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.

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class NumArray
  {
  public:
      NumArray(vector<int> &nums)
      {
          sums.push_back();
          for (int i = ; i < nums.size(); i++)
          {
              sums.push_back(sums[i] + nums[i]);
             values.push_back(nums[i]);
         }
     }

     void update(int i, int val)
     {
         int diff = val - values[i];
         for (int k = i + ; k < sums.size(); k++)
             sums[k] = sums[k] + diff;
         values[i] = val;
     }

     int sumRange(int i, int j)
     {
         return sums[j + ] - sums[i];
     }
 private:
     vector<int> sums;
     vector<int> values;
 };