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Range Addition

Medium

Asked by:

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Topics:

Arrays

You are given an integer length and an array updates where each updates[i] = [startIdxi, endIdxi, inci].

You have an array arr of length length with all zeros. For each updates[i], increment all the elements arr[startIdxi ... endIdxi] (inclusive) by inci.

Return the modified array arr after applying all the updates.

Example 1:

Input: length = 5, updates = [[1,3,2],[2,4,3],[0,2,-2]]
Output: [-2,0,3,5,3]

Example 2:

Input: length = 10, updates = [[2,4,6],[5,6,8],[1,9,-4]]
Output: [0,-4,2,2,2,4,4,-4,-4,-4]

Constraints:

1 <= length <= 10⁵
0 <= updates.length <= 10⁴
0 <= startIdxi < endIdxi < length
-1000 <= inci <= 1000

Try coding it in LeetCode

Range Addition

Medium

Asked by:

15 views

Topics:

Arrays

You are given an integer length and an array updates where each updates[i] = [startIdxi, endIdxi, inci].

You have an array arr of length length with all zeros. For each updates[i], increment all the elements arr[startIdxi ... endIdxi] (inclusive) by inci.

Return the modified array arr after applying all the updates.

Example 1:

Input: length = 5, updates = [[1,3,2],[2,4,3],[0,2,-2]]
Output: [-2,0,3,5,3]

Example 2:

Input: length = 10, updates = [[2,4,6],[5,6,8],[1,9,-4]]
Output: [0,-4,2,2,2,4,4,-4,-4,-4]

Constraints:

1 <= length <= 10⁵
0 <= updates.length <= 10⁴
0 <= startIdxi < endIdxi < length
-1000 <= inci <= 1000

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 length of the array and the number of update operations? How large can 'length' and the number of update operations be?
What are the possible values for the start index, end index, and increment? Can they be negative?
Is the original array initialized with all zeros, or some other default value?
Do the update operations ever overlap (i.e., can multiple operations affect the same index)?
Should I perform input validation (e.g., checking if start index is less than or equal to end index, and that both are within the bounds of the array)? What should I return if the input is invalid?

Brute Force Solution

Approach

The brute force approach directly simulates the addition ranges. It starts with a list of zeros and, for each update, it modifies the affected portion of this list. This is done by stepping through the specified range and adding the given value to each element within that range.

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

Start with a list filled with zeros, representing your initial data.
For each update instruction, identify the starting and ending positions of the range to be modified.
Go through each position within that range, one by one.
At each position, add the update value to the current value already there.
Repeat this process for all update instructions.
The final list contains the updated values after applying all the range additions.

Code Implementation

def range_addition_brute_force(length, updates):
    result_array = [0] * length

    # Iterate through each update instruction
    for update in updates:
        start_index = update[0]
        end_index = update[1]
        increment_value = update[2]

        # Apply the update to the specified range
        for index in range(start_index, end_index + 1):

            # Directly modifying the affected list elements.
            result_array[index] += increment_value

    return result_array

Big(O) Analysis

Time Complexity

O(k*n) – The algorithm starts with an array of size n. Then, for each of the k update operations, it iterates through a subarray of size n in the worst case. Therefore, the outer loop, which represents the number of update operations (k), is multiplied by the inner loop which represents the size of the array (n). This results in a time complexity of O(k*n).

Space Complexity

O(1) – The brute force approach described initializes a list of zeros. However, the algorithm modifies this initial list in place. The provided steps involve identifying range boundaries and updating values directly within the list; there is no mention of creating additional data structures such as temporary lists, hash maps or recursion stacks that scale with the input size. Thus, the auxiliary space used is constant, independent of the size of the input list.

Optimal Solution

Approach

The efficient way to solve this is by tracking the *changes* to values over the given ranges, rather than directly updating each value individually. We leverage these changes to quickly compute the final result. It's like noting down the events, and applying those all at once at the end.

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

For each update, simply add the value to the *start* position of the range.
Also, subtract the same value from the position *just after* the end of the range (if that position exists).
Once you've processed all updates, go through the entire data structure. Keep a running total.
At each position, add the value stored there to the running total. This gives you the correct final value for that position.
This running total now reflects the net effect of all range updates up to that position.

Code Implementation

def range_addition(length_of_array, updates):
    result_array = [0] * length_of_array

    for start_index, end_index, value_to_add in updates:
        # Mark the start of the range with the value
        result_array[start_index] += value_to_add

        if end_index + 1 < length_of_array:
            # Subtract after the end to revert the addition
            result_array[end_index + 1] -= value_to_add

    current_sum = 0
    for index in range(length_of_array):
        # Accumulate the changes to determine final values
        current_sum += result_array[index]
        result_array[index] = current_sum

    return result_array

Big(O) Analysis

Time Complexity

O(n + k) – The algorithm involves iterating through the 'updates' array, which has length 'k', and performing constant-time operations for each update. This step takes O(k) time. Afterwards, it iterates through the result array, which has length 'n', to compute the cumulative sum. This step takes O(n) time. Therefore, the overall time complexity is O(k + n), which can be simplified to O(n + k), assuming k (number of updates) and n (length of array) are independent.

Space Complexity

O(1) – The algorithm uses an array of size N which is part of the input, so we don't count that as auxiliary space. The updates are performed in-place on this input array. Beyond the input, we only store a single running total variable to accumulate values. No other data structures dependent on the size of input are used. Therefore, the auxiliary space complexity is constant.

Edge Cases

Case	How to Handle
Null or empty updates array	Return the original array without modification if the updates array is null or empty.
Updates array with empty update ranges	Skip any updates where the start index is greater than the end index.
Array length of 0	Return an empty array since no updates can be performed on an empty array.
Large array size and numerous updates impacting performance	Use the difference array technique to optimize the update operations to O(1) per update.
Update ranges exceeding array bounds	Clamp the start and end indices of the update ranges to the valid array bounds.
Updates with zero increment values	These updates should be skipped, as they don't affect the array.
Integer overflow when adding increments to array elements	Use a data type with a larger range (e.g., long) to store the array elements and intermediate results to avoid overflow.
Negative update values	The difference array approach handles negative update values correctly by subtracting at the end index.

Null or empty updates array

How to Handle:

Return the original array without modification if the updates array is null or empty.

Updates array with empty update ranges

How to Handle:

Skip any updates where the start index is greater than the end index.

Array length of 0

How to Handle:

Return an empty array since no updates can be performed on an empty array.

Large array size and numerous updates impacting performance

How to Handle:

Use the difference array technique to optimize the update operations to O(1) per update.

Update ranges exceeding array bounds

How to Handle:

Clamp the start and end indices of the update ranges to the valid array bounds.

Updates with zero increment values

How to Handle:

These updates should be skipped, as they don't affect the array.

Integer overflow when adding increments to array elements

How to Handle:

Use a data type with a larger range (e.g., long) to store the array elements and intermediate results to avoid overflow.

Negative update values

How to Handle:

The difference array approach handles negative update values correctly by subtracting at the end index.

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 length of the array and the number of update operations? How large can 'length' and the number of update operations be?
What are the possible values for the start index, end index, and increment? Can they be negative?
Is the original array initialized with all zeros, or some other default value?
Do the update operations ever overlap (i.e., can multiple operations affect the same index)?
Should I perform input validation (e.g., checking if start index is less than or equal to end index, and that both are within the bounds of the array)? What should I return if the input is invalid?

Brute Force Solution

Approach

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

Start with a list filled with zeros, representing your initial data.
For each update instruction, identify the starting and ending positions of the range to be modified.
Go through each position within that range, one by one.
At each position, add the update value to the current value already there.
Repeat this process for all update instructions.
The final list contains the updated values after applying all the range additions.

Code Implementation

def range_addition_brute_force(length, updates):
    result_array = [0] * length

    # Iterate through each update instruction
    for update in updates:
        start_index = update[0]
        end_index = update[1]
        increment_value = update[2]

        # Apply the update to the specified range
        for index in range(start_index, end_index + 1):

            # Directly modifying the affected list elements.
            result_array[index] += increment_value

    return result_array

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

For each update, simply add the value to the *start* position of the range.
Also, subtract the same value from the position *just after* the end of the range (if that position exists).
Once you've processed all updates, go through the entire data structure. Keep a running total.
At each position, add the value stored there to the running total. This gives you the correct final value for that position.
This running total now reflects the net effect of all range updates up to that position.

Code Implementation

def range_addition(length_of_array, updates):
    result_array = [0] * length_of_array

    for start_index, end_index, value_to_add in updates:
        # Mark the start of the range with the value
        result_array[start_index] += value_to_add

        if end_index + 1 < length_of_array:
            # Subtract after the end to revert the addition
            result_array[end_index + 1] -= value_to_add

    current_sum = 0
    for index in range(length_of_array):
        # Accumulate the changes to determine final values
        current_sum += result_array[index]
        result_array[index] = current_sum

    return result_array

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty updates array	Return the original array without modification if the updates array is null or empty.
Updates array with empty update ranges	Skip any updates where the start index is greater than the end index.
Array length of 0	Return an empty array since no updates can be performed on an empty array.
Large array size and numerous updates impacting performance	Use the difference array technique to optimize the update operations to O(1) per update.
Update ranges exceeding array bounds	Clamp the start and end indices of the update ranges to the valid array bounds.
Updates with zero increment values	These updates should be skipped, as they don't affect the array.
Integer overflow when adding increments to array elements	Use a data type with a larger range (e.g., long) to store the array elements and intermediate results to avoid overflow.
Negative update values	The difference array approach handles negative update values correctly by subtracting at the end index.

Null or empty updates array

How to Handle:

Return the original array without modification if the updates array is null or empty.

Updates array with empty update ranges

How to Handle:

Skip any updates where the start index is greater than the end index.

Array length of 0

How to Handle:

Return an empty array since no updates can be performed on an empty array.

Large array size and numerous updates impacting performance

How to Handle:

Use the difference array technique to optimize the update operations to O(1) per update.

Update ranges exceeding array bounds

How to Handle:

Clamp the start and end indices of the update ranges to the valid array bounds.

Updates with zero increment values

How to Handle:

These updates should be skipped, as they don't affect the array.

Integer overflow when adding increments to array elements

How to Handle:

Use a data type with a larger range (e.g., long) to store the array elements and intermediate results to avoid overflow.

Negative update values

How to Handle:

The difference array approach handles negative update values correctly by subtracting at the end index.