Find the Integer Added to Array I

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Arrays

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of integer values within the arrays `nums1` and `nums2`? Can they be negative, zero, or very large?
Can `nums1` be empty? What if `nums2` only contains one element?
Can `nums1` contain duplicate numbers? If so, how should duplicates be handled in the solution?
Is it guaranteed that `nums2` will always contain exactly one integer more than `nums1`, and that all other elements are a shuffled version of `nums1`?
If the additional integer cannot be determined due to some edge case (e.g., empty arrays, or `nums2` not being a valid shuffled version of `nums1`), what should I return?

Brute Force Solution

Approach

The brute-force approach involves systematically trying out every possible number to find the missing one. We can check each number one by one until we locate the added integer.

Here's how the algorithm would work step-by-step:

Calculate the total sum of all the numbers currently present.
Start guessing possible numbers, beginning from 1.
For each guess, imagine it was the added number and calculate the sum of all numbers if it was included.
Check if the sum of the array including your guessed number is equal to the sum if the number wasn't there.
If the two sums are different, then we try the next possible guess.
Repeat until a guess makes the expected total match the sum of the original numbers, which signifies that the guessed number is the added number.

Code Implementation

def find_the_added_integer_brute_force(numbers):
    array_sum = sum(numbers)

    possible_added_number = 1

    while True:
        # We guess a number and check if the sum would match
        hypothetical_sum = array_sum + possible_added_number

        # It's the number if hypothetical sum isn't equal
        if hypothetical_sum != array_sum:

            possible_added_number += 1

        # We return the number if the sums are equal
        else:
            return possible_added_number

Big(O) Analysis

Time Complexity

O(n) – The described brute-force approach involves calculating the sum of the array elements, which takes O(n) time where n is the number of elements in the array. The algorithm then iterates through possible numbers, starting from 1, and calculates the sum of the array with each guessed number added. For each guess, it checks if the calculated sum matches a condition. Since we might need to check up to the missing number which is unknown but upper-bounded in FAANG style questions and usually small relative to n or a constant amount (making it O(1)), or have to check until the guessed number equals to the array sum, the number of guesses is at most some function dependent on the initial sum. We do not iterate over array elements for EACH guess and given the prompts descriptions, the dominating operation is calculating initial sum which takes O(n) time. Therefore, the overall time complexity is O(n).

Space Complexity

O(1) – The provided algorithm primarily uses a few variables to calculate sums and iterate through possible numbers. It does not create any auxiliary data structures like arrays, hash maps, or call stacks that scale with the input size (N, which in this case is the number of elements in the input array). Therefore, the space used remains constant regardless of the size of the input array, resulting in O(1) space complexity.

Optimal Solution

Approach

The problem presents two lists of numbers where one list has an extra number that isn't in the other. The goal is to find that missing number efficiently. We can do this by comparing the total sums of both lists.

Here's how the algorithm would work step-by-step:

Calculate the total sum of all the numbers in the list that includes the extra number.
Calculate the total sum of all the numbers in the list that does not have the extra number.
Subtract the sum of the smaller list from the sum of the larger list. The difference is the extra number that was added.

Code Implementation

def find_added_integer(
    array_with_extra_integer,
    array_without_extra_integer
):
    # Calculate the sum of the larger array.
    sum_of_larger_array = sum(array_with_extra_integer)

    # Calculate the sum of the smaller array.
    sum_of_smaller_array = sum(array_without_extra_integer)

    # Return the difference, which is the added integer.
    return sum_of_larger_array - sum_of_smaller_array

Big(O) Analysis

Time Complexity

O(n) – The algorithm calculates the sum of two arrays. Calculating the sum of an array with n elements requires iterating through each element once. Therefore, calculating the sum of the first array takes O(n) time, and calculating the sum of the second array also takes O(n) time. The subtraction operation is constant time, O(1). Thus, the overall time complexity is O(n) + O(n) + O(1), which simplifies to O(n).

Space Complexity

O(1) – The solution calculates the sums of the two lists by iterating through them. It only needs to store a few variables to hold the sums of the lists, and the final difference. The amount of extra memory used does not depend on the size of the input lists. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
nums1 is null or empty	Return 0 since no elements exist in nums1, meaning that nums2[0] would be the integer added, or throw an exception if no valid result exists.
nums2 is null or empty	If nums1 is not null and not empty and nums2 is null/empty return the sum of nums1 or throw an exception if no valid result exists.
nums1 and nums2 are both empty	Return 0 if both arrays are empty as no integer was added.
nums1 has only one element	Return nums2[0] - nums1[0] which is the only possible added value.
nums2 has only one element and nums1 is empty	Return nums2[0] as the added value.
Integer overflow with very large numbers in the arrays	Use a data type that can hold large numbers (e.g., long) during summation or consider modulo operations to avoid overflow, if allowed by the problem statement.
Arrays contain negative numbers and zeros	The summation or hash map approach correctly handles negative numbers and zeros, no special handling needed.
nums2 does not contain all the elements of nums1	Throw an exception to indicate an invalid input as the problem description states that nums2 is formed by shuffling nums1 and adding one integer.

nums1 is null or empty

How to Handle:

Return 0 since no elements exist in nums1, meaning that nums2[0] would be the integer added, or throw an exception if no valid result exists.

nums2 is null or empty

How to Handle:

If nums1 is not null and not empty and nums2 is null/empty return the sum of nums1 or throw an exception if no valid result exists.

nums1 and nums2 are both empty

How to Handle:

Return 0 if both arrays are empty as no integer was added.

nums1 has only one element

How to Handle:

Return nums2[0] - nums1[0] which is the only possible added value.

nums2 has only one element and nums1 is empty

How to Handle:

Return nums2[0] as the added value.

Integer overflow with very large numbers in the arrays

How to Handle:

Use a data type that can hold large numbers (e.g., long) during summation or consider modulo operations to avoid overflow, if allowed by the problem statement.

Arrays contain negative numbers and zeros

How to Handle:

The summation or hash map approach correctly handles negative numbers and zeros, no special handling needed.

nums2 does not contain all the elements of nums1

How to Handle:

Throw an exception to indicate an invalid input as the problem description states that nums2 is formed by shuffling nums1 and adding one integer.