Range Sum Query - Mutable

Medium

Google

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Topics:

Arrays

You are given an integer array nums. You need to implement a class NumArray that can handle the following types of queries:

Update: Modify the value of an element in the array nums.
Sum Range: Calculate the sum of elements within a specified range (inclusive) in the array nums.

The NumArray class should have the following methods:

NumArray(int[] nums): Initializes the object with the given integer array nums.
void update(int index, int val): Updates the value of nums[index] to val.
int sumRange(int left, int right): Returns the sum of elements from index left to right (inclusive), i.e., nums[left] + nums[left + 1] + ... + nums[right].

Example:

NumArray numArray = new NumArray([1, 3, 5]);
numArray.sumRange(0, 2); // Returns 9 (1 + 3 + 5)
numArray.update(1, 2);   // nums becomes [1, 2, 5]
numArray.sumRange(0, 2); // Returns 8 (1 + 2 + 5)

Constraints:

1 <= nums.length <= 3 * 10^4
-100 <= nums[i] <= 100
0 <= index < nums.length
-100 <= val <= 100
0 <= left <= right < nums.length
The update and sumRange methods will be called at most 3 * 10^4 times.

Design an efficient implementation for the NumArray class, considering both time and space complexity.

Range Sum Query - Mutable

Medium

Google

1 view

Topics:

Arrays

You are given an integer array nums. You need to implement a class NumArray that can handle the following types of queries:

Update: Modify the value of an element in the array nums.
Sum Range: Calculate the sum of elements within a specified range (inclusive) in the array nums.

The NumArray class should have the following methods:

NumArray(int[] nums): Initializes the object with the given integer array nums.
void update(int index, int val): Updates the value of nums[index] to val.
int sumRange(int left, int right): Returns the sum of elements from index left to right (inclusive), i.e., nums[left] + nums[left + 1] + ... + nums[right].

Example:

NumArray numArray = new NumArray([1, 3, 5]);
numArray.sumRange(0, 2); // Returns 9 (1 + 3 + 5)
numArray.update(1, 2);   // nums becomes [1, 2, 5]
numArray.sumRange(0, 2); // Returns 8 (1 + 2 + 5)

Constraints:

1 <= nums.length <= 3 * 10^4
-100 <= nums[i] <= 100
0 <= index < nums.length
-100 <= val <= 100
0 <= left <= right < nums.length
The update and sumRange methods will be called at most 3 * 10^4 times.

Design an efficient implementation for the NumArray class, considering both time and space complexity.

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 values in the array? Can they be negative, zero, or have a specific range?
How frequently will the update and sumRange operations be called relative to each other? Can I assume a certain distribution?
What is the maximum size of the input array? Should I optimize for a very large array?
If `i` or `j` in the `sumRange(i, j)` function are out of bounds (i.e., less than 0 or greater than or equal to the array length), how should I handle that?
Is the `update` operation guaranteed to always receive a valid index within the bounds of the array?

Brute Force Solution

Approach

The brute force method for this problem is straightforward. To find a sum, we will simply add up all the relevant numbers. When a number changes, we will directly update our list of numbers.

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

If someone asks for the sum of a range of numbers, go through each number within that range and add them together.
If someone changes a number in our list, find that number in the list and replace it with the new value.

Code Implementation

class NumArray:

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

    def update(self, index, value):
        # Directly update the number at the given index
        self.numbers_list[index] = value

    def sumRange(self, left_index, right_index):
        range_sum = 0
        # Iterate from left to right index to sum all the elements
        for index in range(left_index, right_index + 1):
            range_sum += self.numbers_list[index]

        return range_sum

Big(O) Analysis

Time Complexity

O(n) – The sumRange operation iterates through the specified range of the input array (nums) of size n, performing a constant-time addition for each element within the range. In the worst-case scenario, the range spans the entire array, requiring n additions. The update operation, on the other hand, involves a direct assignment at a specific index which takes constant time, O(1). Thus, the sumRange operation has a time complexity of O(n), and update has O(1) time complexity. The complexity refers only to the sumRange, based on the provided information.

Space Complexity

O(1) – The provided solution operates directly on the input array. It does not create any auxiliary data structures like new arrays, hash maps, or trees to store intermediate results. It only uses a fixed number of variables for iteration and summation within the given range. Therefore, the auxiliary space complexity is constant, independent of the input size N, where N is the number of elements in the array.

Optimal Solution

Approach

To efficiently manage sums of ranges within a dataset that changes, we use a data structure allowing quick updates and range calculations. The key idea is to precompute and store partial sums in a tree-like structure, enabling us to update individual values and retrieve range sums rapidly.

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

First, build a special tree where each part of the tree represents the sum of a specific segment of the original data.
When you need to update a single value in the original data, find all the segments in the tree that contain that value and update their sums.
When you need to calculate the sum of a range, break that range down into segments that match the segments in the tree.
Then, simply add up the precomputed sums of those matching segments in the tree to get the final range sum. This avoids recalculating the sum from scratch each time.

Code Implementation

class NumArray:

    def __init__(self, nums):
        self.numbers = nums
        self.segment_tree = [0] * (4 * len(nums))
        if nums:
            self.build_segment_tree(0, 0, len(nums) - 1)

    def build_segment_tree(self, segment_tree_index, left, right):
        if left == right:
            self.segment_tree[segment_tree_index] = self.numbers[left]
            return
        
        middle = (left + right) // 2

        self.build_segment_tree(2 * segment_tree_index + 1, left, middle)
        self.build_segment_tree(2 * segment_tree_index + 2, middle + 1, right)
        
        # Each node stores sum of its children
        self.segment_tree[segment_tree_index] = self.segment_tree[2 * segment_tree_index + 1] + self.segment_tree[2 * segment_tree_index + 2]

    def update(self, index, value):
        self.update_segment_tree(0, 0, len(self.numbers) - 1, index, value)
        self.numbers[index] = value

    def update_segment_tree(self, segment_tree_index, left, right, index, value):
        if left == right:
            self.segment_tree[segment_tree_index] = value
            return

        middle = (left + right) // 2

        if index <= middle:
            self.update_segment_tree(2 * segment_tree_index + 1, left, middle, index, value)
        else:
            self.update_segment_tree(2 * segment_tree_index + 2, middle + 1, right, index, value)
        
        # Propagate changes up the tree
        self.segment_tree[segment_tree_index] = self.segment_tree[2 * segment_tree_index + 1] + self.segment_tree[2 * segment_tree_index + 2]

    def sumRange(self, left, right):
        return self.sum_range_segment_tree(0, 0, len(self.numbers) - 1, left, right)

    def sum_range_segment_tree(self, segment_tree_index, left, right, query_left, query_right):
        if query_left <= left and right <= query_right:
            return self.segment_tree[segment_tree_index]

        if right < query_left or query_right < left:
            return 0

        middle = (left + right) // 2

        # Sum up sums of children to calculate range
        return self.sum_range_segment_tree(2 * segment_tree_index + 1, left, middle, query_left, query_right) + self.sum_range_segment_tree(2 * segment_tree_index + 2, middle + 1, right, query_left, query_right)

# Your NumArray object will be instantiated and called as such:
# nums = [1, 3, 5]
# obj = NumArray(nums)
# obj.update(1, 2)
# param_2 = obj.sumRange(0, 2)

Big(O) Analysis

Time Complexity

O(log n) – The time complexity is driven by the height of the tree-like structure used to store partial sums, which is logarithmic with respect to the input size n. Updating a value requires traversing from the leaf node to the root, updating the sums of all affected segments, taking O(log n) time. Similarly, querying the sum of a range involves breaking down the range into O(log n) segments stored in the tree and summing those precomputed values, also taking O(log n) time. Therefore, both update and query operations have a time complexity of O(log n).

Space Complexity

O(N) – The algorithm constructs a tree-like structure to store partial sums, where each node represents the sum of a segment of the original data. In the worst-case scenario, the tree could have a number of nodes proportional to the size of the input array N (for example, in a segment tree implementation). Thus, the auxiliary space required to store this tree is O(N), where N is the number of elements in the original data. No other significant data structures contribute to the space complexity.

Edge Cases

Case	How to Handle
Null or empty input array	The constructor should throw an IllegalArgumentException or return without initializing the segment tree, and update/sumRange should handle the uninitialized state gracefully, potentially returning 0 or throwing an exception.
Array with a single element	The segment tree should be constructed correctly and update/sumRange operations should work as expected on that single element.
Large input array (performance considerations)	The segment tree construction should be O(n) and update/sumRange operations should be O(log n) to ensure scalability; consider using iterative approach for construction to avoid stack overflow with deep recursion.
Array with all identical values	The segment tree construction and operations (update and sumRange) should function correctly; specifically ensure the tree nodes containing sum of the range is updated appropriately.
Array with negative numbers	The segment tree implementation needs to correctly handle negative numbers in the array and during sum calculation; ensure the sum variable type has enough capacity to handle potential negative overflow.
Update index out of bounds	The update function should throw an IndexOutOfBoundsException or return without modifying the array or segment tree to avoid unexpected behavior.
sumRange with left > right	The sumRange function should either return 0, throw an IllegalArgumentException or swap the left and right indices to ensure correct calculation.
Integer overflow during sum calculation	Use `long` data type for storing sum in the segment tree nodes to prevent integer overflow during large range sum computations.