Range Sum Query - Immutable

Easy

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Topics:

ArraysDynamic Programming

Given an integer array nums, design a data structure that efficiently handles multiple queries to calculate the sum of elements within a specified range. You need to implement a class named NumArray with the following functionalities:

Initialization:
- NumArray(int[] nums): Initializes the object with the given integer array nums. This initialization step can perform any necessary pre-computation to optimize subsequent queries.
Range Sum Calculation:
- int sumRange(int left, int right): Calculates and returns the sum of the elements of nums between indices left and right (inclusive), where left <= right.

Example:

NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);
numArray.sumRange(0, 2); // Returns (-2) + 0 + 3 = 1
numArray.sumRange(2, 5); // Returns 3 + (-5) + 2 + (-1) = -1
numArray.sumRange(0, 5); // Returns (-2) + 0 + 3 + (-5) + 2 + (-1) = -3

Constraints:

1 <= nums.length <= 10^4
-10^5 <= nums[i] <= 10^5
0 <= left <= right < nums.length
Maximum 10^4 calls to sumRange will be made.

How would you implement the NumArray class to efficiently handle these range sum queries, considering the constraints and the potential for a large number of queries?

Range Sum Query - Immutable

Easy

Google

3 views

Topics:

ArraysDynamic Programming

Given an integer array nums, design a data structure that efficiently handles multiple queries to calculate the sum of elements within a specified range. You need to implement a class named NumArray with the following functionalities:

Initialization:
- NumArray(int[] nums): Initializes the object with the given integer array nums. This initialization step can perform any necessary pre-computation to optimize subsequent queries.
Range Sum Calculation:
- int sumRange(int left, int right): Calculates and returns the sum of the elements of nums between indices left and right (inclusive), where left <= right.

Example:

NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);
numArray.sumRange(0, 2); // Returns (-2) + 0 + 3 = 1
numArray.sumRange(2, 5); // Returns 3 + (-5) + 2 + (-1) = -1
numArray.sumRange(0, 5); // Returns (-2) + 0 + 3 + (-5) + 2 + (-1) = -3

Constraints:

1 <= nums.length <= 10^4
-10^5 <= nums[i] <= 10^5
0 <= left <= right < nums.length
Maximum 10^4 calls to sumRange will be made.

How would you implement the NumArray class to efficiently handle these range sum queries, considering the constraints and the potential for a large number of queries?

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 the input array `nums`?
Can the values in `nums` be negative, zero, or only positive?
Are the `left` and `right` indices guaranteed to be within the bounds of the array `nums`?
How many times will the `sumRange` function be called after the initial setup? This might influence the choice of precomputation strategy.
Is `left` guaranteed to be less than or equal to `right`?

Brute Force Solution

Approach

Imagine you have a list of numbers, and you want to quickly find the sum of the numbers within a specific range. The brute force approach directly adds up all the numbers within the requested range to find the sum. It's like manually calculating the sum every time you're asked.

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

When you're asked for the sum between a starting point and an ending point, simply start at the number at the starting point.
Add the next number in the list to your running total.
Keep adding each subsequent number until you reach the number at the ending point.
The final total you have is the sum of the numbers within that range.

Code Implementation

class NumArray:

    def __init__(self, numbers):
        self.numbers = numbers

    def sumRange(self, left_index, right_index):
        current_sum = 0
        # Initialize the sum with the element at the starting index
        current_sum = self.numbers[left_index]

        index = left_index + 1

        # Iterate up to the right index to sum the range
        while index <= right_index:
            current_sum += self.numbers[index]
            index += 1

        # Return calculated sum
        return current_sum

Big(O) Analysis

Time Complexity

O(n) – Given an array of n numbers, the range sum query calculates the sum of elements within a specified range [i, j]. The brute force approach iterates through the array from index i to j, adding each element to the running sum. In the worst case, the range could span the entire array (i.e., from 0 to n-1), requiring us to iterate through all n elements. Therefore, the time complexity is directly proportional to the size of the range, which in the worst case is O(n).

Space Complexity

O(1) – The described approach calculates the sum by iterating through the input list directly. It doesn't involve creating any auxiliary data structures like lists, dictionaries, or trees. The only extra memory used is for a few integer variables to track the running total and the current index, and these variables require constant space regardless of the input size, N. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

To quickly find the sum of numbers within a given range, we prepare in advance. We'll create a special list that helps us calculate these sums very fast without recomputing the sum each time we're asked.

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

First, we create a new list that stores the running total as we go through the original numbers.
The first number in our new list is simply the first number from the original list.
Each following number in our new list is the sum of all the numbers up to that point in the original list.
To find the sum of numbers in a range, we subtract the running total before the start of the range from the running total at the end of the range. This immediately gives the sum of the range.
This way, when asked for the sum of a range, we don't need to add up the numbers again, we simply look up the two relevant running totals and subtract them.

Code Implementation

class NumArray:

    def __init__(self, nums):
        self.cumulative_sum = [0] * len(nums)
        if nums:
            self.cumulative_sum[0] = nums[0]

            # Calculate the running totals.
            for index in range(1, len(nums)):
                self.cumulative_sum[index] = self.cumulative_sum[index - 1] + nums[index]

    def sumRange(self, left_index, right_index):
        # Handle the case where left_index is zero.
        if left_index == 0:
            return self.cumulative_sum[right_index]

        # Subtract the cumulative sum before left_index
        # to get the sum of the desired range.
        return self.cumulative_sum[right_index] - self.cumulative_sum[left_index - 1]

Big(O) Analysis

Time Complexity

O(n) – The constructor iterates through the input array of size n once to compute the prefix sums, resulting in O(n) time complexity. The sumRange function performs a constant number of operations (subtraction and array access), so it has O(1) time complexity. Since the constructor is only called once, the overall time complexity is dominated by the constructor's O(n) operation.

Space Complexity

O(N) – The algorithm creates a new list to store the running totals, as described in the plain English explanation. The size of this new list is directly proportional to the number of elements in the original list, which we'll call N. Therefore, the auxiliary space required to store the running totals scales linearly with the input size N. No other significant data structures are used, making the overall auxiliary space complexity O(N).

Edge Cases

Case	How to Handle
Null or empty input array	Throw an IllegalArgumentException or return 0 if the input array is null or empty.
left is negative	Throw an IllegalArgumentException if left is negative as array indices cannot be negative.
right is greater than or equal to the length of the array	Throw an IllegalArgumentException if right is out of bounds.
left is greater than right	Return 0 or swap left and right to enforce left <= right.
Integer overflow when summing large positive numbers	Use long data type for the prefix sum array to prevent integer overflow.
Large input array causing memory issues	The prefix sum approach is efficient in terms of space complexity, but for extremely large arrays, consider memory mapping.
Input array containing very large negative numbers	The prefix sum approach handles negative numbers correctly; however, very large negative numbers might lead to integer underflow if not using long.
Array containing extreme boundary values (Integer.MAX_VALUE, Integer.MIN_VALUE)	Using a long array will prevent potential overflow issues during the prefix sum computation.