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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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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Output:

22:22


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