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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 1: Hash Table

We can use a hash table to store all the prefixes of the numbers in arr1. Then, we traverse all the numbers $x$ in arr2. For each number $x$, we start from the highest bit and gradually decrease, checking whether it exists in the hash table. If it does, we have found a common prefix, and we can update the answer accordingly.

The time complexity is $O(m \times \log M + n \times \log N)$, and the space complexity is $O(m \times \log M)$. Here, $m$ and $n$ are the lengths of arr1 and arr2 respectively, and $M$ and $N$ are the maximum values in arr1 and arr2 respectively.

  • 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;
    }
}

  • class Solution {
public:
    int longestCommonPrefix(vector<int>& arr1, vector<int>& arr2) {
        unordered_set<int> s;
        for (int x : arr1) {
            for (; x; x /= 10) {
                s.insert(x);
            }
        }
        int ans = 0;
        for (int x : arr2) {
            for (; x; x /= 10) {
                if (s.count(x)) {
                    ans = max(ans, (int) log10(x) + 1);
                    break;
                }
            }
        }
        return ans;
    }
};

  • 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


  • 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
}

  • function longestCommonPrefix(arr1: number[], arr2: number[]): number {
    const s: Set<number> = new Set<number>();
    for (let x of arr1) {
        for (; x; x = (x / 10) | 0) {
            s.add(x % 10);
        }
    }
    let ans: number = 0;
    for (let x of arr2) {
        for (; x; x = (x / 10) | 0) {
            if (s.has(x % 10)) {
                ans = Math.max(ans, Math.floor(Math.log10(x)) + 1);
            }
        }
    }
    return ans;
}

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