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++
// 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 digitstring 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 Codeint main(){ string S = "0_:4_"; cout << maximumTime(S); return 0;} |
Java
// Java program for the above approachclass GFG{ // Function to find the maximum// time possible by replacing// each '_' with any digitstatic 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 Codestatic public void main (String []args){ String S = "0_:4_"; maximumTime(S);}}// This code is contributed by AnkThon |
Python3
# Python3 program for the above approach# Function to find the maximum# time possible by replacing# each '_' with any digitdef 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 Codeif __name__ == '__main__': S = "0_:4_" print(maximumTime(S)) # This code is contributed by ipg2016107 |
C#
// C# program for the above approachusing System;class GFG{ // Function to find the maximum// time possible by replacing// each '_' with any digitstatic 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 Codestatic public void Main (){ string S = "0_:4_"; maximumTime(S);}}// This code is contributed by AnkThon |
Javascript
<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
// C program for above approach#include <stdio.h>#include <string.h>// Function to find the maximum// time possible by replacing// each '_' with any digitchar* 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)
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