LeetCode #1363 — HARD

Largest Multiple of Three

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

Given an array of digits digits, return the largest multiple of three that can be formed by concatenating some of the given digits in any order. If there is no answer return an empty string.

Since the answer may not fit in an integer data type, return the answer as a string. Note that the returning answer must not contain unnecessary leading zeros.

Example 1:

Input: digits = [8,1,9]
Output: "981"

Example 2:

Input: digits = [8,6,7,1,0]
Output: "8760"

Example 3:

Input: digits = [1]
Output: ""

Constraints:

1 <= digits.length <= 10⁴
0 <= digits[i] <= 9

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: Given an array of digits digits, return the largest multiple of three that can be formed by concatenating some of the given digits in any order. If there is no answer return an empty string. Since the answer may not fit in an integer data type, return the answer as a string. Note that the returning answer must not contain unnecessary leading zeros.

Baseline thinking

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

Pattern signal: Array · Math · Dynamic Programming · Greedy

Example 1

[8,1,9]

Example 2

[8,6,7,1,0]

Example 3

[1]

Step 02

Core Insight

What unlocks the optimal approach

A number is a multiple of three if and only if its sum of digits is a multiple of three.
Use dynamic programming.
To find the maximum number, try to maximize the number of digits of the number.
Sort the digits in descending order to find the maximum number.

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

Largest constraint values

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 #1363: Largest Multiple of Three
class Solution {
    public String largestMultipleOfThree(int[] digits) {
        Arrays.sort(digits);
        int n = digits.length;
        int[][] f = new int[n + 1][3];
        final int inf = 1 << 30;
        for (var g : f) {
            Arrays.fill(g, -inf);
        }
        f[0][0] = 0;
        for (int i = 1; i <= n; ++i) {
            for (int j = 0; j < 3; ++j) {
                f[i][j] = Math.max(f[i - 1][j], f[i - 1][(j - digits[i - 1] % 3 + 3) % 3] + 1);
            }
        }
        if (f[n][0] <= 0) {
            return "";
        }
        StringBuilder sb = new StringBuilder();
        for (int i = n, j = 0; i > 0; --i) {
            int k = (j - digits[i - 1] % 3 + 3) % 3;
            if (f[i - 1][k] + 1 == f[i][j]) {
                sb.append(digits[i - 1]);
                j = k;
            }
        }
        int i = 0;
        while (i < sb.length() - 1 && sb.charAt(i) == '0') {
            ++i;
        }
        return sb.substring(i);
    }
}

// Accepted solution for LeetCode #1363: Largest Multiple of Three
func largestMultipleOfThree(digits []int) string {
	sort.Ints(digits)
	n := len(digits)
	const inf = 1 << 30
	f := make([][]int, n+1)
	for i := range f {
		f[i] = make([]int, 3)
		for j := range f[i] {
			f[i][j] = -inf
		}
	}
	f[0][0] = 0
	for i := 1; i <= n; i++ {
		for j := 0; j < 3; j++ {
			f[i][j] = max(f[i-1][j], f[i-1][(j-digits[i-1]%3+3)%3]+1)
		}
	}
	if f[n][0] <= 0 {
		return ""
	}
	ans := []byte{}
	for i, j := n, 0; i > 0; i-- {
		k := (j - digits[i-1]%3 + 3) % 3
		if f[i][j] == f[i-1][k]+1 {
			ans = append(ans, byte('0'+digits[i-1]))
			j = k
		}
	}
	i := 0
	for i < len(ans)-1 && ans[i] == '0' {
		i++
	}
	return string(ans[i:])
}

# Accepted solution for LeetCode #1363: Largest Multiple of Three
class Solution:
    def largestMultipleOfThree(self, digits: List[int]) -> str:
        digits.sort()
        n = len(digits)
        f = [[-inf] * 3 for _ in range(n + 1)]
        f[0][0] = 0
        for i, x in enumerate(digits, 1):
            for j in range(3):
                f[i][j] = max(f[i - 1][j], f[i - 1][(j - x % 3 + 3) % 3] + 1)
        if f[n][0] <= 0:
            return ""
        arr = []
        j = 0
        for i in range(n, 0, -1):
            k = (j - digits[i - 1] % 3 + 3) % 3
            if f[i - 1][k] + 1 == f[i][j]:
                arr.append(digits[i - 1])
                j = k
        i = 0
        while i < len(arr) - 1 and arr[i] == 0:
            i += 1
        return "".join(map(str, arr[i:]))

// Accepted solution for LeetCode #1363: Largest Multiple of Three
/**
 * [1363] Largest Multiple of Three
 *
 * Given an array of digits digits, return the largest multiple of three that can be formed by concatenating some of the given digits in any order. If there is no answer return an empty string.
 * Since the answer may not fit in an integer data type, return the answer as a string. Note that the returning answer must not contain unnecessary leading zeros.
 *  
 * Example 1:
 *
 * Input: digits = [8,1,9]
 * Output: "981"
 *
 * Example 2:
 *
 * Input: digits = [8,6,7,1,0]
 * Output: "8760"
 *
 * Example 3:
 *
 * Input: digits = [1]
 * Output: ""
 *
 *  
 * Constraints:
 *
 * 	1 <= digits.length <= 10^4
 * 	0 <= digits[i] <= 9
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/largest-multiple-of-three/
// discuss: https://leetcode.com/problems/largest-multiple-of-three/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    // Credit: https://leetcode.com/problems/largest-multiple-of-three/solutions/3103997/just-a-runnable-solution/
    pub fn largest_multiple_of_three(digits: Vec<i32>) -> String {
        let mut bucket = [0; 10];
        let mut result = Vec::new();
        for digit in digits.iter() {
            bucket[*digit as usize] += 1;
        }
        let (mut cnt_g1, mut cnt_g2) = (0_usize, 0_usize);
        for i in 0..10 {
            if i == 0 || i == 3 || i == 6 || i == 9 {
                for _ in 0..bucket[i as usize] {
                    result.push(i);
                }
                bucket[i as usize] = 0;
            } else if i == 1 || i == 4 || i == 7 {
                cnt_g1 += bucket[i as usize];
            } else {
                cnt_g2 += bucket[i as usize];
            }
        }
        if cnt_g1 % 3 == cnt_g2 % 3 {
            let mut fin_res = digits;
            fin_res.sort_by(|a, b| b.cmp(a));
            if fin_res[0] == 0 {
                return String::from("0");
            }
            return fin_res
                .iter()
                .map(|a| ((*a + '0' as i32) as u8) as char)
                .collect::<String>();
        }
        if cnt_g1 > 0 && cnt_g1 % 3 == 0 && cnt_g2 % 3 == 2 {
            for _ in 0..(cnt_g1 - 1) {
                if bucket[7] > 0 {
                    result.push(7);
                    bucket[7] -= 1;
                } else if bucket[4] > 0 {
                    result.push(4);
                    bucket[4] -= 1;
                } else if bucket[1] > 0 {
                    result.push(1);
                    bucket[1] -= 1;
                }
            }
            for _ in 0..(cnt_g2) {
                if bucket[8] > 0 {
                    result.push(8);
                    bucket[8] -= 1;
                } else if bucket[5] > 0 {
                    result.push(5);
                    bucket[5] -= 1;
                } else if bucket[2] > 0 {
                    result.push(2);
                    bucket[2] -= 1;
                }
            }
        } else if cnt_g2 > 0 && cnt_g2 % 3 == 0 && cnt_g1 % 3 == 2 {
            for _ in 0..(cnt_g2 - 1) {
                if bucket[8] > 0 {
                    result.push(8);
                    bucket[8] -= 1;
                } else if bucket[5] > 0 {
                    result.push(5);
                    bucket[5] -= 1;
                } else if bucket[2] > 0 {
                    result.push(2);
                    bucket[2] -= 1;
                }
            }
            for _ in 0..(cnt_g1) {
                if bucket[7] > 0 {
                    result.push(7);
                    bucket[7] -= 1;
                } else if bucket[4] > 0 {
                    result.push(4);
                    bucket[4] -= 1;
                } else if bucket[1] > 0 {
                    result.push(1);
                    bucket[1] -= 1;
                }
            }
        } else {
            for _ in 0..(cnt_g1 - cnt_g1 % 3) {
                if bucket[7] > 0 {
                    result.push(7);
                    bucket[7] -= 1;
                } else if bucket[4] > 0 {
                    result.push(4);
                    bucket[4] -= 1;
                } else if bucket[1] > 0 {
                    result.push(1);
                    bucket[1] -= 1;
                }
            }
            cnt_g1 %= 3;
            for _ in 0..(cnt_g2 - cnt_g2 % 3) {
                if bucket[8] > 0 {
                    result.push(8);
                    bucket[8] -= 1;
                } else if bucket[5] > 0 {
                    result.push(5);
                    bucket[5] -= 1;
                } else if bucket[2] > 0 {
                    result.push(2);
                    bucket[2] -= 1;
                }
            }
            cnt_g2 %= 3;
            if cnt_g1 + cnt_g2 == 3 {
                if bucket[7] > 0 {
                    result.push(7);
                    bucket[7] -= 1;
                } else if bucket[4] > 0 {
                    result.push(4);
                    bucket[4] -= 1;
                } else if bucket[1] > 0 {
                    result.push(1);
                    bucket[1] -= 1;
                }
                if bucket[8] > 0 {
                    result.push(8);
                    bucket[8] -= 1;
                } else if bucket[5] > 0 {
                    result.push(5);
                    bucket[5] -= 1;
                } else if bucket[2] > 0 {
                    result.push(2);
                    bucket[2] -= 1;
                }
            }
        }
        result.sort_by(|a, b| b.cmp(a));
        if let Some(&v) = result.first() {
            if v == 0 {
                return String::from("0");
            }
        }
        result
            .iter()
            .map(|a| ((*a + '0' as i32) as u8) as char)
            .collect::<String>()
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_1363_example_1() {
        let digits = vec![8, 1, 9];

        let result = "981".to_string();

        assert_eq!(Solution::largest_multiple_of_three(digits), result);
    }

    #[test]
    fn test_1363_example_2() {
        let digits = vec![8, 6, 7, 1, 0];

        let result = "8760".to_string();

        assert_eq!(Solution::largest_multiple_of_three(digits), result);
    }

    #[test]
    fn test_1363_example_3() {
        let digits = vec![1];

        let result = "".to_string();

        assert_eq!(Solution::largest_multiple_of_three(digits), result);
    }
}

// Accepted solution for LeetCode #1363: Largest Multiple of Three
function largestMultipleOfThree(digits: number[]): string {
    digits.sort((a, b) => a - b);
    const n = digits.length;
    const f: number[][] = new Array(n + 1).fill(0).map(() => new Array(3).fill(-Infinity));
    f[0][0] = 0;
    for (let i = 1; i <= n; ++i) {
        for (let j = 0; j < 3; ++j) {
            f[i][j] = Math.max(f[i - 1][j], f[i - 1][(j - (digits[i - 1] % 3) + 3) % 3] + 1);
        }
    }
    if (f[n][0] <= 0) {
        return '';
    }
    const arr: number[] = [];
    for (let i = n, j = 0; i; --i) {
        const k = (j - (digits[i - 1] % 3) + 3) % 3;
        if (f[i - 1][k] + 1 === f[i][j]) {
            arr.push(digits[i - 1]);
            j = k;
        }
    }
    let i = 0;
    while (i < arr.length - 1 && arr[i] === 0) {
        ++i;
    }
    return arr.slice(i).join('');
}

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 × log n)

Space

O(n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

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.