LeetCode 0303. Range Sum Query - Immutable Solution in Java, Python, C++, JavaScript, Go & Rust | Explanation + Code

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0303. 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:

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); // 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 <= 104
  • -105 <= nums[i] <= 105
  • 0 <= left <= right < nums.length
  • At most 104 calls will be made to sumRange.

Solutions

Solution 1: Prefix Sum

We create a prefix sum array s of length n + 1, where s[i] represents the prefix sum of the first i elements, that is, s[i] = ∑_{j=0}^{i-1} nums[j]. Therefore, the sum of the elements between the indices [left, right] can be expressed as s[right + 1] - s[left].

The time complexity for initializing the prefix sum array s is O(n), and the time complexity for querying is O(1). The space complexity is O(n).

PythonJavaC++GoTypeScriptRustJavaScriptPHPCKotlin
class NumArray: def __init__(self, nums: List[int]): self.s = list(accumulate(nums, initial=0)) def sumRange(self, left: int, right: int) -> int: return self.s[right + 1] - self.s[left] # Your NumArray object will be instantiated and called as such: # obj = NumArray(nums) # param_1 = obj.sumRange(left,right)(code-box)

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