LeetCode #949 — MEDIUM

Largest Time for Given Digits

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

Given an array arr of 4 digits, find the latest 24-hour time that can be made using each digit exactly once.

24-hour times are formatted as "HH:MM", where HH is between 00 and 23, and MM is between 00 and 59. The earliest 24-hour time is 00:00, and the latest is 23:59.

Return the latest 24-hour time in "HH:MM" format. If no valid time can be made, return an empty string.

Example 1:

Input: arr = [1,2,3,4]
Output: "23:41"
Explanation: The valid 24-hour times are "12:34", "12:43", "13:24", "13:42", "14:23", "14:32", "21:34", "21:43", "23:14", and "23:41". Of these times, "23:41" is the latest.

Example 2:

Input: arr = [5,5,5,5]
Output: ""
Explanation: There are no valid 24-hour times as "55:55" is not valid.

Constraints:

arr.length == 4
0 <= arr[i] <= 9

Step 01

Brute Force Baseline

Problem summary: Given an array arr of 4 digits, find the latest 24-hour time that can be made using each digit exactly once. 24-hour times are formatted as "HH:MM", where HH is between 00 and 23, and MM is between 00 and 59. The earliest 24-hour time is 00:00, and the latest is 23:59. Return the latest 24-hour time in "HH:MM" format. If no valid time can be made, return an empty string.

Baseline thinking

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

Pattern signal: Array · Backtracking

Example 1

[1,2,3,4]

Example 2

[5,5,5,5]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #949: Largest Time for Given Digits
class Solution {
    public String largestTimeFromDigits(int[] arr) {
        int ans = -1;
        for (int i = 0; i < 4; ++i) {
            for (int j = 0; j < 4; ++j) {
                for (int k = 0; k < 4; ++k) {
                    if (i != j && j != k && i != k) {
                        int h = arr[i] * 10 + arr[j];
                        int m = arr[k] * 10 + arr[6 - i - j - k];
                        if (h < 24 && m < 60) {
                            ans = Math.max(ans, h * 60 + m);
                        }
                    }
                }
            }
        }
        return ans < 0 ? "" : String.format("%02d:%02d", ans / 60, ans % 60);
    }
}

// Accepted solution for LeetCode #949: Largest Time for Given Digits
func largestTimeFromDigits(arr []int) string {
	ans := -1
	for i := 0; i < 4; i++ {
		for j := 0; j < 4; j++ {
			for k := 0; k < 4; k++ {
				if i != j && j != k && i != k {
					h := arr[i]*10 + arr[j]
					m := arr[k]*10 + arr[6-i-j-k]
					if h < 24 && m < 60 {
						ans = max(ans, h*60+m)
					}
				}
			}
		}
	}
	if ans < 0 {
		return ""
	}
	return fmt.Sprintf("%02d:%02d", ans/60, ans%60)
}

# Accepted solution for LeetCode #949: Largest Time for Given Digits
class Solution:
    def largestTimeFromDigits(self, arr: List[int]) -> str:
        cnt = [0] * 10
        for v in arr:
            cnt[v] += 1
        for h in range(23, -1, -1):
            for m in range(59, -1, -1):
                t = [0] * 10
                t[h // 10] += 1
                t[h % 10] += 1
                t[m // 10] += 1
                t[m % 10] += 1
                if cnt == t:
                    return f'{h:02}:{m:02}'
        return ''

// Accepted solution for LeetCode #949: Largest Time for Given Digits
struct Solution;

use std::fmt;

#[derive(Debug, PartialEq, Eq, Clone)]
struct Time {
    hour: i32,
    minute: i32,
}

impl Time {
    fn new(hour: i32, minute: i32) -> Self {
        Time { hour, minute }
    }
    fn is_valid(&self) -> bool {
        self.hour < 24 && self.minute < 60
    }
    fn from_digits(a: &[i32]) -> Self {
        Self::new(a[0] * 10 + a[1], a[2] * 10 + a[3])
    }
    fn to_minutes(&self) -> i32 {
        self.hour * 60 + self.minute
    }
    fn to_digits(&self) -> Vec<i32> {
        vec![
            self.hour / 10,
            self.hour % 10,
            self.minute / 10,
            self.minute % 10,
        ]
    }
}

impl fmt::Display for Time {
    fn fmt(&self, f: &mut fmt::Formatter<'_>) -> fmt::Result {
        let a = self.to_digits();
        write!(f, "{}{}:{}{}", a[0], a[1], a[2], a[3])
    }
}

use std::cmp::Ordering;

impl PartialOrd for Time {
    fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
        Some(self.to_minutes().cmp(&other.to_minutes()))
    }
}

impl Solution {
    fn backtrack(a: &mut Vec<i32>, index: usize, max: &mut Option<Time>) {
        let n = a.len();
        if index == n {
            let time = Time::from_digits(a);
            if time.is_valid() {
                if let Some(max_time) = max {
                    if time > *max_time {
                        *max = Some(time)
                    }
                } else {
                    *max = Some(time)
                }
            }
        } else {
            for i in index..n {
                a.swap(index, i);
                Self::backtrack(a, index + 1, max);
                a.swap(index, i);
            }
        }
    }
    fn largest_time_from_digits(mut a: Vec<i32>) -> String {
        let mut max: Option<Time> = None;
        Self::backtrack(&mut a, 0, &mut max);
        if let Some(max_time) = max {
            max_time.to_string()
        } else {
            "".to_string()
        }
    }
}

#[test]
fn test() {
    let a = vec![1, 2, 3, 4];
    let res = "23:41".to_string();
    assert_eq!(Solution::largest_time_from_digits(a), res);
    let a = vec![5, 5, 5, 5];
    let res = "".to_string();
    assert_eq!(Solution::largest_time_from_digits(a), res);
    let a = vec![0, 0, 0, 0];
    let res = "00:00".to_string();
    assert_eq!(Solution::largest_time_from_digits(a), res);
}

// Accepted solution for LeetCode #949: Largest Time for Given Digits
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #949: Largest Time for Given Digits
// class Solution {
//     public String largestTimeFromDigits(int[] arr) {
//         int ans = -1;
//         for (int i = 0; i < 4; ++i) {
//             for (int j = 0; j < 4; ++j) {
//                 for (int k = 0; k < 4; ++k) {
//                     if (i != j && j != k && i != k) {
//                         int h = arr[i] * 10 + arr[j];
//                         int m = arr[k] * 10 + arr[6 - i - j - k];
//                         if (h < 24 && m < 60) {
//                             ans = Math.max(ans, h * 60 + m);
//                         }
//                     }
//                 }
//             }
//         }
//         return ans < 0 ? "" : String.format("%02d:%02d", ans / 60, ans % 60);
//     }
// }

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(n!)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

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.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.