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Latest Time You Can Obtain After Replacing Characters
The Latest Time You Can Obtain After Replacing Characters sub-task is applied to an input string, in which the string is represented as a 12-hour format time when the maximum number of characters are replaced by '?'.
In a 12-hour format time, "HH:MM” where HH is an element from the set {00, 01, …, 10, 11} and MM is also an element from the set {00, 01, …, 59}. The earliest possible time is 00:00, and the latest time is 11:59.Problem Statement
In this problem statement, the goal is to replace all "?" characters in the string s with digits to form the latest valid time in the format "HH:MM".
Example Scenario 1
Input: s="03:??"
Output: 03:59
Replace ? with 5 and 9 to get the latest valid time 03:59.
Example Scenario 2
Input: s="0?:3?"
Output: 09:39
Replace ? with 9 and 9 to get the latest valid time 09:39.
Example Scenario 3
Input: s="1?:22"
Output: 11:22
Replace ? with 1 to get the latest valid time 11:22.
Time Complexity
The time complexity of the Latest Time You Can Obtain After Replacing Characters is O(1).
To solve this problem in various programming languages, use the following methods.
- Greedy Approach
- Brute Force Approach
Greedy Approach
Example
Following is the Greedy Approach involves replacing each '?' in the time string with the maximum possible digit that still forms a valid 12-hour format time. This ensures that we get the latest possible time.
#include <iostream>
#include <string>
using namespace std;
string latestTime(string s) {
if (s[0] == '?') s[0] = (s[1] <= '1' || s[1] == '?') ? '1' : '0';
if (s[1] == '?') s[1] = (s[0] == '1') ? '1' : '9';
if (s[3] == '?') s[3] = '5';
if (s[4] == '?') s[4] = '9';
return s;
}
int main() {
string s = "1?:?0";
cout << "Latest time = " << latestTime(s) << endl;
return 0;
}
Output
Latest time = 11:50
public class LatestTime {
public static String latestTime(String s) {
char[] time = s.toCharArray();
if (time[0] == '?') time[0] = (time[1] <= '1' || time[1] == '?') ? '1' : '0';
if (time[1] == '?') time[1] = (time[0] == '1') ? '1' : '9';
if (time[3] == '?') time[3] = '5';
if (time[4] == '?') time[4] = '9';
return new String(time);
}
public static void main(String[] args) {
String s = "1?:?0";
System.out.println("Latest time = " + latestTime(s));
}
}
Output
Latest time = 11:50
def latest_time(s):
time = list(s)
if time[0] == '?':
time[0] = '1' if time[1] <= '1' or time[1] == '?' else '0'
if time[1] == '?':
time[1] = '1' if time[0] == '1' else '9'
if time[3] == '?':
time[3] = '5'
if time[4] == '?':
time[4] = '9'
return ''.join(time)
s = "1?:?0"
print(f"Latest time = {latest_time(s)}")
Output
Latest time = 11:50
Brute Force Approach
Example
The Brute Force Approach is used to generate all possible valid times in the 12-hour format and checking which one matches the given pattern with '?' characters.
#include <iostream>
#include <string>
using namespace std;
bool isValidTime(const string& time) {
int hours = stoi(time.substr(0, 2));
int minutes = stoi(time.substr(3, 2));
return hours >= 0 && hours <= 11 && minutes >= 0 && minutes <= 59;
}
string latestTime(string s) {
for (int h = 11; h >= 0; --h) {
for (int m = 59; m >= 0; --m) {
string hh = (h < 10 ? "0" : "") + to_string(h);
string mm = (m < 10 ? "0" : "") + to_string(m);
string candidate = hh + ":" + mm;
bool match = true;
for (int i = 0; i < 5; ++i) {
if (s[i] != '?' && s[i] != candidate[i]) {
match = false;
break;
}
}
if (match) return candidate;
}
}
return "";
}
int main() {
string s = "1?:?0";
cout << "Latest time = " << latestTime(s) << endl;
return 0;
}
Output
Latest time = 11:50
public class LatestTime {
public static boolean isValidTime(String time) {
int hours = Integer.parseInt(time.substring(0, 2));
int minutes = Integer.parseInt(time.substring(3, 5));
return hours >= 0 && hours <= 11 && minutes >= 0 && minutes <= 59;
}
public static String latestTime(String s) {
for (int h = 11; h >= 0; --h) {
for (int m = 59; m >= 0; --m) {
String hh = (h < 10 ? "0" : "") + h;
String mm = (m < 10 ? "0" : "") + m;
String candidate = hh + ":" + mm;
boolean match = true;
for (int i = 0; i < 5; ++i) {
if (s.charAt(i) != '?' && s.charAt(i) != candidate.charAt(i)) {
match = false;
break;
}
}
if (match) return candidate;
}
}
return "";
}
public static void main(String[] args) {
String s = "1?:?0";
System.out.println("Latest time = " + latestTime(s));
}
}
Output
Latest time = 11:50
import re
def is_valid_time(time):
hours, minutes = map(int, time.split(':'))
return 0 <= hours <= 11 and 0 <= minutes <= 59
def latest_time(s):
for hours in range(11, -1, -1):
for minutes in range(59, -1, -1):
hh = f"{hours:02d}"
mm = f"{minutes:02d}"
candidate = f"{hh}:{mm}"
match = True
for i in range(5):
if s[i] != '?' and s[i] != candidate[i]:
match = False
break
if match:
return candidate
return ""
s = "1?:?0"
print(f"Latest time: {latest_time(s)}")
Output
Latest time = 11:50
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