LeetCode #2437 — EASY

Number of Valid Clock Times

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

The Problem

Problem Statement

You are given a string of length 5 called time, representing the current time on a digital clock in the format "hh:mm". The earliest possible time is "00:00" and the latest possible time is "23:59".

In the string time, the digits represented by the ? symbol are unknown, and must be replaced with a digit from 0 to 9.

Return an integer answer, the number of valid clock times that can be created by replacing every ? with a digit from 0 to 9.

Example 1:

Input: time = "?5:00"
Output: 2
Explanation: We can replace the ? with either a 0 or 1, producing "05:00" or "15:00". Note that we cannot replace it with a 2, since the time "25:00" is invalid. In total, we have two choices.

Example 2:

Input: time = "0?:0?"
Output: 100
Explanation: Each ? can be replaced by any digit from 0 to 9, so we have 100 total choices.

Example 3:

Input: time = "??:??"
Output: 1440
Explanation: There are 24 possible choices for the hours, and 60 possible choices for the minutes. In total, we have 24 * 60 = 1440 choices.

Constraints:

time is a valid string of length 5 in the format "hh:mm".
"00" <= hh <= "23"
"00" <= mm <= "59"
Some of the digits might be replaced with '?' and need to be replaced with digits from 0 to 9.

Step 01

Brute Force Baseline

Problem summary: You are given a string of length 5 called time, representing the current time on a digital clock in the format "hh:mm". The earliest possible time is "00:00" and the latest possible time is "23:59". In the string time, the digits represented by the ? symbol are unknown, and must be replaced with a digit from 0 to 9. Return an integer answer, the number of valid clock times that can be created by replacing every ? with a digit from 0 to 9.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

"?5:00"

Example 2

"0?:0?"

Example 3

"??:??"

Core Insight

What unlocks the optimal approach

Brute force all possible clock times.
Checking if a clock time is valid can be done with Regex.

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 #2437: Number of Valid Clock Times
class Solution {
    public int countTime(String time) {
        int ans = 0;
        for (int h = 0; h < 24; ++h) {
            for (int m = 0; m < 60; ++m) {
                String s = String.format("%02d:%02d", h, m);
                int ok = 1;
                for (int i = 0; i < 5; ++i) {
                    if (s.charAt(i) != time.charAt(i) && time.charAt(i) != '?') {
                        ok = 0;
                        break;
                    }
                }
                ans += ok;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2437: Number of Valid Clock Times
func countTime(time string) int {
	ans := 0
	for h := 0; h < 24; h++ {
		for m := 0; m < 60; m++ {
			s := fmt.Sprintf("%02d:%02d", h, m)
			ok := 1
			for i := 0; i < 5; i++ {
				if s[i] != time[i] && time[i] != '?' {
					ok = 0
					break
				}
			}
			ans += ok
		}
	}
	return ans
}

# Accepted solution for LeetCode #2437: Number of Valid Clock Times
class Solution:
    def countTime(self, time: str) -> int:
        def check(s: str, t: str) -> bool:
            return all(a == b or b == '?' for a, b in zip(s, t))

        return sum(
            check(f'{h:02d}:{m:02d}', time) for h in range(24) for m in range(60)
        )

// Accepted solution for LeetCode #2437: Number of Valid Clock Times
impl Solution {
    pub fn count_time(time: String) -> i32 {
        let mut ans = 0;

        for i in 0..24 {
            for j in 0..60 {
                let mut ok = true;
                let t = format!("{:02}:{:02}", i, j);

                for (k, ch) in time.chars().enumerate() {
                    if ch != '?' && ch != t.chars().nth(k).unwrap() {
                        ok = false;
                    }
                }

                if ok {
                    ans += 1;
                }
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #2437: Number of Valid Clock Times
function countTime(time: string): number {
    let ans = 0;
    for (let h = 0; h < 24; ++h) {
        for (let m = 0; m < 60; ++m) {
            const s = `${h}`.padStart(2, '0') + ':' + `${m}`.padStart(2, '0');
            let ok = 1;
            for (let i = 0; i < 5; ++i) {
                if (s[i] !== time[i] && time[i] !== '?') {
                    ok = 0;
                    break;
                }
            }
            ans += ok;
        }
    }
    return ans;
}

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(24 × 60)

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.