Welcome to Subscribe On Youtube

3131. Find the Integer Added to Array I

Description

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 that nums1 can become equal to nums2 by adding x to each element of nums1.

Solutions

Solution 1: Calculate Minimum Difference

We can find the minimum value of each array, then return the difference between the two minimum values.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

  • 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());
    }
};

  • class Solution:
    def addedInteger(self, nums1: List[int], nums2: List[int]) -> int:
        return min(nums2) - min(nums1)


  • 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);
}

All Problems

All Solutions