LeetCode #1736 — EASY

Latest Time by Replacing Hidden Digits

Build confidence with an intuition-first walkthrough focused on greedy fundamentals.

The Problem

Problem Statement

You are given a string time in the form of hh:mm, where some of the digits in the string are hidden (represented by ?).

The valid times are those inclusively between 00:00 and 23:59.

Return the latest valid time you can get from time by replacing the hidden digits.

Example 1:

Input: time = "2?:?0"
Output: "23:50"
Explanation: The latest hour beginning with the digit '2' is 23 and the latest minute ending with the digit '0' is 50.

Example 2:

Input: time = "0?:3?"
Output: "09:39"

Example 3:

Input: time = "1?:22"
Output: "19:22"

Constraints:

time is in the format hh:mm.
It is guaranteed that you can produce a valid time from the given string.

Step 01

Brute Force Baseline

Problem summary: You are given a string time in the form of hh:mm, where some of the digits in the string are hidden (represented by ?). The valid times are those inclusively between 00:00 and 23:59. Return the latest valid time you can get from time by replacing the hidden digits.

Baseline thinking

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

Pattern signal: Greedy

Example 1

"2?:?0"

Example 2

"0?:3?"

Example 3

"1?:22"

Core Insight

What unlocks the optimal approach

Trying out all possible solutions from biggest to smallest would fit in the time limit.
To check if the solution is okay, you need to find out if it's valid and matches every character

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 #1736: Latest Time by Replacing Hidden Digits
class Solution {
    public String maximumTime(String time) {
        char[] t = time.toCharArray();
        if (t[0] == '?') {
            t[0] = t[1] >= '4' && t[1] <= '9' ? '1' : '2';
        }
        if (t[1] == '?') {
            t[1] = t[0] == '2' ? '3' : '9';
        }
        if (t[3] == '?') {
            t[3] = '5';
        }
        if (t[4] == '?') {
            t[4] = '9';
        }
        return new String(t);
    }
}

// Accepted solution for LeetCode #1736: Latest Time by Replacing Hidden Digits
func maximumTime(time string) string {
	t := []byte(time)
	if t[0] == '?' {
		if t[1] >= '4' && t[1] <= '9' {
			t[0] = '1'
		} else {
			t[0] = '2'
		}
	}
	if t[1] == '?' {
		if t[0] == '2' {
			t[1] = '3'
		} else {
			t[1] = '9'
		}
	}
	if t[3] == '?' {
		t[3] = '5'
	}
	if t[4] == '?' {
		t[4] = '9'
	}
	return string(t)
}

# Accepted solution for LeetCode #1736: Latest Time by Replacing Hidden Digits
class Solution:
    def maximumTime(self, time: str) -> str:
        t = list(time)
        if t[0] == '?':
            t[0] = '1' if '4' <= t[1] <= '9' else '2'
        if t[1] == '?':
            t[1] = '3' if t[0] == '2' else '9'
        if t[3] == '?':
            t[3] = '5'
        if t[4] == '?':
            t[4] = '9'
        return ''.join(t)

// Accepted solution for LeetCode #1736: Latest Time by Replacing Hidden Digits
struct Solution;

impl Solution {
    fn maximum_time(time: String) -> String {
        let time: Vec<char> = time.chars().collect();
        let mut max = 0;
        for i in 0..24 {
            'outer: for j in 0..60 {
                let t = i * 60 + j;
                let s = format!("{:02}:{:02}", i, j);
                let s: Vec<char> = s.chars().collect();
                for i in 0..5 {
                    if time[i] != '?' && time[i] != s[i] {
                        continue 'outer;
                    }
                }
                max = max.max(t);
            }
        }
        format!("{:02}:{:02}", max / 60, max % 60)
    }
}

#[test]
fn test() {
    let time = "2?:?0".to_string();
    let res = "23:50".to_string();
    assert_eq!(Solution::maximum_time(time), res);
    let time = "0?:3?".to_string();
    let res = "09:39".to_string();
    assert_eq!(Solution::maximum_time(time), res);
    let time = "1?:22".to_string();
    let res = "19:22".to_string();
    assert_eq!(Solution::maximum_time(time), res);
}

// Accepted solution for LeetCode #1736: Latest Time by Replacing Hidden Digits
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1736: Latest Time by Replacing Hidden Digits
// class Solution {
//     public String maximumTime(String time) {
//         char[] t = time.toCharArray();
//         if (t[0] == '?') {
//             t[0] = t[1] >= '4' && t[1] <= '9' ? '1' : '2';
//         }
//         if (t[1] == '?') {
//             t[1] = t[0] == '2' ? '3' : '9';
//         }
//         if (t[3] == '?') {
//             t[3] = '5';
//         }
//         if (t[4] == '?') {
//             t[4] = '9';
//         }
//         return new String(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

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.