3491. Phone Number Prefix 🔒

Description

You are given a string array numbers that represents phone numbers. Return true if no phone number is a prefix of any other phone number; otherwise, return false.

Example 1:

Input: numbers = ["1","2","4","3"]

Output: true

Explanation:

No number is a prefix of another number, so the output is true.

Example 2:

Input: numbers = ["001","007","15","00153"]

Output: false

Explanation:

The string "001" is a prefix of the string "00153". Thus, the output is false.

Constraints:

2 <= numbers.length <= 50
1 <= numbers[i].length <= 50
All numbers contain only digits '0' to '9'.

Solutions

Solution 1: Sorting + Prefix Checking

We can first sort the array \(\textit{numbers}\) based on the length of strings. Then, we iterate through each string \(\textit{s}\) in the array and check if there is any previous string \(\textit{t}\) that is a prefix of \(\textit{s}\). If such a string exists, it means there is a string that is a prefix of another string, so we return \(\textit{false}\). If we have checked all strings and haven't found any prefix relationships, we return \(\textit{true}\).

The time complexity is \(O(n^2 \times m + n \times \log n)\), and the space complexity is \(O(m + \log n)\), where \(n\) is the length of the array \(\textit{numbers}\), and \(m\) is the average length of strings in the array \(\textit{numbers}\).

Python3JavaC++GoTypeScript

class Solution:
    def phonePrefix(self, numbers: List[str]) -> bool:
        numbers.sort(key=len)
        for i, s in enumerate(numbers):
            if any(s.startswith(t) for t in numbers[:i]):
                return False
        return True

class Solution {
    public boolean phonePrefix(String[] numbers) {
        Arrays.sort(numbers, (a, b) -> Integer.compare(a.length(), b.length()));
        for (int i = 0; i < numbers.length; i++) {
            String s = numbers[i];
            for (int j = 0; j < i; j++) {
                if (s.startsWith(numbers[j])) {
                    return false;
                }
            }
        }
        return true;
    }
}

#include <ranges>

class Solution {
public:
    bool phonePrefix(vector<string>& numbers) {
        ranges::sort(numbers, [](const string& a, const string& b) {
            return a.size() < b.size();
        });
        for (int i = 0; i < numbers.size(); i++) {
            if (ranges::any_of(numbers | views::take(i), [&](const string& t) {
                    return numbers[i].starts_with(t);
                })) {
                return false;
            }
        }
        return true;
    }
};

func phonePrefix(numbers []string) bool {
    sort.Slice(numbers, func(i, j int) bool {
        return len(numbers[i]) < len(numbers[j])
    })
    for i, s := range numbers {
        for _, t := range numbers[:i] {
            if strings.HasPrefix(s, t) {
                return false
            }
        }
    }
    return true
}

function phonePrefix(numbers: string[]): boolean {
    numbers.sort((a, b) => a.length - b.length);
    for (let i = 0; i < numbers.length; i++) {
        for (let j = 0; j < i; j++) {
            if (numbers[i].startsWith(numbers[j])) {
                return false;
            }
        }
    }
    return true;
}

3491. Phone Number Prefix 🔒

Description

Solutions

Solution 1: Sorting + Prefix Checking

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