3329. Find The Length Of The Longest Common Prefix¶

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3329. Find the Length of the Longest Common Prefix

Medium

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¶

class Solution {
    public int longestCommonPrefix(int[] arr1, int[] arr2) {
        int n = arr1.length;
        int m = arr2.length;
        HashSet<String> set = new HashSet<>();
        for (int i = 0; i < n; i++) {
            String current = "";
            current += arr1[i];
            String to_put = "";
            for (int j = 0; j < current.length(); j++) {
                to_put += current.charAt(j);
                set.add(to_put);
            }
        }

        int maxi = 0;
        for (int i = 0; i < m; i++) {
            String current = "";
            current += arr2[i];
            String to_check = "";
            for (int j = 0; j < current.length(); j++) {
                to_check += current.charAt(j);
                if (set.contains(to_check)) maxi = Math.max(maxi, to_check.length());
            }
        }
        return maxi;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here