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The simplest way to solve this problem is to iterate over each pair of integers from arr1 and arr2. For each pair, convert the integers to strings and compare character by character to find the common prefix. Keep a running maximum of the lengths of these prefixes.
Since we are comparing each pair, the time complexity is O(n * m * k), where n is the length of arr1, m is the length of arr2, and k is the length of the longest number when converted to a string. Space complexity is O(1) as we're only storing a few additional variables.
Time complexity: O(n * m * k), Space complexity: O(1)
1#include <stdio.h>
2#include <string.h>
3
4int commonPrefixLength(char *a, char *b) {
5 int i = 0;
6 while (a[i] && b[i] && a[i] == b[i]) {
7 i++;
8 }
9 return i;
10}
11
12int longestCommonPrefix(int* arr1, int arr1Size, int* arr2, int arr2Size) {
13 int maxLength = 0;
14 for (int i = 0; i < arr1Size; i++) {
15 for (int j = 0; j < arr2Size; j++) {
16 char str1[10], str2[10];
17 sprintf(str1, "%d", arr1[i]);
18 sprintf(str2, "%d", arr2[j]);
19 int currLength = commonPrefixLength(str1, str2);
20 if (currLength > maxLength) {
21 maxLength = currLength;
22 }
23 }
24 }
25 return maxLength;
26}
27
28int main() {
29 int arr1[] = {1, 10, 100};
30 int arr2[] = {1000};
31 printf("%d\n", longestCommonPrefix(arr1, 3, arr2, 1));
32 return 0;
33}This C solution utilizes a nested loop to iterate through each combination of numbers from arr1 and arr2. For each combination, it finds the common prefix length using a helper function that compares the string representations of the numbers. The maximum prefix length found is returned.
This optimized approach leverages sorting of the input arrays to reduce comparisons. After sorting, only consecutive elements can potentially have a longer common prefix with elements of the other array than already found max. This minimizes the number of comparisons, improving efficiency over the brute force approach.
Time complexity may approach O(n + m) in the best-case scenario where common prefixes are found early, but in the worst case it remains approximately O(n * m). The space complexity is still O(1).
Average Time complexity: O(n log n + m log m + n + m), Space complexity: O(1)
This Python solution features sorting of input arrays for efficient prefix checks. Only necessary comparisons are executed, speeding up computation relative to the unsorted method.