comments | difficulty | edit_url | rating | source | tags | |
---|---|---|---|---|---|---|
true |
Easy |
1160 |
Weekly Contest 395 Q1 |
|
You are given two arrays of equal length, nums1
and nums2
.
Each element in nums1
has been increased (or decreased in the case of negative) by an integer, represented by the variable x
.
As a result, nums1
becomes equal to nums2
. Two arrays are considered equal when they contain the same integers with the same frequencies.
Return the integer x
.
Example 1:
Input: nums1 = [2,6,4], nums2 = [9,7,5]
Output: 3
Explanation:
The integer added to each element of nums1
is 3.
Example 2:
Input: nums1 = [10], nums2 = [5]
Output: -5
Explanation:
The integer added to each element of nums1
is -5.
Example 3:
Input: nums1 = [1,1,1,1], nums2 = [1,1,1,1]
Output: 0
Explanation:
The integer added to each element of nums1
is 0.
Constraints:
1 <= nums1.length == nums2.length <= 100
0 <= nums1[i], nums2[i] <= 1000
- The test cases are generated in a way that there is an integer
x
such thatnums1
can become equal tonums2
by addingx
to each element ofnums1
.
We can find the minimum value of each array, then return the difference between the two minimum values.
The time complexity is
class Solution:
def addedInteger(self, nums1: List[int], nums2: List[int]) -> int:
return min(nums2) - min(nums1)
class Solution {
public int addedInteger(int[] nums1, int[] nums2) {
return Arrays.stream(nums2).min().getAsInt() - Arrays.stream(nums1).min().getAsInt();
}
}
class Solution {
public:
int addedInteger(vector<int>& nums1, vector<int>& nums2) {
return *min_element(nums2.begin(), nums2.end()) - *min_element(nums1.begin(), nums1.end());
}
};
func addedInteger(nums1 []int, nums2 []int) int {
return slices.Min(nums2) - slices.Min(nums1)
}
function addedInteger(nums1: number[], nums2: number[]): number {
return Math.min(...nums2) - Math.min(...nums1);
}