/*
 * 303. Range Sum Query - Immutable
 * Difficulty: Easy
 * 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
 * 
 *  
 * 
 * Constraints:
 * 
 * 	• 1 <= nums.length <= 10^4
 * 
 * 	• -10^5 <= nums[i] <= 10^5
 * 
 * 	• 0 <= left <= right < nums.length
 * 
 * 	• At most 10^4 calls will be made to sumRange.
*/

/**
 * @param {number[]} nums
 */
class NumArray {
    constructor(nums) {
      this.prefix = [0];
      for (let n of nums){
        this.prefix.push(this.prefix[this.prefix.length - 1] + n);
      }
    }
    /**
     * @param {number} left
     * @param {number} right
     * @return {number}
     */
    sumRange(left, right) {
        return this.prefix[right + 1] - this.prefix[left];
    }
}


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