Given a string S representing a time in 24 hours format, with '_' placed at positions of some digits, the task is to find the maximum time possible by replacing the characters '_' with any digit.
Examples:
Input: S = “0_:4_”
Output: 09:39
Explanation: Replacing the characters S[1] and S[4] with '9' modifies the string to "09:49", which is the maximum time possible.Input: S = “__:__”
Output: 23:59
Approach: The given problem can be solved by greedily selecting the digits for each '_' present in the string. Follow the steps below to solve the problem:
- If the character S[0] is equal to '_' and S[1] is either '_' or is less than 4, then assign '2' to S[0]. Otherwise, assign '1' to S[0].
- If the character S[1] is equal to '_' and S[0] is '2', then assign '3' to S[1]. Otherwise, assign '9' to S[1].
- If the character S[3] is equal to '_', then assign '5' to S[3].
- If the character S[4] is equal to '_', then assign '9' to S[4].
- After completing the above steps, print the modified string S.
Below is the implementation of the above approach:
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the maximum
// time possible by replacing
// each '_' with any digit
string maximumTime(string s)
{
// If the first character is '_'
if (s[0] == '_') {
// If s[1] is '_' or
// s[1] is less than 4
if ((s[1] == '_')
|| (s[1] >= '0'
&& s[1] < '4')) {
// Update s[0] as 2
s[0] = '2';
}
// Otherwise, update s[0] = 1
else {
s[0] = '1';
}
}
// If s[1] is equal to '_'
if (s[1] == '_') {
// If s[0] is equal to '2'
if (s[0] == '2') {
s[1] = '3';
}
// Otherwise
else {
s[1] = '9';
}
}
// If S[3] is equal to '_'
if (s[3] == '_') {
s[3] = '5';
}
// If s[4] is equal to '_'
if (s[4] == '_') {
s[4] = '9';
}
// Return the modified string
return s;
}
// Driver Code
int main()
{
string S = "0_:4_";
cout << maximumTime(S);
return 0;
}
// Java program for the above approach
class GFG{
// Function to find the maximum
// time possible by replacing
// each '_' with any digit
static void maximumTime(String str)
{
char []s = str.toCharArray();
// If the first character is '_'
if (s[0] == '_')
{
// If s[1] is '_' or
// s[1] is less than 4
if ((s[1] == '_') ||
(s[1] >= '0' && s[1] < '4'))
{
// Update s[0] as 2
s[0] = '2';
}
// Otherwise, update s[0] = 1
else
{
s[0] = '1';
}
}
// If s[1] is equal to '_'
if (s[1] == '_')
{
// If s[0] is equal to '2'
if (s[0] == '2')
{
s[1] = '3';
}
// Otherwise
else
{
s[1] = '9';
}
}
// If S[3] is equal to '_'
if (s[3] == '_')
{
s[3] = '5';
}
// If s[4] is equal to '_'
if (s[4] == '_')
{
s[4] = '9';
}
// Print the modified string
for(int i = 0; i < s.length; i++)
System.out.print(s[i]);
}
// Driver Code
static public void main (String []args)
{
String S = "0_:4_";
maximumTime(S);
}
}
// This code is contributed by AnkThon
# Python3 program for the above approach
# Function to find the maximum
# time possible by replacing
# each '_' with any digit
def maximumTime(s):
s = list(s)
# If the first character is '_'
if (s[0] == '_'):
# If s[1] is '_' or
# s[1] is less than 4
if ((s[1] == '_') or (s[1] >= '0' and
s[1] < '4')):
# Update s[0] as 2
s[0] = '2'
# Otherwise, update s[0] = 1
else:
s[0] = '1'
# If s[1] is equal to '_'
if (s[1] == '_'):
# If s[0] is equal to '2'
if (s[0] == '2'):
s[1] = '3'
# Otherwise
else:
s[1] = '9'
# If S[3] is equal to '_'
if (s[3] == '_'):
s[3] = '5'
# If s[4] is equal to '_'
if (s[4] == '_'):
s[4] = '9'
# Return the modified string
s = ''.join(s)
return s
# Driver Code
if __name__ == '__main__':
S = "0_:4_"
print(maximumTime(S))
# This code is contributed by ipg2016107
// C# program for the above approach
using System;
class GFG{
// Function to find the maximum
// time possible by replacing
// each '_' with any digit
static void maximumTime(string str)
{
char []s = str.ToCharArray();
// If the first character is '_'
if (s[0] == '_')
{
// If s[1] is '_' or
// s[1] is less than 4
if ((s[1] == '_') ||
(s[1] >= '0' && s[1] < '4'))
{
// Update s[0] as 2
s[0] = '2';
}
// Otherwise, update s[0] = 1
else
{
s[0] = '1';
}
}
// If s[1] is equal to '_'
if (s[1] == '_')
{
// If s[0] is equal to '2'
if (s[0] == '2')
{
s[1] = '3';
}
// Otherwise
else
{
s[1] = '9';
}
}
// If S[3] is equal to '_'
if (s[3] == '_')
{
s[3] = '5';
}
// If s[4] is equal to '_'
if (s[4] == '_')
{
s[4] = '9';
}
// Print the modified string
for(int i = 0; i < s.Length; i++)
Console.Write(s[i]);
}
// Driver Code
static public void Main ()
{
string S = "0_:4_";
maximumTime(S);
}
}
// This code is contributed by AnkThon
<script>
// javascript program for the above approach
// Function to find the maximum
// time possible by replacing
// each '_' with any digit
function maximumTime(str)
{
var s = str.split("");
// If the first character is '_'
if (s[0] == '_')
{
// If s[1] is '_' or
// s[1] is less than 4
if ((s[1] == '_') ||
(s[1] >= '0' && s[1] < '4'))
{
// Update s[0] as 2
s[0] = '2';
}
// Otherwise, update s[0] = 1
else
{
s[0] = '1';
}
}
// If s[1] is equal to '_'
if (s[1] == '_')
{
// If s[0] is equal to '2'
if (s[0] == '2')
{
s[1] = '3';
}
// Otherwise
else
{
s[1] = '9';
}
}
// If S[3] is equal to '_'
if (s[3] == '_')
{
s[3] = '5';
}
// If s[4] is equal to '_'
if (s[4] == '_')
{
s[4] = '9';
}
// Print the modified string
for(var i = 0; i < s.length; i++)
document.write(s[i]);
}
// Driver Code
var S = "0_:4_";
maximumTime(S);
// This code is contributed by bunnyram19.
</script>
// C program for above approach
#include <stdio.h>
#include <string.h>
// Function to find the maximum
// time possible by replacing
// each '_' with any digit
char* maximumTime(char s[])
{
// If the first character is '_'
if (s[0] == '_') {
// If s[1] is '_' or
// s[1] is less than 4
if ((s[1] == '_') || (s[1] >= '0' && s[1] < '4')) {
// Update s[0] as 2
s[0] = '2';
}
else { // Otherwise, update s[0] = 1
s[0] = '1';
}
}
// If s[1] is equal to '_'
if (s[1] == '_') {
// If s[0] is equal to '2'
if (s[0] == '2') {
s[1] = '3';
}
// otherwise
else {
s[1] = '9';
}
}
// If S[3] is equal to '_'
if (s[3] == '_') {
s[3] = '5';
}
// If s[4] is equal to '_'
if (s[4] == '_') {
s[4] = '9';
}
return s; // Return the modified string
}
int main()
{
char S[] = "0_:4_";
printf("%s", maximumTime(S));
return 0;
}
// This code is contributed by Tapesh (tapeshdua420)
Output
09:49
Time Complexity: O(1)
Auxiliary Space: O(1)