Binary Indexed Tree-307. 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:

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

 

---

 

 

 

class NumArray
 2 {
 3 public:
 4     NumArray(vector<int> &nums) 
 5     {
 6         sums.push_back(0);
 7         for (int i = 0; i < nums.size(); i++)
 8         {
 9             sums.push_back(sums[i] + nums[i]);
10             values.push_back(nums[i]);
11         }
12     }
13 
14     void update(int i, int val) 
15     {
16         int diff = val - values[i];
17         for (int k = i + 1; k < sums.size(); k++)
18             sums[k] = sums[k] + diff;
19         values[i] = val;
20     }
21 
22     int sumRange(int i, int j) 
23     {
24         return sums[j + 1] - sums[i];
25     }
26 private:
27     vector<int> sums;
28     vector<int> values;
29 };