LeetCode #3114 — EASY

Latest Time You Can Obtain After Replacing Characters

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

Problem Statement

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.

Example 1:

Input: s = "1?:?4"

Output: "11:54"

Explanation: The latest 12-hour format time we can achieve by replacing "?" characters is "11:54".

Example 2:

Input: s = "0?:5?"

Output: "09:59"

Explanation: The latest 12-hour format time we can achieve by replacing "?" characters is "09:59".

Constraints:

s.length == 5
s[2] is equal to the character ":".
All characters except s[2] are digits or "?" characters.
The input is generated such that there is at least one time between "00:00" and "11:59" that you can obtain after replacing the "?" characters.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"1?:?4"

Example 2

"0?:5?"

Core Insight

What unlocks the optimal approach

Try using a brute force approach.
Iterate over all possible times that can be generated from the string and find the latest one.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3114: Latest Time You Can Obtain After Replacing Characters
class Solution {
    public String findLatestTime(String s) {
        for (int h = 11;; h--) {
            for (int m = 59; m >= 0; m--) {
                String t = String.format("%02d:%02d", h, m);
                boolean ok = true;
                for (int i = 0; i < s.length(); i++) {
                    if (s.charAt(i) != '?' && s.charAt(i) != t.charAt(i)) {
                        ok = false;
                        break;
                    }
                }
                if (ok) {
                    return t;
                }
            }
        }
    }
}

// Accepted solution for LeetCode #3114: Latest Time You Can Obtain After Replacing Characters
func findLatestTime(s string) string {
	for h := 11; ; h-- {
		for m := 59; m >= 0; m-- {
			t := fmt.Sprintf("%02d:%02d", h, m)
			ok := true
			for i := 0; i < len(s); i++ {
				if s[i] != '?' && s[i] != t[i] {
					ok = false
					break
				}
			}
			if ok {
				return t
			}
		}
	}
}

# Accepted solution for LeetCode #3114: Latest Time You Can Obtain After Replacing Characters
class Solution:
    def findLatestTime(self, s: str) -> str:
        for h in range(11, -1, -1):
            for m in range(59, -1, -1):
                t = f"{h:02d}:{m:02d}"
                if all(a == b for a, b in zip(s, t) if a != "?"):
                    return t

// Accepted solution for LeetCode #3114: Latest Time You Can Obtain After Replacing Characters
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3114: Latest Time You Can Obtain After Replacing Characters
// class Solution {
//     public String findLatestTime(String s) {
//         for (int h = 11;; h--) {
//             for (int m = 59; m >= 0; m--) {
//                 String t = String.format("%02d:%02d", h, m);
//                 boolean ok = true;
//                 for (int i = 0; i < s.length(); i++) {
//                     if (s.charAt(i) != '?' && s.charAt(i) != t.charAt(i)) {
//                         ok = false;
//                         break;
//                     }
//                 }
//                 if (ok) {
//                     return t;
//                 }
//             }
//         }
//     }
// }

// Accepted solution for LeetCode #3114: Latest Time You Can Obtain After Replacing Characters
function findLatestTime(s: string): string {
    for (let h = 11; ; h--) {
        for (let m = 59; m >= 0; m--) {
            const t: string = `${h.toString().padStart(2, '0')}:${m.toString().padStart(2, '0')}`;
            let ok: boolean = true;
            for (let i = 0; i < s.length; i++) {
                if (s[i] !== '?' && s[i] !== t[i]) {
                    ok = false;
                    break;
                }
            }
            if (ok) {
                return t;
            }
        }
    }
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(1)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.