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+/*
+ * 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)
+ */