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This approach utilizes the fact that both arrays are sorted. Using two pointers, one for each array, we traverse the arrays to find the smallest common value:
Time Complexity: O(n + m), where n and m are the lengths of the two arrays. We only traverse each array once.
Space Complexity: O(1), no additional space is used apart from a few variables.
1using System;
2
3public class Program
4{
5 public static int FindMinCommonValue(int[] nums1, int[] nums2)
6 {
7 int i = 0, j = 0;
8 while (i < nums1.Length && j < nums2.Length)
9 {
10 if (nums1[i] == nums2[j])
11 {
12 return nums1[i];
13 }
14 else if (nums1[i] < nums2[j])
{
i++;
}
else
{
j++;
}
}
return -1;
}
public static void Main()
{
int[] nums1 = {1, 2, 3};
int[] nums2 = {2, 4};
Console.WriteLine(FindMinCommonValue(nums1, nums2));
}
}This C# solution similarly utilizes the two-pointer approach to find the smallest common value. It's concise and efficient, iterating over the integer arrays using simple control structures.
This approach uses a HashSet to store the elements of the smaller array, providing a quick way to check for common elements:
Time Complexity: O(n log n + m), mainly due to qsort and possibly bsearch in the worst case scenario per element in the second array.
Space Complexity: O(n), additional space used for sorting elements.
This Python solution uses Python's set data structure to conveniently test for the presence of elements in constant time. We insert elements of one array into a set and test elements of the other array against it.