Find the Integer Added to Array II

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Arrays

You are given two integer arrays nums1 and nums2.

From nums1 two elements have been removed, and all other elements have 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 minimum possible integer x that achieves this equivalence.

Example 1:

Input: nums1 = [4,20,16,12,8], nums2 = [14,18,10]

Output: -2

Explanation:

After removing elements at indices [0,4] and adding -2, nums1 becomes [18,14,10].

Example 2:

Input: nums1 = [3,5,5,3], nums2 = [7,7]

Output: 2

Explanation:

After removing elements at indices [0,3] and adding 2, nums1 becomes [7,7].

Constraints:

3 <= nums1.length <= 200
nums2.length == nums1.length - 2
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 removing two elements and 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 are the possible ranges for the integer values within the input array?
Can the input array be empty or null?
Is it guaranteed that exactly one integer was added, or could there be zero or multiple additions?
If no integer was added such that the condition is met, what should I return?
What data type should I return for the added integer? (e.g., int, long)

Brute Force Solution

Approach

The brute force approach involves examining every possible integer. We will systematically test each number to see if it solves our puzzle. This is done by directly substituting it to check if it fits.

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

Start by considering a possible number, starting with a small value such as 1.
Pretend this number was added to the original collection of numbers.
Calculate the sum of all the numbers in the modified collection (the original numbers plus the added one).
Check if the calculated sum matches the required total sum.
If the sum matches, you've found the added number.
If the sum does not match, discard this possibility and try a different number. Maybe the next number, 2.
Continue this process, trying out different numbers, until you find one that produces the required total sum when added to the original collection.

Code Implementation

def find_the_integer_added_to_array_ii(original_numbers, required_total_sum):

    possible_number = 1

    while True:

        # Create the modified list with the possible added number
        modified_numbers = original_numbers + [possible_number]

        # Calculate the sum of the modified list
        calculated_sum = sum(modified_numbers)

        # If the sum matches, we found the added number
        if calculated_sum == required_total_sum:
            return possible_number

        # Try the next possible number
        possible_number += 1

Big(O) Analysis

Time Complexity

O(k) – The brute force approach iterates through possible integers (let's denote the number of iterations as k) until it finds the integer that, when added to the original array, produces the target sum. For each potential integer, it calculates the sum of the array plus that integer in O(n) time, where n is the size of the array. However, the problem description only specifies that we are testing each number. The critical factor is how many numbers we test. We test k possible integer numbers, so the outer loop determines the complexity. Therefore the time complexity is O(k), where k is the number of tested integers.

Space Complexity

O(1) – The provided brute force approach iteratively tests integer values. It involves calculating the sum of the input array plus a single candidate integer. Only a few constant space variables are needed: one to store the candidate integer being tested, one to accumulate the sum, and potentially a variable to store the size of the input array. The memory used is independent of the size of the input array, N. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

This problem is about finding a missing number in a collection after a number has been added. The clever trick is to compare the sums of both collections of numbers to find the difference, which will tell us the added number.

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

First, find the total of all the numbers in the original collection.
Next, find the total of all the numbers in the new collection after the addition.
Subtract the original total from the new total. The result is the number that was added.

Code Implementation

def find_added_integer(original_collection, new_collection):
    original_total = 0
    new_total = 0

    # Calculate the sum of the original
    # collection before addition.
    for number in original_collection:
        original_total += number

    # Calculate the sum of the new collection
    # after addition.
    for number in new_collection:
        new_total += number

    #The added number is the difference between the new and original totals
    added_number = new_total - original_total

    return added_number

Big(O) Analysis

Time Complexity

O(n) – The provided solution involves iterating through the original collection once to calculate its sum, which takes O(n) time where n is the number of elements in the original collection. Similarly, it iterates through the new collection once to calculate its sum, also taking O(n) time. Finally, it performs a single subtraction operation. Therefore, the dominant factor is the two O(n) operations, and the overall time complexity is O(n).

Space Complexity

O(1) – The provided algorithm calculates the sum of each collection (original and new). It only stores the sums of the original and new collections in constant space. These sums are stored in variables that do not depend on the size of the input collections. Therefore, the auxiliary space complexity is constant, or O(1).

Edge Cases

Case	How to Handle
Null input array	Throw an IllegalArgumentException or return an empty list indicating invalid input.
Empty input array	Return an empty list immediately as there's nothing to process.
Input array with only one element	Return an empty list because we need at least two elements to find the added integer.
Array with all identical values	The solution should handle duplicate numbers correctly to prevent incorrect counts or infinite loops.
Large input array causing potential memory issues	Ensure that the algorithm's space complexity is reasonable or consider alternative data structures for efficient memory usage.
Array contains negative numbers	The solution should handle negative numbers in the array correctly during the addition or comparison process.
Integer overflow during addition of large numbers	Use a data type with a larger range (e.g., long) or implement overflow checking during addition.
Array contains zeros	The presence of zero should not cause division by zero errors or other unexpected behavior in the core logic.

Null input array

How to Handle:

Throw an IllegalArgumentException or return an empty list indicating invalid input.

Empty input array

How to Handle:

Return an empty list immediately as there's nothing to process.

Input array with only one element

How to Handle:

Return an empty list because we need at least two elements to find the added integer.

Array with all identical values

How to Handle:

The solution should handle duplicate numbers correctly to prevent incorrect counts or infinite loops.

Large input array causing potential memory issues

How to Handle:

Ensure that the algorithm's space complexity is reasonable or consider alternative data structures for efficient memory usage.

Array contains negative numbers

How to Handle:

The solution should handle negative numbers in the array correctly during the addition or comparison process.

Integer overflow during addition of large numbers

How to Handle:

Use a data type with a larger range (e.g., long) or implement overflow checking during addition.

Array contains zeros

How to Handle:

The presence of zero should not cause division by zero errors or other unexpected behavior in the core logic.