Sum of Distances

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Arrays

You are given a 0-indexed integer array nums. There exists an array arr of length nums.length, where arr[i] is the sum of |i - j| over all j such that nums[j] == nums[i] and j != i. If there is no such j, set arr[i] to be 0.

Return the array arr.

Example 1:

Input: nums = [1,3,1,1,2]
Output: [5,0,3,4,0]
Explanation: 
When i = 0, nums[0] == nums[2] and nums[0] == nums[3]. Therefore, arr[0] = |0 - 2| + |0 - 3| = 5. 
When i = 1, arr[1] = 0 because there is no other index with value 3.
When i = 2, nums[2] == nums[0] and nums[2] == nums[3]. Therefore, arr[2] = |2 - 0| + |2 - 3| = 3. 
When i = 3, nums[3] == nums[0] and nums[3] == nums[2]. Therefore, arr[3] = |3 - 0| + |3 - 2| = 4. 
When i = 4, arr[4] = 0 because there is no other index with value 2.

Example 2:

Input: nums = [0,5,3]
Output: [0,0,0]
Explanation: Since each element in nums is distinct, arr[i] = 0 for all i.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁹

Note: This question is the same as 2121: Intervals Between Identical Elements.

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 values that each number in the input array can have? Can they be negative or zero?
How large can the input array be? What is the expected order of magnitude of the array size?
Are there any duplicate numbers in the input array? If so, how should they be handled when calculating distances?
Can you provide an example input and the expected output to ensure I understand the distance calculation correctly?
Is the input array guaranteed to be non-empty?

Brute Force Solution

Approach

To find the sum of distances, we will look at each location one by one. For each location, we'll calculate its distance to every other location and then add up all those distances.

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

Pick the first location.
Calculate the distance between that location and every other location in the set.
Add up all the distances you just calculated. This gives you the total distance for that first location.
Now, pick the second location.
Again, calculate the distance between this second location and every other location.
Add up all those distances to get the total distance for the second location.
Repeat this process for every location in the set.
Finally, add up all the total distances you calculated for each location. This final sum is the answer.

Code Implementation

def sum_of_distances_brute_force(locations):
    total_sum_of_distances = 0

    for current_location_index in range(len(locations)):
        current_location_total_distance = 0

        # Iterate through all other locations to calculate distance
        for other_location_index in range(len(locations)):
            # Skip calculating distance to itself
            if current_location_index == other_location_index:
                continue

            distance = abs(locations[current_location_index] - locations[other_location_index])
            current_location_total_distance += distance

        # Accumulate total distance for each location
        total_sum_of_distances += current_location_total_distance

    return total_sum_of_distances

Big(O) Analysis

Time Complexity

O(n^2) – The algorithm iterates through each of the n locations. For each location, it calculates the distance to every other location, which involves another loop of approximately n operations. Thus, for each of the n locations, we perform n distance calculations. This results in a total of n * n operations. Therefore, the time complexity is O(n^2).

Space Complexity

O(1) – The provided plain English explanation calculates distances between locations one by one and sums them up. It does not explicitly mention the use of any auxiliary data structures like arrays, hash maps, or any other temporary storage that grows with the input size N (where N is the number of locations). The calculations appear to be performed in place, using only a few variables to store the current location index, other location index and running sums, and temporary distance results, all of which take constant space. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

To efficiently calculate the sum of distances, avoid redundant calculations by reusing previously computed information. We'll precompute sums from the beginning and end of the data to quickly determine distances without iterating repeatedly.

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

First, create a running total of all the numbers from the start up to each position.
Next, create another running total but this time going from the end backwards up to each position.
Now, for each number in the original data, use the precomputed totals to easily find the sum of distances to all other numbers.
Calculate each individual distance efficiently using the formula derived from precomputed running totals.
Finally, add up all the individual distances calculated in the previous step to arrive at the final answer, which is the total sum of distances.

Code Implementation

def sum_of_distances(data):
    array_length = len(data)
    prefix_sum = [0] * array_length
    suffix_sum = [0] * array_length

    prefix_sum[0] = data[0]
    for index in range(1, array_length):
        prefix_sum[index] = prefix_sum[index - 1] + data[index]

    suffix_sum[array_length - 1] = data[array_length - 1]
    for index in range(array_length - 2, -1, -1):
        suffix_sum[index] = suffix_sum[index + 1] + data[index]

    total_distance = 0

    # Calculating total distance for each element.
    for index in range(array_length):

        left_distance = 0
        if index > 0:
            # Calculating distance to all elements to the left.
            left_distance = data[index] * index - prefix_sum[index - 1]

        right_distance = 0
        if index < array_length - 1:
            # Calculating distance to all elements to the right.
            right_distance = (suffix_sum[index + 1] - data[index] * (array_length - 1 - index))

        total_distance += left_distance + right_distance

    return total_distance

Big(O) Analysis

Time Complexity

O(n) – The algorithm involves creating two prefix sum arrays, each requiring iteration through the input array of size n once. Then, it iterates through the input array a third time to calculate the sum of distances using the precomputed prefix sums. Each of these iterations contributes linearly to the time complexity. Thus, the overall time complexity is O(n).

Space Complexity

O(N) – The algorithm creates two running total arrays: one from the start and one from the end. Each of these arrays stores N numbers, where N is the size of the original input data. Therefore, the auxiliary space required is proportional to the size of the input data. This results in a space complexity of O(N).

Edge Cases

Case	How to Handle
Empty input array	Return an empty array or an array of zeros with the same size as the input, depending on the specific requirements.
Input array with only one element	Return an array of zeros with the same size as the input, since the distance to itself is zero.
Integer overflow when calculating sums of distances	Use a data type with larger capacity, like long, to store intermediate sums.
Large input array causing time limit exceed	Optimize the algorithm to achieve a lower time complexity (e.g., from O(n^2) to O(n)).
Input array contains negative numbers	The algorithm should handle negative numbers correctly as they contribute to distances.
Input array contains duplicate numbers	Ensure the algorithm considers all instances of duplicates when calculating distances.
Input array contains very large numbers	Be mindful of potential integer overflow and consider using a larger data type, like long or double, for calculations and storage.
Input array with all identical numbers	All distances will be zero in this case, and the algorithm should correctly return an array of zeros.

Empty input array

How to Handle:

Return an empty array or an array of zeros with the same size as the input, depending on the specific requirements.

Input array with only one element

How to Handle:

Return an array of zeros with the same size as the input, since the distance to itself is zero.

Integer overflow when calculating sums of distances

How to Handle:

Use a data type with larger capacity, like long, to store intermediate sums.

Large input array causing time limit exceed

How to Handle:

Optimize the algorithm to achieve a lower time complexity (e.g., from O(n^2) to O(n)).

Input array contains negative numbers

How to Handle:

The algorithm should handle negative numbers correctly as they contribute to distances.

Input array contains duplicate numbers

How to Handle:

Ensure the algorithm considers all instances of duplicates when calculating distances.

Input array contains very large numbers

How to Handle:

Be mindful of potential integer overflow and consider using a larger data type, like long or double, for calculations and storage.

Input array with all identical numbers

How to Handle:

All distances will be zero in this case, and the algorithm should correctly return an array of zeros.