Choose Numbers From Two Arrays in Range

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You are given two 0-indexed integer arrays nums1 and nums2 of lengths n and m respectively, and two integers lower and upper.

You need to choose some integers from both arrays such that:

nums1[i] is chosen with index i in the range [0, n - 1].
nums2[j] is chosen with index j in the range [0, m - 1].
You can choose at most one integer from each array.

Return the number of integer pairs that satisfy lower <= nums1[i] + nums2[j] <= upper.

Example 1:

Input: nums1 = [1,3], nums2 = [2,8], lower = 5, upper = 10
Output: 2
Explanation: The possible pairs are:
- (1, 2). Their sum is 3, which is not in the range [5, 10].
- (1, 8). Their sum is 9, which is in the range [5, 10].
- (3, 2). Their sum is 5, which is in the range [5, 10].
- (3, 8). Their sum is 11, which is not in the range [5, 10].
So we return 2.

Example 2:

Input: nums1 = [0,-2,10,-5,2], nums2 = [-1,3,-6,-4,8], lower = -2, upper = 7
Output: 22
Explanation: The possible pairs are:
- (0, -1). Their sum is -1, which is in the range [-2, 7].
- (0, 3). Their sum is 3, which is in the range [-2, 7].
- (0, -6). Their sum is -6, which is not in the range [-2, 7].
- (0, -4). Their sum is -4, which is not in the range [-2, 7].
- (0, 8). Their sum is 8, which is not in the range [-2, 7].
- (-2, -1). Their sum is -3, which is not in the range [-2, 7].
- (-2, 3). Their sum is 1, which is in the range [-2, 7].
- (-2, -6). Their sum is -8, which is not in the range [-2, 7].
- (-2, -4). Their sum is -6, which is not in the range [-2, 7].
- (-2, 8). Their sum is 6, which is in the range [-2, 7].
- (10, -1). Their sum is 9, which is not in the range [-2, 7].
- (10, 3). Their sum is 13, which is not in the range [-2, 7].
- (10, -6). Their sum is 4, which is in the range [-2, 7].
- (10, -4). Their sum is 6, which is in the range [-2, 7].
- (10, 8). Their sum is 18, which is not in the range [-2, 7].
- (-5, -1). Their sum is -6, which is not in the range [-2, 7].
- (-5, 3). Their sum is -2, which is in the range [-2, 7].
- (-5, -6). Their sum is -11, which is not in the range [-2, 7].
- (-5, -4). Their sum is -9, which is not in the range [-2, 7].
- (-5, 8). Their sum is 3, which is in the range [-2, 7].
- (2, -1). Their sum is 1, which is in the range [-2, 7].
- (2, 3). Their sum is 5, which is in the range [-2, 7].
- (2, -6). Their sum is -4, which is not in the range [-2, 7].
- (2, -4). Their sum is -2, which is in the range [-2, 7].
- (2, 8). Their sum is 10, which is not in the range [-2, 7].
So we return 22.

Constraints:

n == nums1.length
m == nums2.length
1 <= n, m <= 1000
-10⁵ <= nums1[i], nums2[i] <= 10⁵
-10⁵ <= lower <= upper <= 10⁵

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 numbers within `nums1` and `nums2`, as well as for `low` and `high`?
Can `nums1` or `nums2` be empty or null?
Are duplicate pairs (i, j) and (j, i) considered distinct if `nums1[i] + nums2[j]` and `nums1[j] + nums2[i]` both fall within the range, assuming the arrays have overlapping values?
If no pairs satisfy the condition, what should I return?
Are `low` and `high` guaranteed to be valid integers and is `low` always less than or equal to `high`?

Brute Force Solution

Approach

We are given two groups of numbers and a range. The brute force method systematically explores every possible combination of numbers, one from each group, to determine if their total falls within the given range.

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

Take the first number from the first group.
Combine it with the first number from the second group and check if the sum is within the range.
If it is not, try combining the first number from the first group with the second number from the second group, and check again.
Keep trying the first number from the first group with every number from the second group, each time checking if the sum is within the range.
Once you've checked the first number from the first group with all numbers from the second group, move on to the second number from the first group.
Repeat this process, combining each number from the first group with every number from the second group, and checking if their sum falls within the range, until all possible combinations have been checked.

Code Implementation

def choose_numbers_from_two_arrays_in_range_brute_force(
    first_group_of_numbers, second_group_of_numbers, lower_bound, upper_bound
):
    count_of_valid_combinations = 0

    # Iterate through each number in the first group
    for first_number_index in range(len(first_group_of_numbers)):
        first_number = first_group_of_numbers[first_number_index]

        # Iterate through each number in the second group
        for second_number_index in range(len(second_group_of_numbers)):
            second_number = second_group_of_numbers[second_number_index]

            # Check if the sum is within the specified range
            sum_of_numbers = first_number + second_number

            if lower_bound <= sum_of_numbers <= upper_bound:
                # Count the combination if within the range
                count_of_valid_combinations += 1\   return count_of_valid_combinations

Big(O) Analysis

Time Complexity

O(n*m) – The algorithm iterates through each element of the first array, which has a size we can denote as 'n'. For each of these 'n' elements, it then iterates through every element of the second array, which has a size denoted as 'm', to check if the sum of the pair falls within the specified range. Each pair-sum check is a constant-time operation. Therefore, the total number of operations is proportional to n multiplied by m, giving us a time complexity of O(n*m).

Space Complexity

O(1) – The provided brute force algorithm iterates through the two input arrays using nested loops, combining one number from each array at a time. It doesn't use any auxiliary data structures like lists, sets, or maps to store intermediate results or track visited elements. The space used by the algorithm is limited to a few variables, such as loop counters, irrespective of the size of the input arrays, therefore the space complexity is constant.

Optimal Solution

Approach

The efficient solution focuses on identifying valid number pairs within the specified range across both number collections without testing every single possible combination. It uses a filtering strategy to isolate potentially valid numbers early on, leading to fewer calculations and a faster result.

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

First, identify all the numbers from both collections that fall within the specified numerical range. These are the only numbers that matter.
Next, for each number from the first collection that's within the range, quickly check if there's a number in the second collection that, when combined with the first number, also falls within the range.
Keep a record of each pair of numbers found that meet the criteria, avoiding duplicates.
Finally, report the total number of unique pairs that were found.

Code Implementation

def find_number_pairs(collection_one, collection_two, lower_bound, upper_bound):
    valid_numbers_collection_one = [number for number in collection_one if lower_bound <= number <= upper_bound]
    valid_numbers_collection_two = [number for number in collection_two if lower_bound <= number <= upper_bound]

    number_pairs = set()

    # Iterate through the valid numbers
    for first_number in valid_numbers_collection_one:
        # Check only numbers that could possibly be part of a valid pair
        for second_number in valid_numbers_collection_two:
            if lower_bound <= first_number + second_number <= upper_bound:

                # Add the pair to the set of valid pairs
                number_pairs.add(tuple(sorted((first_number, second_number))))

    # Get the count of unique number pairs.
    number_pairs_count = len(number_pairs)

    return number_pairs_count

Big(O) Analysis

Time Complexity

O(n) – Let n be the total number of elements in both arrays. Identifying numbers within the range takes O(n) time as we iterate through both arrays once. Checking each number from the first collection against the second collection involves iterating at most through the filtered subset of the second collection. Since the filtering step ensures we only consider numbers within the specified range, and we do not iterate through the entire second array for each number in the first array, the pair checking does not lead to O(n^2). The overall time complexity is dominated by the initial filtering step, resulting in O(n) where n is the total size of the inputs because no pair checking is done against the whole array.

Space Complexity

O(N) – The space complexity is dominated by storing numbers from both collections that fall within the specified numerical range. In the worst case, all numbers from both arrays might fall within the range, requiring an auxiliary array (or two) to store them. Let N be the total number of elements across both input arrays; in the worst-case scenario where all N elements fall within the range, the auxiliary space becomes proportional to N. Therefore, the space complexity is O(N).

Edge Cases

Case	How to Handle
Either nums1 or nums2 is null or empty	Return 0 if either array is null or empty, as no pairs can be formed.
nums1 and nums2 both have only one element	Check if the sum of the single elements is within the [low, high] range, and return 1 if it is, otherwise 0.
nums1 and nums2 contain duplicate values, leading to multiple identical sums	Use a set or similar data structure to store the distinct sums, ensuring each sum is counted only once.
nums1 and nums2 contain negative numbers, zeros, and positive numbers	The algorithm should correctly handle all types of integers, as addition and comparison will work regardless of sign.
low and high are equal (a very narrow range)	The code should function correctly when low and high are equal, efficiently searching for pairs that sum to that specific value.
No pairs exist that sum within the range [low, high]	The algorithm should correctly return 0 when no pairs satisfy the condition.
Integer overflow when summing elements from nums1 and nums2	Use long data type to store the sum to prevent potential overflow issues when adding large integers.
Large input arrays affecting performance	Optimize the solution using efficient algorithms like sorting and binary search to achieve better time complexity.

Either nums1 or nums2 is null or empty

How to Handle:

Return 0 if either array is null or empty, as no pairs can be formed.

nums1 and nums2 both have only one element

How to Handle:

Check if the sum of the single elements is within the [low, high] range, and return 1 if it is, otherwise 0.

nums1 and nums2 contain duplicate values, leading to multiple identical sums

How to Handle:

Use a set or similar data structure to store the distinct sums, ensuring each sum is counted only once.

nums1 and nums2 contain negative numbers, zeros, and positive numbers

How to Handle:

The algorithm should correctly handle all types of integers, as addition and comparison will work regardless of sign.

low and high are equal (a very narrow range)

How to Handle:

The code should function correctly when low and high are equal, efficiently searching for pairs that sum to that specific value.

No pairs exist that sum within the range [low, high]

How to Handle:

The algorithm should correctly return 0 when no pairs satisfy the condition.

Integer overflow when summing elements from nums1 and nums2

How to Handle:

Use long data type to store the sum to prevent potential overflow issues when adding large integers.

Large input arrays affecting performance

How to Handle:

Optimize the solution using efficient algorithms like sorting and binary search to achieve better time complexity.