This page explains Java solution to problem Range Sum Query - Mutable
using Binary Indexed Tree
data structure.
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
.
Input: nums = [1, 3, 5]
Output:
sumRange(0, 2) -> 9
update(1, 2)
sumRange(0, 2) -> 8
package com.vc.medium;
class RangeSumQueryMutable {
private int[] arr;
private int[] tree;
private int n;
public RangeSumQueryMutable(int[] arr) {
this.arr = arr;
this.n = arr.length;
tree = new int[n + 1];
for(int i = 0; i < n; i++) {
set(i, arr[i]);
}
}
//Add element to each of it's child
private void set(int index, int val) {
for(int i = index + 1; i <= n; i = getNext(i)) tree[i] += val;
}
public void update(int i, int val) {
int delta = val - arr[i];
arr[i] = val;
set(i, delta);
}
private int sum(int index) {
int sum = 0;
for(int i = index; i > 0; i = getParent(i)) {
sum += tree[i];
}
return sum;
}
public int sumRange(int i, int j) {
return sum(j + 1) - sum(i);
}
/**
Java uses two's complement to represent binary number
Two's complement is invert all the digits and add one
so for e.g. we have
5 => 101
-5 => 010 + 001 => 011
5 & -5 should give us last set bit
5 & -5 => 101 & 011 => 001
And we have to remove last set bit in a number to get to it's parent
*/
private int getParent(int x) {
return x - (x & -x);
}
private int getNext(int x) {
return x + (x & -x);
}
}
/**
* 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);
*/
O(N * log N) to Generate Binary Indexed Tree using Array
O(log N) to Update element in an input array & Binary Indexed Tree
O(log N) to Sum elements in an input range Where
N is total number of elements in an input array
O(N) Where
N is total number of elements in an input array