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3043. 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

Similar Questions:

• Longest Common Prefix (Easy)

Hints:

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

Solution 1.

// OJ: https://leetcode.com/problems/find-the-length-of-the-longest-common-prefix
// Author: github.com/lzl124631x
// Time: O(NW)
// Space: O(NW)
class Solution {
public:
    int longestCommonPrefix(vector<int>& A, vector<int>& B) {
        unordered_set<string> st;
        int len = 0;
        for (int n : A) {
            auto s = to_string(n);
            for (int len = s.size(); len >= 1; --len) {
                st.insert(s.substr(0, len));
            }
            len = max(len, (int)s.size());
        }
        vector<string> sb;
        for (int n : B) sb.push_back(to_string(n));
        for (; len > 0; --len) {
            for (auto &s : sb) {
                s = s.substr(0, len);
                if (s.size() == len && st.count(s)) return len;
            }
        }
        return 0;
    }
};