- A Merge Sort Tree (MST) is a data structure that allows for efficient querying of the number of elements less than or equal to a certain value in a range of an array. It is built using a divide-and-conquer approach similar to the merge sort algorithm, where the array is recursively divided into two halves and a tree is constructed to represent the sorted subarrays.
- A Policy Based Data Structure (PBD) is a data structure that allows for efficient updates, such as insertion and deletion, without the need to rebuild the entire structure. Examples of PBDs include persistent segment trees, persistent arrays, and persistent Union-Find trees.
By using a Policy Based Data Structure as the underlying data structure, we can perform point updates on the array, which is not possible with a regular Merge Sort Tree. This makes the Merge Sort Tree with point update using Policy Based Data Structure a useful tool for solving various range query problems such as counting the number of elements less than or equal to a certain value in a given range of an array.
Approach: The algorithm for the Merge Sort Tree with point update using Policy Based Data Structure is as follows:
- Initialize an array of integers and a Policy Based Data Structure (PBD) such as a persistent segment tree.
- Create a class MST (Merge Sort Tree) that takes in the array a and the PBD.
- Implement a function build that takes in the left and right indices of the range of the array to be considered and returns the root of the PBD. This function will recursively divide the array into two halves and use the PBD to keep track of the number of elements less than or equal to a certain value in the current range.
- Implement a function query that takes in the current root of the PBD, the left and right indices of the current range, the left and right indices of the query range, and a value x. The function will return the number of elements less than or equal to x in the query range by using the PBD to query the relevant subtrees.
- Implement a function update that takes in the index of the element to be updated and its new value. This function will rebuild the entire MST with the updated value.
- In the main function, use the query function to get the number of elements less than or equal to a certain value in a range of the array and use the update function to update an element in the array.
Below is the implementation for the above approach:
C++
// C++ code for the above approach: #include <bits/stdc++.h> using namespace std;
// Persistent Segment Tree class PST {
public:
struct Node {
int val;
int lc;
int rc;
Node(int v = 0, int l = -1, int r = -1)
: val(v), lc(l), rc(r)
{
}
};
vector<Node> t;
PST() { t.push_back({}); }
int update(int cur, int l, int r, int idx, int diff)
{
int new_node = t.size();
// Creates a new version of node
t.push_back(t[cur]);
t[new_node].val += diff;
if (l == r) {
return new_node;
}
int mid = (l + r) >> 1;
if (idx <= mid) {
t[new_node].lc
= update(t[cur].lc, l, mid, idx, diff);
}
else {
t[new_node].rc
= update(t[cur].rc, mid + 1, r, idx, diff);
}
return new_node;
}
int query(int cur, int l, int r, int ql, int qr)
{
if (ql <= l && r <= qr) {
return t[cur].val;
}
int mid = (l + r) >> 1;
int res = 0;
if (ql <= mid) {
res += query(t[cur].lc, l, mid, ql, qr);
}
if (mid < qr) {
res += query(t[cur].rc, mid + 1, r, ql, qr);
}
return res;
}
}; // Merge Sort Tree with point update class MST {
public:
vector<int> a;
PST pst;
vector<int> root;
int n;
MST(vector<int> v)
{
a = v;
n = v.size();
root.push_back(pst.update(0, 0, n - 1, v[0], 1));
for (int i = 1; i < n; i++) {
root.push_back(
pst.update(root.back(), 0, n - 1, v[i], 1));
}
}
int build(int l, int r)
{
if (l == r) {
return pst.update(0, 0, n - 1, a[l], 1);
}
int mid = (l + r) >> 1;
int left = build(l, mid);
int right = build(mid + 1, r);
return merge(left, right, l, r);
}
int merge(int left, int right, int l, int r)
{
int cur = root.size();
root.push_back(0);
int mid = (l + r) >> 1;
for (int i = l, j = mid + 1; i <= mid || j <= r;) {
if (j > r || (i <= mid && a[i] < a[j])) {
root[cur] = pst.update(root[cur], 0, n - 1,
a[i], 1);
i++;
}
else {
root[cur] = pst.update(root[cur], 0, n - 1,
a[j], 1);
j++;
}
}
return cur;
}
int query(int cur, int l, int r, int ql, int qr, int x)
{
if (ql <= l && r <= qr) {
return pst.query(root[cur], 0, n - 1, 0, x);
}
int mid = (l + r) >> 1;
int res = 0;
if (ql <= mid) {
res += query(cur - 1, l, mid, ql, qr, x);
}
if (mid < qr) {
res += query(cur - 1, mid + 1, r, ql, qr, x);
}
return res;
}
void update(int idx, int new_val)
{
a[idx] = new_val;
root.push_back(build(0, n - 1));
}
}; // Drivers code int main()
{ vector<int> v = { 1, 4, 2, 3, 5 };
MST mst(v);
// Query for values less than or
// equal to 3 in the range [1, 3]
cout << mst.query(v.size() - 1, 0, v.size() - 1, 1, 3,
3)
<< endl;
// Update a[2] = 5
mst.update(2, 5);
// Query for values less than or equal
// to 3 in the range [1, 3]
cout << mst.query(v.size() - 1, 0, v.size() - 1, 1, 3,
3)
<< endl;
return 0;
} |
Output
2 2
- The first query in the main function is asking for the number of elements less than or equal to 3 in the range [1, 3] of the original array. The original array is [1, 4, 2, 3, 5], so the values in the range [1, 3] are [4, 2, 3] and there are 2 elements (2 and 3) that are less than or equal to 3.
- The second query is done after the value of a[2] is updated to 5. Now the array is [1, 4, 5, 3, 5], so the values in the range [1, 3] are [4, 5, 3] and there are still 2 elements (4 and 3) that are less than or equal to 3.
Complexity analysis:
- In this implementation, the MST uses the PST to build a tree for each version of the array after updates, and the PST is used to maintain the values in each node and the left and right children of each node.
- The MST’s ‘query’ method has a time complexity of O(log n) as it uses a divide and conquer approach to find the answer in the range.
- The ‘update’ method has a time complexity of O(n*log n) as it rebuilds the entire tree from scratch after updating the value of a single element.
- Overall the time complexity of this implementation is O(n*log n) for updates and O(log n) for queries.
- And the auxiliary space is O(n*log n) as it stores all the versions of the tree.
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