Range Sum Query - Immutable

Description

Given an integer array nums, handle multiple queries of the following type:

1. Calculate the sum of the elements of nums between indices left and right inclusive where left <= right.

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]).

Example 1:

<strong>Input</strong>
["NumArray", "sumRange", "sumRange", "sumRange"]
[[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]
<strong>Output</strong>
[null, 1, -1, -3]

<strong>Explanation</strong>
NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);
numArray.sumRange(0, 2); // return (-2) + 0 + 3 = 1
numArray.sumRange(2, 5); // return 3 + (-5) + 2 + (-1) = -1
numArray.sumRange(0, 5); // return (-2) + 0 + 3 + (-5) + 2 + (-1) = -3

Constraints:

• 1 <= nums.length <= 10<sup>4</sup>
• -10<sup>5</sup> <= nums[i] <= 10<sup>5</sup>
• 0 <= left <= right < nums.length
• At most 10<sup>4</sup> calls will be made to sumRange.

Code

Time Complexity: $O()$, Space Complexity: $O()$

Source

• Range Sum Query - Immutable - LeetCode