303. Range Sum Query - Immutable

Problem

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.

Solution

class NumArray {
    private int[] sums;

    public NumArray(int[] nums) {
        sums = new int[nums.length];
        for (int i = 0; i < nums.length; i++) {
            if (i == 0) {
                sums[i] = nums[i];
            } else {
                sums[i] = sums[i - 1] + nums[i];
            }
        }
    }

    public int sumRange(int i, int j) {
        if (i == 0) {
            return sums[j];
        }
        return sums[j] - sums[i - 1];
    }
}

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

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).