Latest Time You Can Obtain After Replacing Characters

Latest Time You Can Obtain After Replacing Characters - Problem

String Enumeration Easy

You are given a string s representing a 12-hour format time where some of the digits (possibly none) are replaced with a "?".

12-hour times are formatted as "HH:MM", where HH is between 00 and 11, and MM is between 00 and 59.

The earliest 12-hour time is 00:00, and the latest is 11:59.

You have to replace all the "?" characters in s with digits such that the time we obtain by the resulting string is a valid 12-hour format time and is the latest possible.

Return the resulting string.

Input & Output

Example 1 — Basic Case

$ Input: s = "1?:?9"

› Output: "11:59"

💡 Note: Replace first '?' with '1' to get maximum hour 11, and second '?' with '5' to get maximum minute 59. Result: 11:59 is the latest valid time.

Example 2 — All Question Marks

$ Input: s = "??:??"

› Output: "11:59"

💡 Note: All positions are flexible. Choose maximum valid digits: hour = 11 (max hour), minute = 59 (max minute). Result: 11:59.

Example 3 — Hour Constraint

$ Input: s = "0?:??"

› Output: "09:59"

💡 Note: First hour digit is 0, so second can be at most 9 (giving 09). Minutes can be maximized to 59. Result: 09:59.

Constraints

s.length == 5
s[2] == ':'
All other characters in s are digits or '?'

Visualization

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Asked in

G Google 15 a Amazon 12

The key insight is to use a greedy approach, choosing the maximum valid digit for each position. The greedy approach works optimally because each position has clear constraints. Time: O(1), Space: O(1).

Common Approaches

✓ Binary Search Answer

⏱️ Time: N/A Space: N/A

Brute Force - Try All Combinations

⏱️ Time: O(10^k) Space: O(1)

Replace each '?' with digits 0-9, generate all combinations, validate each time, and return the latest valid one. This involves checking all possible replacements systematically.

Greedy Approach - Position by Position

⏱️ Time: O(1) Space: O(1)

Work through each position systematically, choosing the maximum valid digit that maintains time validity. This greedy approach ensures we get the latest possible time without generating all combinations.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char* solution(char* s) {
    char* result = (char*)malloc(6);
    strcpy(result, s);
    
    // Replace ? in first hour digit (position 0)
    if (result[0] == '?') {
        if (result[1] == '?' || result[1] <= '1') {
            result[0] = '1';  // Can be 1 if second digit allows
        } else {
            result[0] = '0';  // Must be 0 if second digit is > 1
        }
    }
    
    // Replace ? in second hour digit (position 1)
    if (result[1] == '?') {
        if (result[0] == '1') {
            result[1] = '1';  // 11 is max for 12-hour format
        } else {
            result[1] = '9';  // 09 is max when first digit is 0
        }
    }
    
    // Replace ? in first minute digit (position 3)
    if (result[3] == '?') {
        result[3] = '5';  // Max first minute digit is 5 (59 minutes max)
    }
    
    // Replace ? in second minute digit (position 4)
    if (result[4] == '?') {
        result[4] = '9';  // Max second minute digit is 9
    }
    
    return result;
}

int main() {
    char s[100001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    char* result = solution(s);
    printf("%s\n", result);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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