp + ":0" + rev_hs; 
        else :
            res = res + temp + ":" + rev_hs; 
      
    else :
          
        # Increment hours 
        h += 1
  
        # Reverse of the hour after incrementing 1 
        rev_h = (h % 10) * 10 + ((h % 100) - (h % 10)) //10
          
        rev_hs = str(rev_h); 
  
        # 0 is added if HH < 10 
        if (h < 10) :
            temp = "0"
              
        temp = temp + str(h); 
  
        # 0 is added if rev_h < 10 
        if (rev_h < 10) :
            res = res + temp + ":0" + rev_hs; 
        else :
            res = res + temp + ":" + rev_hs; 
      
    return res; 
  
  
# Driver code 
if __name__ == "__main__"
  
    s = "21:12"
    n = len(s); 
  
    print(getTime(s, n)); 
  
    # This code is contributed by AnkitRai01

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PHP

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<?php
  
//PHP implementation of the approach
   
// Function to return the required time
function getTime( $s, $n)
{
    // To store the resultant time
    $res="";
   
    // Hours are stored in h as integer
    $h = intval($s.substr(0, 2));
   
    // Minutes are stored in m as integer
    $m = intval($s.substr(3, 2));
   
    // Reverse of h
    $rev_h = ($h % 10) * 10 + (($h % 100) - ($h % 10)) / 10;
   
    // Reverse of h as a string
    $rev_hs = strval($rev_h);
   
    if ($h == 23 && $m >= 32) {
        $res = "-1";
    }
   
    // If MM < reverse of (HH)
    else if ($m < $rev_h) {
        $temp="";
   
        // 0 is added if HH < 10
        if ($h < 10)
            $temp = "0";
        $temp = $temp . strval($h);
   
        // 0 is added if rev_h < 10
        if ($rev_h < 10)
            $res = $res . $temp . ":0" . $rev_hs;
        else
            $res = $res. $temp .":" . $rev_hs;
    }
    else {
   
        // Increment hours
        $h++;
   
        // Reverse of the hour after incrementing 1
        $rev_h = ($h % 10) * 10 + (($h % 100) - ($h % 10)) / 10;
        $rev_hs = strval($rev_h);
   
        $temp="";
   
        // 0 is added if HH < 10
        if ($h < 10)
            $temp = "0";
        $temp = $temp . strval($h);
   
        // 0 is added if rev_h < 10
        if ($rev_h < 10)
            $res = $res . $temp . ":0" . $rev_hs;
        else
            $res = $res . $temp . ":" . $rev_hs;
    }
    return $res;
}
   
// Driver code
  
    $s = "21:12";
    $n = strlen($s);
   
    echo getTime($s, $n);
   
    return 0;
  
// This code is contributed by ChitraNayal
?>

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Output:

22:22


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