LeetCode #3043 — MEDIUM

Find the Length of the Longest Common Prefix

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

The Problem

Problem Statement

You are given two arrays with positive integers arr1 and arr2.

A prefix of a positive integer is an integer formed by one or more of its digits, starting from its leftmost digit. For example, 123 is a prefix of the integer 12345, while 234 is not.

A common prefix of two integers a and b is an integer c, such that c is a prefix of both a and b. For example, 5655359 and 56554 have common prefixes 565 and 5655 while 1223 and 43456 do not have a common prefix.

You need to find the length of the longest common prefix between all pairs of integers (x, y) such that x belongs to arr1 and y belongs to arr2.

Return the length of the longest common prefix among all pairs. If no common prefix exists among them, return 0.

Example 1:

Input: arr1 = [1,10,100], arr2 = [1000]
Output: 3
Explanation: There are 3 pairs (arr1[i], arr2[j]):
- The longest common prefix of (1, 1000) is 1.
- The longest common prefix of (10, 1000) is 10.
- The longest common prefix of (100, 1000) is 100.
The longest common prefix is 100 with a length of 3.

Example 2:

Input: arr1 = [1,2,3], arr2 = [4,4,4]
Output: 0
Explanation: There exists no common prefix for any pair (arr1[i], arr2[j]), hence we return 0.
Note that common prefixes between elements of the same array do not count.

Constraints:

1 <= arr1.length, arr2.length <= 5 * 10⁴
1 <= arr1[i], arr2[i] <= 10⁸

Step 01

Brute Force Baseline

Problem summary: You are given two arrays with positive integers arr1 and arr2. A prefix of a positive integer is an integer formed by one or more of its digits, starting from its leftmost digit. For example, 123 is a prefix of the integer 12345, while 234 is not. A common prefix of two integers a and b is an integer c, such that c is a prefix of both a and b. For example, 5655359 and 56554 have common prefixes 565 and 5655 while 1223 and 43456 do not have a common prefix. You need to find the length of the longest common prefix between all pairs of integers (x, y) such that x belongs to arr1 and y belongs to arr2. Return the length of the longest common prefix among all pairs. If no common prefix exists among them, return 0.

Baseline thinking

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

Pattern signal: Array · Hash Map · Trie

Example 1

[1,10,100]
[1000]

Example 2

[1,2,3]
[4,4,4]

Core Insight

What unlocks the optimal approach

Put all the possible prefixes of each element in <code>arr1</code> into a HashSet.
For all the possible prefixes of each element in <code>arr2</code>, check if it exists in the HashSet.

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 #3043: Find the Length of the Longest Common Prefix
class Solution {
    public int longestCommonPrefix(int[] arr1, int[] arr2) {
        Set<Integer> s = new HashSet<>();
        for (int x : arr1) {
            for (; x > 0; x /= 10) {
                s.add(x);
            }
        }
        int ans = 0;
        for (int x : arr2) {
            for (; x > 0; x /= 10) {
                if (s.contains(x)) {
                    ans = Math.max(ans, String.valueOf(x).length());
                    break;
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3043: Find the Length of the Longest Common Prefix
func longestCommonPrefix(arr1 []int, arr2 []int) (ans int) {
	s := map[int]bool{}
	for _, x := range arr1 {
		for ; x > 0; x /= 10 {
			s[x] = true
		}
	}
	for _, x := range arr2 {
		for ; x > 0; x /= 10 {
			if s[x] {
				ans = max(ans, int(math.Log10(float64(x)))+1)
				break
			}
		}
	}
	return
}

# Accepted solution for LeetCode #3043: Find the Length of the Longest Common Prefix
class Solution:
    def longestCommonPrefix(self, arr1: List[int], arr2: List[int]) -> int:
        s = set()
        for x in arr1:
            while x:
                s.add(x)
                x //= 10
        ans = 0
        for x in arr2:
            while x:
                if x in s:
                    ans = max(ans, len(str(x)))
                    break
                x //= 10
        return ans

// Accepted solution for LeetCode #3043: Find the Length of the Longest Common Prefix
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3043: Find the Length of the Longest Common Prefix
// class Solution {
//     public int longestCommonPrefix(int[] arr1, int[] arr2) {
//         Set<Integer> s = new HashSet<>();
//         for (int x : arr1) {
//             for (; x > 0; x /= 10) {
//                 s.add(x);
//             }
//         }
//         int ans = 0;
//         for (int x : arr2) {
//             for (; x > 0; x /= 10) {
//                 if (s.contains(x)) {
//                     ans = Math.max(ans, String.valueOf(x).length());
//                     break;
//                 }
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3043: Find the Length of the Longest Common Prefix
function longestCommonPrefix(arr1: number[], arr2: number[]): number {
    const s: Set<number> = new Set<number>();
    for (let x of arr1) {
        for (; x; x = Math.floor(x / 10)) {
            s.add(x);
        }
    }
    let ans: number = 0;
    for (let x of arr2) {
        for (; x; x = Math.floor(x / 10)) {
            if (s.has(x)) {
                ans = Math.max(ans, Math.floor(Math.log10(x)) + 1);
            }
        }
    }
    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(m × log M + n × log N)

Space

O(m × log M)

Approach Breakdown

HASH SET

O(N × L) time

O(N × L) space

Store all N words in a hash set. Each insert/lookup hashes the entire word of length L, giving O(L) per operation. Prefix queries require checking every stored word against the prefix — O(N × L) per prefix search. Space is O(N × L) for storing all characters.

TRIE

O(L) time

O(N × L) space

Each operation (insert, search, prefix) takes O(L) time where L is the word length — one node visited per character. Total space is bounded by the sum of all stored word lengths. Tries win over hash sets when you need prefix matching: O(L) prefix search vs. checking every stored word.

Shortcut: One node per character → O(L) per operation. Prefix queries are what make tries worthwhile.

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.