Related Articles

# Maximize the missing values in given time in HH:MM format

• Last Updated : 23 Jul, 2021

Given a string S representing the time in the 24-hour format “HH:MM” such that some digits are represented by ‘?’, the task is to replace ‘?’ with any possible digits such that the resultant time is the maximum possible time.

Examples:

Input: S = “?4:5?”
Output: 14:59
Explanation:
After replacing the first and the second ‘?’ with the digits 1 and 9, modifies the given time to “14:59”, which is maximum among all the possible time that can be made by replacing ‘?’.

Input: S = “0?:??”
Output: 09:59

Approach: The given problem can be solved by traversing the given string S and replace the ‘?’ in such a way that the substring before the character ‘:’ lies over the range [0, 23] and the substring after the ‘:’ must be at most 59 and print the maximum time obtained. Follow the below steps to solve the given problem:

• If the value of the character S at the index 0 is ‘?’ and the character at the index 1 is ‘3’ or ‘?’, then update the value of S[0] as ‘2’. Otherwise, update the value of S[0] as ‘1’.
• If the value of the character S at the index 1 is ‘?’ and the character at the index 0 is not ‘2’, then update the value of S[1] as ‘9’. Otherwise, update the value of S[1] as ‘3’.
• If the value of the character S at the index 3 is ‘?’, then update the value of S[3] as ‘5’.
• If the value of the character S at the index 4 is ‘?’, then update the value of S[4] as ‘9’.
• After completing the above steps, print the value of the string S as the resultant time.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;``// Function to find the maximum time``// by replacing '?' by any digits``void` `maxTime(string s)``{``    ``// Convert the string to the``    ``// character array` `    ``// If the 0th index is '?'``    ``if` `(s[0] == ``'?'``) {``        ``if` `(s[1] <= ``'3'` `|| s[1] == ``'?'``)``            ``s[0] = ``'2'``;``        ``else``            ``s[0] = ``'1'``;``    ``}` `    ``// If the 1st index is '?'``    ``if` `(s[1] == ``'?'``) {``        ``if` `(s[0] != ``'2'``) {``            ``s[1] = 9;``        ``}``        ``else``            ``s[1] = 3;``    ``}` `    ``// If the 3rd index is '?'``    ``if` `(s[3] == ``'?'``)``        ``s[3] = ``'5'``;` `    ``// If the 4th index is '?'``    ``if` `(s[4] == ``'?'``)``        ``s[4] = ``'9'``;` `    ``// Return new string``    ``cout << s << endl;``}` `// Driver Code``int` `main()``{``    ``string S = ``"?4:5?"``;``    ``maxTime(S);``    ``return` `0;``}``// This code is contributed by Potta Lokesh`

## Java

 `// Java program for the above approach``public` `class` `Main {` `    ``// Function to find the maximum time``    ``// by replacing '?' by any digits``    ``public` `static` `void` `maxTime(String S)``    ``{``      ` `        ``// Convert the string to the``        ``// character array``        ``char``[] s = S.toCharArray();` `        ``// If the 0th index is '?'``        ``if` `(s[``0``] == ``'?'``) {``            ``if` `(s[``1``] <= ``'3'` `|| s[``1``] == ``'?'``)``                ``s[``0``] = ``'2'``;``            ``else``                ``s[``0``] = ``'1'``;``        ``}` `        ``// If the 1st index is '?'``        ``if` `(s[``1``] == ``'?'``) {``            ``if` `(s[``0``] != ``'2'``) {``                ``s[``1``] = ``9``;``            ``}``            ``else``                ``s[``1``] = ``3``;``        ``}` `        ``// If the 3rd index is '?'``        ``if` `(s[``3``] == ``'?'``)``            ``s[``3``] = ``'5'``;` `        ``// If the 4th index is '?'``        ``if` `(s[``4``] == ``'?'``)``            ``s[``4``] = ``'9'``;` `        ``// Return new string``        ``System.out.println(``            ``new` `String(s));``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String S = ``"?4:5?"``;``        ``maxTime(S);``    ``}``}` `// This code is contributed by lokeshpotta20.`

## Python3

 `# Python3 program for the above approach` `# Function to find the maximum time``# by replacing '?' by any digits``def` `maxTime(s):``    ` `    ``# Convert the string to the``    ``# character array` `    ``# If the 0th index is '?'``    ``s ``=` `list``(s)``    ` `    ``if` `(s[``0``] ``=``=` `'?'``):``        ``if` `(s[``1``] <``=` `'3'` `or` `s[``1``] ``=``=` `'?'``):``            ``s[``0``] ``=` `'2'``        ``else``:``            ``s[``0``] ``=` `'1'``    ` `    ``# If the 1st index is '?'``    ``if` `(s[``1``] ``=``=` `'?'``):``        ``if` `(s[``0``] !``=` `'2'``):``            ``s[``1``] ``=` `9``        ``else``:``            ``s[``1``] ``=` `3` `    ``# If the 3rd index is '?'``    ``if` `(s[``3``] ``=``=` `'?'``):``        ``s[``3``] ``=` `'5'` `    ``# If the 4th index is '?'``    ``if` `(s[``4``] ``=``=` `'?'``):``        ``s[``4``] ``=` `'9'` `    ``# Return new string``    ``print``("".join(s))` `# Driver Code``S ``=` `"?4:5?"``maxTime(S)` `# This code is contributed by _saurabh_jaiswal`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;``class` `gfg {` `    ``// Function to find the maximum time``    ``// by replacing '?' by any digits``    ``public` `static` `void` `maxTime(String S)``    ``{``      ` `        ``// Convert the string to the``        ``// character array``        ``char``[] s = S.ToCharArray();` `        ``// If the 0th index is '?'``        ``if` `(s[0] == ``'?'``) {``            ``if` `(s[1] <= ``'3'` `|| s[1] == ``'?'``)``                ``s[0] = ``'2'``;``            ``else``                ``s[0] = ``'1'``;``        ``}` `        ``// If the 1st index is '?'``        ``if` `(s[1] == ``'?'``) {``            ``if` `(s[0] != ``'2'``) {``                ``s[1] = ``'9'``;``            ``}``            ``else``                ``s[1] = ``'3'``;``        ``}` `        ``// If the 3rd index is '?'``        ``if` `(s[3] == ``'?'``)``            ``s[3] = ``'5'``;` `        ``// If the 4th index is '?'``        ``if` `(s[4] == ``'?'``)``            ``s[4] = ``'9'``;` `        ``// Return new string``        ``Console.Write(``new` `String(s));``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``String S = ``"?4:5?"``;``        ``maxTime(S);``    ``}``}` `// This code is contributed by shivanisinghss2110`

## Javascript

 ``
Output:
`14:59`

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up