'''Leetcode- https://leetcode.com/problems/range-sum-query-immutable/ '''
'''
Given an integer array nums, handle multiple queries of the following type:

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
'''

class NumArray:

    def __init__(self, nums):
        self.sums = nums[:]
        for i in range(1, len(self.sums)):
            self.sums[i] = self.sums[i-1] + nums[i]

    def sumRange(self, left: int, right: int):
         if left == 0:
            return self.sums[right]
         else:
            return self.sums[right] - self.sums[left-1]
        
        

#T:O(n)
#S:O(n)