3043. Find the Length of the Longest Common Prefix

Description

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

 

Solutions

Solution: Trie

• Time complexity: O(n1*Max(arr1[i].length)+n2*Max(arr2[i].length))
• Space complexity: O(n1*Max(arr1[i].length))

 

JavaScript

/**
 * @param {number[]} arr1
 * @param {number[]} arr2
 * @return {number}
 */
const longestCommonPrefix = function (arr1, arr2) {
  const trie = new Map();

  for (const num of arr1) {
    let node = trie;

    for (const char of `${num}`) {
      if (!node.has(char)) {
        node.set(char, new Map());
      }
      node = node.get(char);
    }
  }
  let result = 0;

  for (const num of arr2) {
    const str = `${num}`;
    let node = trie;

    for (const [index, char] of str.entries()) {

      if (!node.has(char)) break;
      node = node.get(char);
      result = Math.max(index + 1, result);
    }
  }
  return result;
};