Phone Number Prefix

Phone Number Prefix - Problem

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.

A prefix is a substring that appears at the beginning of another string. For example, "123" is a prefix of "12345", but "234" is not a prefix of "12345".

Input & Output

Example 1 — Has Prefix

$ Input: numbers = ["123", "12", "456"]

› Output: false

💡 Note: "12" is a prefix of "123", so return false

Example 2 — No Prefix

$ Input: numbers = ["123", "456", "789"]

› Output: true

💡 Note: No number is a prefix of any other number

Example 3 — Single Digit Prefix

$ Input: numbers = ["1", "12", "123"]

› Output: false

💡 Note: "1" is a prefix of both "12" and "123"

Constraints

1 ≤ numbers.length ≤ 1000
1 ≤ numbers[i].length ≤ 20
numbers[i] consists of digits only

Visualization

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Asked in

G Google 45 a Amazon 38 M Microsoft 32

The key insight is that after sorting the phone numbers lexicographically, any prefix relationship can only exist between adjacent elements. Best approach is sort first then check adjacent pairs. Time: O(n log n), Space: O(1).

Common Approaches

✓ Brute Force Comparison

⏱️ Time: O(n² × m) Space: O(1)

For each phone number, check if it's a prefix of any other phone number by comparing it with all other numbers in the array.

Sort Then Check Adjacent

⏱️ Time: O(n log n + n × m) Space: O(1)

After sorting the phone numbers lexicographically, any prefix relationship can only exist between adjacent elements, significantly reducing comparisons needed.

Brute Force Comparison — Algorithm Steps

Iterate through each phone number
For each number, check against all other numbers
Return false if any number is a prefix of another

Visualization

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Step-by-Step Walkthrough

Compare First

Check if first number is prefix of others

Compare Second

Check if second number is prefix of others

Result

Return false if any prefix found

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

bool solution(char numbers[][20], int n) {
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (i != j && strstr(numbers[j], numbers[i]) == numbers[j]) {
                return false;
            }
        }
    }
    return true;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    char numbers[100][20];
    int count = 0;
    char *token = strtok(line, "[],\"");
    while (token != NULL) {
        if (strlen(token) > 0 && strcmp(token, " ") != 0) {
            strcpy(numbers[count], token);
            count++;
        }
        token = strtok(NULL, "[],\"");
    }
    
    bool result = solution(numbers, count);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n² × m)

n² comparisons between all pairs, each comparison takes O(m) where m is average string length

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Comparison — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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