Range Sum Query

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

Example:

Given nums = [-2, 0, 3, -5, 2, -1]

sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3

Note:

1. You may assume that the array does not change.
2. There are many calls to sumRange function.

TIME LIMITED EXCEEDED VERSION

class NumArray {
    private:
    vector<int> arr;
public:
    NumArray(vector<int> &nums) {
        arr.clear();
        for(int i = 0; i < nums.size(); i++)
            arr.push_back(nums[i]);
    }

    int sumRange(int i, int j) {
        i = i < 0 ? 0 : i;
        j = j >= arr.size() ? arr.size() - 1 : j;
        
        int sum = 0;
        for(int k = i; k <= j; k++)
            sum += arr[k];
        return sum;
    }
};


// Your NumArray object will be instantiated and called as such:
// NumArray numArray(nums);
// numArray.sumRange(0, 1);
// numArray.sumRange(1, 2);

Runtime: 588ms

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

    int sumRange(int i, int j) {
        return arr[j + 1] - arr[i];
    }
private: 
    vector<int> arr;
};


// Your NumArray object will be instantiated and called as such:
// NumArray numArray(nums);
// numArray.sumRange(0, 1);
// numArray.sumRange(1, 2);