RangeSumQuery-Immutable

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

Implement the NumArray class:

NumArray(int[] nums) Initializes the object with the integer array nums.
int sumRange(int i, int j) Return the sum of the elements of the nums array in the range [i, j] inclusive (i.e., sum(nums[i], nums[i + 1], ... , nums[j]))
 
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 1 ((-2) + 0 + 3)
numArray.sumRange(2, 5); // return -1 (3 + (-5) + 2 + (-1)) 
numArray.sumRange(0, 5); // return -3 ((-2) + 0 + 3 + (-5) + 2 + (-1))
 

Constraints:
0 <= nums.length <= 104
-105 <= nums[i] <= 105
0 <= i <= j < nums.length
At most 104 calls will be made to sumRange.

class NumArray {
    int arr[];

    public NumArray(int[] nums) 
    {
        arr = new int[nums.length];
        arr = nums;
    }
    
    public int sumRange(int i, int j) 
    {
        int sum = 0;
        for(int x = i; x <= j; x++)
            sum+= arr[x];
        return sum;
    }
}

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

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