Range Sum Query – Immutable Leetcode Problem 303 [Python Solution]

In this post, we’ll tackle the “Range Sum Query – Immutable” Leetcode problem, a classic question for coding interviews.

We’ll explore a Python solution that optimizes query processing.

By the end of this guide, you’ll understand the problem, the efficient solution, its reasoning, and its time and space complexities.

If you’re ready to level up your coding skills, let’s dive in!

Problem Overview

The “Range Sum Query – Immutable” problem is all about efficiently handling multiple queries of the following type:

• Calculate the sum of the elements of a given array nums between indices left and right, inclusive, where left is less than or equal to right.

Here’s the task:

Implement the NumArray class:

• NumArray(int[] nums): Initializes the object with the integer array nums.
• int sumRange(int left, int right): Returns the sum of the elements of nums between indices left and right, inclusive (i.e., nums[left] + nums[left + 1] + ... + nums[right]).

Let’s illustrate this with an example:

Example 1:

Input
["NumArray", "sumRange", "sumRange", "sumRange"]
[[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]

Output
[null, 1, -1, -3]

Explanation:

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
• At most 10^4 calls will be made to sumRange.

Now that we understand the problem, let’s proceed with our Python solution.

Efficient Python Code Solution

To efficiently solve this problem, we’ll use a class called NumArray.

This class will store the prefix sums of the input array, allowing us to perform query operations in constant time.

Here’s the Python code for this solution:

class NumArray:
    def __init__(self, nums: List[int]):
        self.prefix = []
        cur = 0
        for n in nums:
            cur += n
            self.prefix.append(cur)

    def sumRange(self, left: int, right: int) -> int:
        r = self.prefix[right]
        l = self.prefix[left - 1] if left > 0 else 0
        return r - l

In the constructor __init__, we initialize the prefix array by computing the prefix sums of the input array nums.

This step ensures that we can efficiently retrieve the sum of any subarray using the prefix sums.

In the sumRange method, we retrieve the prefix sum for the right index and, if necessary, the prefix sum for the left index minus one.

This allows us to calculate the sum of the desired subarray efficiently and return it.

This solution guarantees constant-time complexity for query operations and linear time complexity for initializing the NumArray object.

Now, let’s dive into the reasoning behind our approach.

Reasoning Behind Our Approach

Our approach is based on the concept of prefix sums.

Prefix sums are an essential tool for solving various range sum problems efficiently.

Here’s the reasoning behind our solution:

1. Prefix Sums: We create a prefix array to store the prefix sums of the input array.

Each element prefix[i] will represent the sum of elements from index 0 to index i in the input array.

By precomputing these values, we can avoid repetitive calculations when handling queries.

2. Query Processing: To compute the sum of a subarray, we only need two prefix sum values: one for the right index and, if applicable, one for the left index minus one.

We subtract the left sum from the right sum to obtain the sum of the desired subarray.

This process ensures that we can handle queries in constant time, irrespective of the query’s range.

3. Efficiency: While the initialization of the prefix array takes linear time, this cost is incurred only once during the object’s creation.

Subsequent query operations are extremely efficient, as they involve simple array lookups and subtraction.

4. Edge Cases Handling: We also consider edge cases where the left index is at the beginning of the array.

In such cases, there is no need to subtract the left sum, as it would be zero.

If the left index is not at the array’s start, we ensure that the left sum is computed correctly by checking for index validity.

By employing these principles, our solution provides an elegant and efficient way to handle range sum queries.

Time and Space Complexity

Let’s analyze the time and space complexity of our solution:

• Time Complexity:
  • Initializing the NumArray object (__init__ method): O(n), where n is the length of the input array nums.
  • Handling a query (sumRange method): O(1).

The query operation is performed in constant time, as it involves basic array lookups and subtraction.

Overall, our solution is very efficient in handling queries, making it suitable for scenarios with a large number of queries.

• Space Complexity:
  • Additional space for the prefix array: O(n).

We store the prefix sums, which requires an array of the same length as the input array.

In conclusion, our solution efficiently handles range sum queries with a constant-time complexity for each query operation.

It uses extra space to store the prefix array, but this space cost is reasonable given the improved query processing performance.

Conclusion

In this guide, we explored the “Range Sum Query – Immutable” problem, which involves efficiently calculating the sum of elements in a given subarray.

We introduced an efficient Python solution that leverages prefix sums to achieve constant-time query processing.

By precomputing the prefix sums and subtracting the appropriate values, we can quickly resolve queries of any range.

Our approach is not only efficient but also elegant, making it suitable for a wide range of applications.

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For the full problem description and more details, check out the Range Sum Query – Immutable on LeetCode.

Happy coding!