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303. 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, $s[i] = \sum_{j=0}^{i-1} nums[j]$. 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 $O(n)$.

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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)

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class NumArray {
    private int[] s;

    public NumArray(int[] nums) {
        int n = nums.length;
        s = new int[n + 1];
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
    }

    public int sumRange(int left, int right) {
        return s[right + 1] - s[left];
    }
}

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray obj = new NumArray(nums);
 * int param_1 = obj.sumRange(left,right);
 */

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class NumArray {
public:
    NumArray(vector<int>& nums) {
        int n = nums.size();
        s.resize(n + 1);
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
    }

    int sumRange(int left, int right) {
        return s[right + 1] - s[left];
    }

private:
    vector<int> s;
};

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray* obj = new NumArray(nums);
 * int param_1 = obj->sumRange(left,right);
 */

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type NumArray struct {
    s []int
}

func Constructor(nums []int) NumArray {
    n := len(nums)
    s := make([]int, n+1)
    for i, v := range nums {
        s[i+1] = s[i] + v
    }
    return NumArray{s}
}

func (this *NumArray) SumRange(left int, right int) int {
    return this.s[right+1] - this.s[left]
}

/**
 * Your NumArray object will be instantiated and called as such:
 * obj := Constructor(nums);
 * param_1 := obj.SumRange(left,right);
 */

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class NumArray {
    private s: number[];

    constructor(nums: number[]) {
        const n = nums.length;
        this.s = Array(n + 1).fill(0);
        for (let i = 0; i < n; ++i) {
            this.s[i + 1] = this.s[