Prerequisites: MO’s algorithm, SQRT Decomposition
Given an array arr[] of N elements and two integers A to B, the task is to answer Q queries each having two integers L and R. For each query, find the number of elements in the subarray arr[L, R] which lies within the range A to B (inclusive).
Examples:
Input: arr[] = {3, 4, 6, 2, 7, 1}, A = 1, B = 6, query = {0, 4}
Output: 4
Explanation:
All 3, 4, 6, 2 lies within 1 to 6 in the subarray {3, 4, 6, 2}
Therefore, the count of such elements is 4.Input: arr[] = {0, 1, 2, 3, 4, 5, 6, 7}, A = 1, B = 5, query = {3, 5}
Output: 3
Explanation:
All the elements 3, 4 and 5 lies within the range 1 to 5 in the subarray {3, 4, 5}.
Therefore, the count of such elements is 3.
Approach: The idea is to use MO’s algorithm to pre-process all queries so that result of one query can be used in the next query. Below is the illustration of the steps:
- Group the queries into mutiple chunks where each chunk contains the values of starting range in (0 to √N – 1), (√N to 2x√N – 1) and so on. Sort the queries within a chunk in incresing order of R.
- Process all queries one by one in a way that every query uses result computed in the previous query.
- Maintain the frequency array that will count the frequency of arr[i] as they appear in the range [L, R].
For example:
arr[] = [3, 4, 6, 2, 7, 1], L = 0, R = 4 and A = 1, B = 6
Initially frequency array is initialized to 0 i.e freq[]=[0….0]
Step 1: Add arr[0] and increment its frequency as freq[arr[0]]++
i.e freq[3]++ and freq[]=[0, 0, 0, 1, 0, 0, 0, 0]Step 2: Add arr[1] and increment freq[arr[1]]++
i.e freq[4]++ and freq[]=[0, 0, 0, 1, 1, 0, 0, 0]Step 3: Add arr[2] and increment freq[arr[2]]++
i.e freq[6]++ and freq[]=[0, 0, 0, 1, 1, 0, 1, 0]Step 4: Add arr[3] and increment freq[arr[3]]++
i.e freq[2]++ and freq[]=[0, 0, 1, 1, 1, 0, 1, 0]Step 5: Add arr[4] and increment freq[arr[4]]++
i.e freq[7]++ and freq[]=[0, 0, 1, 1, 1, 0, 1, 1]
Step 6: Now we need to find the numbers of elements between A and B.
Step 7: The answer is equal to
To calculate the sum in Step 7, we cannot do iteration because that would lead to O(N) time complexity per query. So we will use square root decomposition technique to find the sum, whose time complexity is O(√N) per query.
Below is the implementation of the above approach:
// C++ implementation to find the // values in the range A to B // in a subarray of L to R #include <bits/stdc++.h> using namespace std;
#define MAX 100001 #define SQRSIZE 400 // Variable to represent block size. // This is made global so compare() // of sort can use it. int query_blk_sz;
// Structure to represent a query range struct Query {
int L;
int R;
}; // Frequency array // to keep count of elements int frequency[MAX];
// Array which contains the frequency // of a particular block int blocks[SQRSIZE];
// Block size int blk_sz;
// Function used to sort all queries // so that all queries of the same // block are arranged together and // within a block, queries are sorted // in increasing order of R values. bool compare(Query x, Query y)
{ if (x.L / query_blk_sz
!= y.L / query_blk_sz)
return (x.L / query_blk_sz
< y.L / query_blk_sz);
return x.R < y.R;
} // Function used to get the block // number of current a[i] i.e ind int getblocknumber(int ind)
{ return (ind) / blk_sz;
} // Function to get the answer // of range [0, k] which uses the // sqrt decompostion technique int getans(int A, int B)
{ int ans = 0;
int left_blk, right_blk;
left_blk = getblocknumber(A);
right_blk = getblocknumber(B);
// If left block is equal to rigth block
// then we can traverse that block
if (left_blk == right_blk) {
for (int i = A; i <= B; i++)
ans += frequency[i];
}
else {
// Traversing first block in
// range
for (int i = A;
i < (left_blk + 1) * blk_sz;
i++)
ans += frequency[i];
// Traversing completely overlapped
// blocks in range
for (int i = left_blk + 1;
i < right_blk; i++)
ans += blocks[i];
// Traversing last block in range
for (int i = right_blk * blk_sz;
i <= B; i++)
ans += frequency[i];
}
return ans;
} void add(int ind, int a[])
{ // Increment the frequency of a[ind]
// in the frequency array
frequency[a[ind]]++;
// Get the block number of a[ind]
// to update the result in blocks
int block_num = getblocknumber(a[ind]);
blocks[block_num]++;
} void remove(int ind, int a[])
{ // Decrement the frequency of
// a[ind] in the frequency array
frequency[a[ind]]--;
// Get the block number of a[ind]
// to update the result in blocks
int block_num = getblocknumber(a[ind]);
blocks[block_num]--;
} void queryResults(int a[], int n,
Query q[], int m,
int A, int B)
{ // Initialize the block size
// for queries
query_blk_sz = sqrt(m);
// Sort all queries so that queries
// of same blocks are arranged
// together.
sort(q, q + m, compare);
// Initialize current L,
// current R and current result
int currL = 0, currR = 0;
for (int i = 0; i < m; i++) {
// L and R values of the
// current range
int L = q[i].L, R = q[i].R;
// Add Elements of current
// range
while (currR <= R) {
add(currR, a);
currR++;
}
while (currL > L) {
add(currL - 1, a);
currL--;
}
// Remove element of previous
// range
while (currR > R + 1)
{
remove(currR - 1, a);
currR--;
}
while (currL < L) {
remove(currL, a);
currL++;
}
printf("%d\n", getans(A, B));
}
} // Driver code int main()
{ int arr[] = { 3, 4, 6, 2, 7, 1 };
int N = sizeof(arr) / sizeof(arr[0]);
int A = 1, B = 6;
blk_sz = sqrt(N);
Query Q[] = { { 0, 4 } };
int M = sizeof(Q) / sizeof(Q[0]);
// Answer the queries
queryResults(arr, N, Q, M, A, B);
return 0;
} |
4
Time Complexity: O(Q*√N)
Space Complexity: O(N)