# 307. Range Sum Query - Mutable

### Problem:

```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.
```

https://en.wikipedia.org/wiki/Fenwick_tree

### Solutions:

``````public class NumArray {
private int[] sums;
private int[] nums;
public NumArray(int[] nums) {
this.nums = nums;
sums = new int[nums.length];
int sum = 0;
for (int i = 0; i < nums.length; i ++) {
sum += nums[i];
sums[i] = sum;
}
}

void update(int i, int val) {
int diff = val - nums[i];
nums[i] = val;
for (int x = i; x < nums.length; x ++) {
sums[x] += diff;
}
}

public int sumRange(int i, int j) {
return sums[j] - sums[i] + nums[i];
}
}

// 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);
``````
``````public class NumArray {
int[] sums;
int[] nums;
public NumArray(int[] nums) {
sums = new int[nums.length + 1];
this.nums = new int[nums.length];
for (int i = 0; i < nums.length; i ++) {
update(i, nums[i]);
}
}

public void update(int i, int val) {
int diff = val - nums[i];
int j = i + 1;
while (j < sums.length) {
sums[j] += diff;
j = j + (j&(-j));
}
nums[i] = val;
}
private int getSum(int i) {
int sum = 0;
while (i > 0) {
sum += sums[i];
i = i - (i&(-i));
}
return sum;
}

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

/**
* Your NumArray object will be instantiated and called as such:
* NumArray obj = new NumArray(nums);
* obj.update(i,val);
* int param_2 = obj.sumRange(i,j);
*/
``````