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Find the Distance Value Between Two Arrays

Easy

Asked by:

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Topics:

ArraysBinary Search

Given two integer arrays arr1 and arr2, and the integer d, return the distance value between the two arrays.

The distance value is defined as the number of elements arr1[i] such that there is not any element arr2[j] where |arr1[i]-arr2[j]| <= d.

Example 1:

Input: arr1 = [4,5,8], arr2 = [10,9,1,8], d = 2
Output: 2
Explanation: 
For arr1[0]=4 we have: 
|4-10|=6 > d=2 
|4-9|=5 > d=2 
|4-1|=3 > d=2 
|4-8|=4 > d=2 
For arr1[1]=5 we have: 
|5-10|=5 > d=2 
|5-9|=4 > d=2 
|5-1|=4 > d=2 
|5-8|=3 > d=2
For arr1[2]=8 we have:
|8-10|=2 <= d=2
|8-9|=1 <= d=2
|8-1|=7 > d=2
|8-8|=0 <= d=2

Example 2:

Input: arr1 = [1,4,2,3], arr2 = [-4,-3,6,10,20,30], d = 3
Output: 2

Example 3:

Input: arr1 = [2,1,100,3], arr2 = [-5,-2,10,-3,7], d = 6
Output: 1

Constraints:

1 <= arr1.length, arr2.length <= 500
-1000 <= arr1[i], arr2[j] <= 1000
0 <= d <= 100

Try coding it in LeetCode

Find the Distance Value Between Two Arrays

Easy

Asked by:

24 views

Topics:

ArraysBinary Search

Given two integer arrays arr1 and arr2, and the integer d, return the distance value between the two arrays.

The distance value is defined as the number of elements arr1[i] such that there is not any element arr2[j] where |arr1[i]-arr2[j]| <= d.

Example 1:

Input: arr1 = [4,5,8], arr2 = [10,9,1,8], d = 2
Output: 2
Explanation: 
For arr1[0]=4 we have: 
|4-10|=6 > d=2 
|4-9|=5 > d=2 
|4-1|=3 > d=2 
|4-8|=4 > d=2 
For arr1[1]=5 we have: 
|5-10|=5 > d=2 
|5-9|=4 > d=2 
|5-1|=4 > d=2 
|5-8|=3 > d=2
For arr1[2]=8 we have:
|8-10|=2 <= d=2
|8-9|=1 <= d=2
|8-1|=7 > d=2
|8-8|=0 <= d=2

Example 2:

Input: arr1 = [1,4,2,3], arr2 = [-4,-3,6,10,20,30], d = 3
Output: 2

Example 3:

Input: arr1 = [2,1,100,3], arr2 = [-5,-2,10,-3,7], d = 6
Output: 1

Constraints:

1 <= arr1.length, arr2.length <= 500
-1000 <= arr1[i], arr2[j] <= 1000
0 <= d <= 100

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 constraints on the size of `arr1` and `arr2`?
Can the integer values in `arr1`, `arr2`, and `d` be negative, zero, or very large?
Are duplicate values allowed within `arr1` and `arr2`? If so, how should I handle them?
If either `arr1` or `arr2` is empty, what should the return value be?
Could you provide an example where the distance value is zero?

Brute Force Solution

Approach

We need to find how many numbers from one list are 'far enough' away from all numbers in another list. The brute force approach is like checking every possible pairing between the two lists to see if the distance requirement is met.

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

Take the first number from the first list.
Compare that number to every single number in the second list.
If the difference between the number from the first list and any number in the second list is too small (not 'far enough' apart), then move on to the next number in the first list.
If the difference between the number from the first list and all the numbers in the second list is big enough ('far enough' apart from all), then remember that the number from the first list meets our criteria.
Repeat this process for every number in the first list.
Finally, count how many numbers from the first list met the criteria of being 'far enough' away from all the numbers in the second list.

Code Implementation

def find_the_distance_value(first_array, second_array, distance):
    valid_numbers_count = 0

    for first_array_number in first_array:
        is_valid = True

        # Compare the number from the first array with all numbers from the second array
        for second_array_number in second_array:

            # If absolute difference is less than or equal to distance, then the number is not valid
            if abs(first_array_number - second_array_number) <= distance:
                is_valid = False
                break

        # If, after comparing with all numbers from the second array, the number is still valid
        if is_valid:
            valid_numbers_count += 1

    return valid_numbers_count

Big(O) Analysis

Time Complexity

O(n*m) – The algorithm iterates through each of the 'n' elements in the first array (arr1). For each element in arr1, it iterates through all 'm' elements in the second array (arr2) to check if the absolute difference between the current element in arr1 and any element in arr2 is less than or equal to 'd'. Therefore, for each of the 'n' elements in arr1, we perform 'm' comparisons. Consequently, the total number of operations is proportional to n * m, which gives us a time complexity of O(n*m).

Space Complexity

O(1) – The described brute force approach does not use any auxiliary data structures like lists, maps, or sets to store intermediate results. It only utilizes a few constant space variables such as counters and boolean flags to track the comparisons, regardless of the sizes of the input arrays. Therefore, the auxiliary space complexity is constant and independent of the input size. This results in O(1) space complexity.

Optimal Solution

Approach

The efficient way to solve this problem is to check each number in the first group against the numbers in the second group in an organized way. Sorting the second group allows us to quickly find the closest number to each number in the first group, so we don't need to check every single pair.

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

First, arrange the numbers in the second group from smallest to largest. This makes it easier to quickly find the closest number.
Now, take each number from the first group one at a time.
For each number from the first group, find the number in the second group that's closest to it. Since the second group is sorted, you can quickly narrow down your search.
Check if the closest number is within the specified distance. If it is, then we know this number from the first group should not be counted.
If the closest number is further than the allowed distance, we count this number from the first group because it meets the criteria.
Repeat this process for all numbers in the first group and total up the count of numbers that met the criteria.

Code Implementation

def findTheDistanceValue(array_one, array_two, distance):
    array_two.sort()
    distance_value = 0

    for number_one in array_one:
        closest_distance = float('inf')

        # Find closest number in array_two using binary search principle
        left_pointer = 0
        right_pointer = len(array_two) - 1

        while left_pointer <= right_pointer:
            middle_pointer = (left_pointer + right_pointer) // 2
            absolute_difference = abs(number_one - array_two[middle_pointer])
            closest_distance = min(closest_distance, absolute_difference)

            # Adjust search range based on value at middle_pointer
            if array_two[middle_pointer] < number_one:
                left_pointer = middle_pointer + 1
            else:
                right_pointer = middle_pointer - 1

        # Increment distance_value if no element is within distance
        if closest_distance > distance:
            distance_value += 1

    return distance_value

Big(O) Analysis

Time Complexity

O(n log m) – First, the algorithm sorts arr2, which takes O(m log m) time, where m is the length of arr2. Then, for each of the n elements in arr1, a binary search is performed on arr2 to find the closest element, which takes O(log m) time per element. This binary search is repeated n times, resulting in a total time complexity of O(n log m) for the search operations. Since the sorting operation O(m log m) only happens once, and assuming n is large enough relative to m such that n log m dominates m log m, the overall time complexity is O(n log m).

Space Complexity

O(1) – The provided solution sorts the second array in place, so it doesn't use additional space proportional to the input size for sorting. We only need to store a few constant-size variables such as the distance value, loop counters, and potentially a variable to keep track of the count. Therefore, the algorithm's space complexity is constant, independent of the size of the input arrays.

Edge Cases

Case	How to Handle
Both arr1 and arr2 are empty	Return 0, as there are no elements in arr1 to satisfy the condition.
arr1 is empty, arr2 is not empty	Return 0, as there are no elements in arr1 to check.
arr2 is empty, arr1 is not empty	Return the length of arr1, as the condition is trivially satisfied for all elements in arr1.
d is 0	Check for exact matches between elements of arr1 and arr2.
d is a large value (close to max int)	Ensure that the absolute difference calculation doesn't cause integer overflow.
arr1 and arr2 contain the same element multiple times, d is small	Ensure each element of arr1 is only counted once if no element in arr2 satisfies condition.
arr1 and arr2 contain negative numbers	The absolute difference calculation must handle negative numbers correctly.
Large arrays with a small d, many elements in arr1 close to elements in arr2	Consider sorting arr2 to improve search efficiency using binary search.

Both arr1 and arr2 are empty

How to Handle:

Return 0, as there are no elements in arr1 to satisfy the condition.

arr1 is empty, arr2 is not empty

How to Handle:

Return 0, as there are no elements in arr1 to check.

arr2 is empty, arr1 is not empty

How to Handle:

Return the length of arr1, as the condition is trivially satisfied for all elements in arr1.

d is 0

How to Handle:

Check for exact matches between elements of arr1 and arr2.

d is a large value (close to max int)

How to Handle:

Ensure that the absolute difference calculation doesn't cause integer overflow.

arr1 and arr2 contain the same element multiple times, d is small

How to Handle:

Ensure each element of arr1 is only counted once if no element in arr2 satisfies condition.

arr1 and arr2 contain negative numbers

How to Handle:

The absolute difference calculation must handle negative numbers correctly.

Large arrays with a small d, many elements in arr1 close to elements in arr2

How to Handle:

Consider sorting arr2 to improve search efficiency using binary search.

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 constraints on the size of `arr1` and `arr2`?
Can the integer values in `arr1`, `arr2`, and `d` be negative, zero, or very large?
Are duplicate values allowed within `arr1` and `arr2`? If so, how should I handle them?
If either `arr1` or `arr2` is empty, what should the return value be?
Could you provide an example where the distance value is zero?

Brute Force Solution

Approach

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

Take the first number from the first list.
Compare that number to every single number in the second list.
If the difference between the number from the first list and any number in the second list is too small (not 'far enough' apart), then move on to the next number in the first list.
If the difference between the number from the first list and all the numbers in the second list is big enough ('far enough' apart from all), then remember that the number from the first list meets our criteria.
Repeat this process for every number in the first list.
Finally, count how many numbers from the first list met the criteria of being 'far enough' away from all the numbers in the second list.

Code Implementation

def find_the_distance_value(first_array, second_array, distance):
    valid_numbers_count = 0

    for first_array_number in first_array:
        is_valid = True

        # Compare the number from the first array with all numbers from the second array
        for second_array_number in second_array:

            # If absolute difference is less than or equal to distance, then the number is not valid
            if abs(first_array_number - second_array_number) <= distance:
                is_valid = False
                break

        # If, after comparing with all numbers from the second array, the number is still valid
        if is_valid:
            valid_numbers_count += 1

    return valid_numbers_count

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, arrange the numbers in the second group from smallest to largest. This makes it easier to quickly find the closest number.
Now, take each number from the first group one at a time.
For each number from the first group, find the number in the second group that's closest to it. Since the second group is sorted, you can quickly narrow down your search.
Check if the closest number is within the specified distance. If it is, then we know this number from the first group should not be counted.
If the closest number is further than the allowed distance, we count this number from the first group because it meets the criteria.
Repeat this process for all numbers in the first group and total up the count of numbers that met the criteria.

Code Implementation

def findTheDistanceValue(array_one, array_two, distance):
    array_two.sort()
    distance_value = 0

    for number_one in array_one:
        closest_distance = float('inf')

        # Find closest number in array_two using binary search principle
        left_pointer = 0
        right_pointer = len(array_two) - 1

        while left_pointer <= right_pointer:
            middle_pointer = (left_pointer + right_pointer) // 2
            absolute_difference = abs(number_one - array_two[middle_pointer])
            closest_distance = min(closest_distance, absolute_difference)

            # Adjust search range based on value at middle_pointer
            if array_two[middle_pointer] < number_one:
                left_pointer = middle_pointer + 1
            else:
                right_pointer = middle_pointer - 1

        # Increment distance_value if no element is within distance
        if closest_distance > distance:
            distance_value += 1

    return distance_value

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Both arr1 and arr2 are empty	Return 0, as there are no elements in arr1 to satisfy the condition.
arr1 is empty, arr2 is not empty	Return 0, as there are no elements in arr1 to check.
arr2 is empty, arr1 is not empty	Return the length of arr1, as the condition is trivially satisfied for all elements in arr1.
d is 0	Check for exact matches between elements of arr1 and arr2.
d is a large value (close to max int)	Ensure that the absolute difference calculation doesn't cause integer overflow.
arr1 and arr2 contain the same element multiple times, d is small	Ensure each element of arr1 is only counted once if no element in arr2 satisfies condition.
arr1 and arr2 contain negative numbers	The absolute difference calculation must handle negative numbers correctly.
Large arrays with a small d, many elements in arr1 close to elements in arr2	Consider sorting arr2 to improve search efficiency using binary search.

Both arr1 and arr2 are empty

How to Handle:

Return 0, as there are no elements in arr1 to satisfy the condition.

arr1 is empty, arr2 is not empty

How to Handle:

Return 0, as there are no elements in arr1 to check.

arr2 is empty, arr1 is not empty

How to Handle:

Return the length of arr1, as the condition is trivially satisfied for all elements in arr1.

d is 0

How to Handle:

Check for exact matches between elements of arr1 and arr2.

d is a large value (close to max int)

How to Handle:

Ensure that the absolute difference calculation doesn't cause integer overflow.

arr1 and arr2 contain the same element multiple times, d is small

How to Handle:

Ensure each element of arr1 is only counted once if no element in arr2 satisfies condition.

arr1 and arr2 contain negative numbers

How to Handle:

The absolute difference calculation must handle negative numbers correctly.

Large arrays with a small d, many elements in arr1 close to elements in arr2

How to Handle:

Consider sorting arr2 to improve search efficiency using binary search.