Range Sum Query - Mutable

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

The update(i, val) function modifies nums by updating the element at index i to val.

Example:

Given nums = [1, 3, 5]

sumRange(0, 2) -> 9 update(1, 2) sumRange(0, 2) -> 8

Note:

The array is only modifiable by the update function. You may assume the number of calls to update and sumRange function is distributed evenly.

Solution

public class NumArray {

      int[] nums;
    int[] BIT;
    int n;


    public NumArray(int[] nums) {
        this.nums = nums;
        n = nums.length;
        BIT = new int[n + 1];
        for (int i = 0; i < nums.length; i++) {
            init(i, nums[i]);
        }
    }

    private void init(int i, int val) {
        i++;
        while (i <= n) {
            BIT[i] += val;
            i += (i & -i);
        }
    }

    void update(int i, int val) {
        int diff = val - nums[i];
        nums[i] = val;
        init(i, diff);
    }

    private int sum(int i) {
        i++;
        int sum = 0;
        while (i > 0) {
            sum += BIT[i];
            i -= (i & -i);
        }
        return sum;
    }

    public int sumRange(int i, int j) {
        return sum(j) - sum(i - 1);
    }
}


// Your NumArray object will be instantiated and called as such:
// NumArray numArray = new NumArray(nums);
// numArray.sumRange(0, 1);
// numArray.update(1, 10);
// numArray.sumRange(1, 2);