303. Range Sum Query - Immutable

Problem Description

The LeetCode problem presents a common scenario in data processing — computing the sum of a subarray, which is a contiguous segment of an array. You are provided with an integer array nums and are expected to handle multiple queries asking for the sum of elements between two indices, left and right (both inclusive). To efficiently answer these queries, a data structure or algorithm is needed that can quickly calculate the sum of any given range in nums.


For solving this problem, a key observation is that repeatedly computing the sum of a range of elements directly from the array can be time-consuming, especially if the array is large or if there are many queries. To optimize this, a common approach is to use a technique called prefix sum.

The prefix sum array is an auxiliary array where each element at index i stores the sum of all elements from the start of the original array up to index i. By preprocessing the input array into a prefix sum array, we can calculate the sum of any subarray in constant time. The sum of elements between indices left and right can be found by subtracting the prefix sum at left - 1 from the prefix sum at right. This works because the prefix sum at right includes the total sum up to right, and if we subtract the sum up to left - 1, we are left with the sum from left to right.

In this solution, Python's accumulate function from the itertools module is used to create the prefix sum array easily. This function takes an iterable, in this case, nums, and returns a new iterable yielding accumulated sums. The additional initial=0 parameter ensures that the 0th index of the resulting prefix sum array (self.s) is 0, which is helpful for handling cases where left is 0.

By preparing this prefix sum array (self.s) during the initialization of the NumArray class, we ensure that each sumRange query can be answered in constant time by simply calculating self.s[right + 1] - self.s[left], leading to an efficient solution for the problem at hand.

Learn more about Prefix Sum patterns.

Solution Approach

The provided Python code implements an efficient solution to the subarray sum problem by using the prefix sum technique. The prefix sum array is a powerful tool in algorithm design to solve range sum queries, reducing time complexity from potentially O(n) per query to O(1) per query after an initial preprocessing step.

Here's a step-by-step explanation of the code:

Class Definition:

  • NumArray is a class that takes an array and processes it to potentially answer many range sum queries.

__init__ Method:

  • self.s: An instance variable that holds the prefix sum array.
  • accumulate(nums, initial=0): A call to Python's accumulate function, which constructs the prefix sum array from the input nums.
    • The accumulate function takes an iterable and returns an iterable with the accumulated values.
    • The initial=0 parameter is important as it prefixes the resulted iterable with 0, giving us the flexibility to handle the sumRange query accurately even when the left index is 0.

sumRange Method:

  • The sumRange function computes the sum of elements in the range [left, right] by returning self.s[right + 1] - self.s[left].
    • The reason for right + 1 is because the prefix sum array is one element longer than the original array (initial=0 has been added at the start), and sums are stored at one index ahead.

By using a prefix sum array, we trade off some space (O(n) additional space for the auxiliary array) and preprocessing time (O(n) time to construct the prefix sum array) for a massive gain in query time, reducing it to O(1) per query.


  1. Compute the prefix sums of the input array nums and store it in self.s.
  2. When the sumRange is called with left and right indices, return the sum for the specific range by the difference of prefix sums, which represents the sum of elements inclusively between left and right.

Data Structures:

  • A list self.s, which is essentially the auxiliary prefix sum array.

Using these concepts, the class NumArray allows for the fast computation of any given sumRange query, which is particularly useful for scenarios where there will be a large number of these queries on a pre-defined array where the contents do not change.

In summary, the implementation uses the prefix sum pattern to initialize a structure with O(n) complexity, but then allows each sum query to be answered in O(1) time, highlighting an effective trade-off for query-intensive use cases.

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Example Walkthrough

Here is a small example to illustrate the solution approach using a hypothetical array and a few queries:

Let's consider the following array: nums = [3, 0, 1, 4, 2]

  1. We initiate our NumArray object with this array which triggers the creation of the prefix sum array (self.s). The accumulate function cumulatively adds up each value in nums while including an initial 0 at the start. The resulting prefix sum array would look like:

    self.s: [0, 3, 3, 4, 8, 10]


    • Index 0: Initial value, 0.
    • Index 1: Sum up to nums[0] which is 3 (0+3).
    • Index 2: Sum up to nums[1] which is 3 (3+0 since nums[1] is 0).
    • Index 3: Sum up to nums[2] which is 4 (3+1).
    • Index 4: Sum up to nums[3] which is 8 (4+4).
    • Index 5: Sum up to nums[4] which is 10 (8+2).
  2. Suppose we want to know the sum from index 1 to 3 in the nums array. We use the sumRange method and provide the indices to it:

    sumRange(1, 3) should return 0 + 1 + 4 = 5.

    Using the prefix sum array self.s:

    • We take the value at right + 1 which is self.s[3 + 1] = 8
    • We subtract the value at left which is self.s[1] = 3
    • The result is 8 - 3 = 5, which matches the expected output.
  3. Let's say we have another query asking for the sum from the start up to index 2, that's sumRange(0, 2):

    sumRange(0, 2) should return 3 + 0 + 1 = 4.

    Using the prefix sum method:

    • Value at right + 1 is self.s[2 + 1] = 4
    • Value at left is self.s[0] = 0 (since left is 0, it naturally includes no numbers)
    • The result is 4 - 0 = 4, as expected.

These examples demonstrate how by initializing the prefix sum array once, we're able to answer multiple sumRange queries efficiently, each in constant time, without the need to re-calculate sums directly from the nums array. This becomes particularly powerful when dealing with a high volume of sum range queries on an unchanging array.

Solution Implementation

1from itertools import accumulate
3class NumArray:
4    def __init__(self, nums: List[int]):
5        # Pre-calculate the cumulative sum of the array.
6        # The 'initial=0' makes sure the sum starts from index 0 for easier calculations.
7        self.cumulative_sum = list(accumulate(nums, initial=0))
9    def sumRange(self, left: int, right: int) -> int:
10        # Calculate the sum of elements between 'left' and 'right'
11        # by subtracting the sum up to 'left' from the sum up to 'right + 1'.
12        return self.cumulative_sum[right + 1] - self.cumulative_sum[left]
15# Example of usage:
16# numArray = NumArray(nums)
17# sum = numArray.sumRange(left, right)
1class NumArray {
2    // The sum array stores the cumulative sum from the beginning up to the current index.
3    private int[] sumArray;
5    // Constructor that computes the cumulative sum of the numbers array.
6    public NumArray(int[] nums) {
7        int n = nums.length;
8        sumArray = new int[n + 1]; // Initialized with an extra element to handle the sum from 0 to ith index.
10        // Accumulate the sum of elements so that sumArray[i] holds the sum up to nums[i-1].
11        for (int i = 0; i < n; i++) {
12            sumArray[i + 1] = sumArray[i] + nums[i];
13        }
14    }
16    // Method to compute sum of elements within the range [left, right] both inclusive.
17    public int sumRange(int left, int right) {
18        // The sum of elements in range [left, right] is computed by subtracting the cumulative sum up to 'left' from the sum up to 'right + 1'.
19        return sumArray[right + 1] - sumArray[left];
20    }
24 * Usage example:
25 * 
26 * NumArray obj = new NumArray(nums);
27 * int sum = obj.sumRange(left, right);
28 */
1#include <vector>
3class NumArray {
5    // Prefix sum array to store the accumulated sum from the beginning up to each index.
6    std::vector<int> prefixSum;
9    // Constructor that initializes the prefix sum array using the input 'nums' array.
10    NumArray(std::vector<int>& nums) {
11        int size = nums.size();
12        prefixSum.resize(size + 1); // Resizing with an extra element to handle the zero prefix sum.
13        prefixSum[0] = 0; // Initialize the zero-th index with 0 for the prefix sum.
15        // Calculate the prefix sum by adding the current element to the accumulated sum.
16        for (int i = 0; i < size; ++i) {
17            prefixSum[i + 1] = prefixSum[i] + nums[i];
18        }
19    }
21    // Function to calculate the sum of the elements in the range [left, right] in the 'nums' array.
22    int sumRange(int left, int right) {
23        // Return the difference between the prefix sums to get the range sum.
24        return prefixSum[right + 1] - prefixSum[left];
25    }
29 * Usage:
30 * std::vector<int> nums = { ... };
31 * NumArray* obj = new NumArray(nums);
32 * int sum = obj->sumRange(left, right);
33 * ...
34 * delete obj; // Don't forget to deallocate the memory when done.
35 */
1// Global variable to store the sum of elements up to each index.
2let sumArray: number[] = [];
5 * Initialize the sumArray with the prefix sum of the given nums array.
6 * @param nums - The input array of numbers.
7 */
8function createNumArray(nums: number[]): void {
9    const n = nums.length;
10    sumArray = new Array(n + 1).fill(0);
11    for (let i = 0; i < n; ++i) {
12        sumArray[i + 1] = sumArray[i] + nums[i];
13    }
17 * Calculates the sum of elements within the range [left, right] in the array.
18 * @param left - The starting index of the range (inclusive).
19 * @param right - The ending index of the range (inclusive).
20 * @returns The sum of elements within the range [left, right].
21 */
22function sumRange(left: number, right: number): number {
23    return sumArray[right + 1] - sumArray[left];
26// Example of usage:
27// createNumArray([1, 2, 3, 4]);
28// console.log(sumRange(1, 3)); // Output would be 9, which is the sum of elements [2, 3, 4].

Time and Space Complexity

The provided code implements a class NumArray that precomputes the cumulative sum of an array to efficiently find the sum of elements in a given range.

Time Complexity

  • __init__ Method: The initial sum computation is done with accumulate, which processes each element once to create a cumulative sum list. This operation has a time complexity of O(n), where n is the number of elements in the list nums.

  • sumRange Method: This method computes the sum in constant time by subtracting the accumulated sum at the left index from the accumulated sum at the right + 1 index. The time complexity for each sumRange query is O(1).

Space Complexity

  • The space complexity of the NumArray class is primarily determined by the cumulative sum list self.s. Since this list has one more element than the input list (due to initial=0), the space complexity is O(n), where n is the number of elements in the input list nums.

Overall, the preprocessing step (__init__ method) requires O(n) time, and each sumRange query can be answered in O(1) time, with a space complexity of O(n).

Learn more about how to find time and space complexity quickly using problem constraints.

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