How do you solve Largest Time for Given Digits on LeetCode?

Largest Time for Given Digits (LeetCode #949) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: Sorting helps in quickly finding the largest possible time.

What is the brute force solution for Largest Time for Given Digits?

The brute force approach for Largest Time for Given Digits is Brute Force, with O(24 * 60) = O(1) time complexity and O(1) space complexity. The brute force approach generates all possible permutations of the four digits and checks which ones can form valid 24-hour times. This is straightforward but inefficient as it examines every combination. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Largest Time for Given Digits?

Largest Time for Given Digits uses the following concepts: Array, String, Backtracking, Enumeration. The recommended patterns to study are: Backtracking, Enumeration.

What are the interview tips for Largest Time for Given Digits?

Always validate your output against edge cases. Explain your thought process clearly while coding. Consider both time and space complexities in your solution.

What are common mistakes when solving Largest Time for Given Digits?

Not checking for valid hour and minute ranges. Failing to consider all permutations or combinations.

#949

Largest Time for Given Digits

Medium

ArrayStringBacktrackingEnumerationBacktrackingEnumeration

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(24 * 60) = O(1)	O(1)
Space	O(1)	O(1)

💡

Intuition

Time O(1)Space O(1)

The optimal solution avoids generating all permutations by directly constructing valid times from the digits. It checks combinations that can form valid hours and minutes, significantly reducing unnecessary checks.

⚙️

Algorithm

3 steps

1Step 1: Sort the input array to facilitate the formation of the largest time.
2Step 2: Iterate through all combinations of the first two digits as hours and the last two as minutes.
3Step 3: Validate the constructed time and keep track of the largest valid time found.

solution.py12 lines

1# Full working Python code
2import itertools
3
4def largestTimeFromDigits(arr):
5    arr.sort(reverse=True)
6    for i in range(24):
7        for j in range(60):
8            hh = f'{i:02}'
9            mm = f'{j:02}'
10            if all(d in arr for d in map(int, hh + mm)):
11                return f'{hh}:{mm}'
12    return ""

ℹ

Complexity note: The time complexity is constant because we are iterating through a fixed number of possible times (1440). The space complexity is constant as well since we only store a few variables.

1Sorting helps in quickly finding the largest possible time.
2Validating time combinations reduces unnecessary checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.