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feat: solve No.3043
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# 3043. Find the Length of the Longest Common Prefix

- Difficulty: Medium.
- Related Topics: Array, Hash Table, String, Trie.
- Similar Questions: Longest Common Prefix, Longest Common Suffix Queries.

## Problem

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 a common prefix `565` 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`



## Solution

```javascript
/**
* @param {number[]} arr1
* @param {number[]} arr2
* @return {number}
*/
var longestCommonPrefix = function(arr1, arr2) {
var trie = {};
for (var i = 0; i < arr1.length; i++) {
var str = `${arr1[i]}`;
var map = trie;
for (var j = 0; j < str.length; j++) {
if (!map[str[j]]) {
map[str[j]] = {};
}
map = map[str[j]];
}
}
var max = 0;
for (var i = 0; i < arr2.length; i++) {
var str = `${arr2[i]}`;
var map = trie;
var len = 0;
for (var j = 0; j < str.length; j++) {
if (!map[str[j]]) {
break;
}
len += 1;
map = map[str[j]];
}
max = Math.max(max, len);
}
return max;
};
```

**Explain:**

Trie.

**Complexity:**

* Time complexity : O(n * m).
* Space complexity : O(n * m).

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