Number of elements less than or equal to a number in a subarray : MO’s Algorithm


Given an array arr of size N and Q queries of the form L, R and X, the task is to print the number of elements less than or equal to X in the subarray represented by L to R.

Prerequisites: MO’s Algorithm, Sqrt Decomposition

Examples:

Input: 
arr[] = {2, 3, 4, 5}
Q = {{0, 3, 5}, {0, 2, 2}}
Output:
 4
 1
Explanation:
Number of elements less than or equal to 5
in arr[0..3] is 4 (all elements)

Number of elements less than or equal to 2 
in arr[0..2] is 1 (only 2)

Approach:
The idea of MO’s algorithm is to pre-process all queries so that result of one query can be used in the next query.
Let arr[0…n-1] be input array and Q[0..m-1] be an array of queries.

  1. Sort all queries in a way that queries with L values from 0 to √n – 1 are put together, then all queries from √n to 2×√n – 1, and so on. All queries within a block are sorted in increasing order of R values.
  2. Process all queries one by one in a way that every query uses result computed in the previous query.
  3. We will maintain the frequency array that will count the frequency of arr[i] as they appear in the range [L, R].
  4. For example: arr[]=[3, 4, 6, 2, 7, 1], L=0, R=4 and X=5



    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 less than or equal to X(here X=5).

    Step 7– The answer is equal to \sum_{i=0}^{X} freq[i]

    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 sqrt decomposition technique to find the sum whose time complexity is O(√n) per query.

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// C++ program to answer queries to 
// count number of elements smaller 
// than or equal to x.
#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;
    int x;
};
  
// 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 k)
{
    int ans = 0;
    int left_blk, right_blk;
    left_blk = 0;
    right_blk = getblocknumber(k);
  
    // If left block is equal to
    // rigth block then we can traverse
    // that block
    if (left_blk == right_blk) {
        for (int i = 0; i <= k; i++)
            ans += frequency[i];
    }
    else {
        // Traversing first block in 
        // range
        for (int i = 0; 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 <= k; 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)
{
  
    // 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,
                x = q[i].x;
  
        // 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("query[%d, %d, %d] : %d\n"
               L, R, x, getans(x));
    }
}
// Driver code
int main()
{
  
    int arr[] = { 2, 0, 3, 1, 4, 2, 5, 11 };
    int N = sizeof(arr) / sizeof(arr[0]);
  
    blk_sz = sqrt(N);
    Query Q[] = { { 0, 2, 2 }, { 0, 3, 5 },
                { 5, 7, 10 } };
  
    int M = sizeof(Q) / sizeof(Q[0]);
      
    // Answer the queries
    queryResults(arr, N, Q, M);
    return 0;
}

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Output:

query[0, 2, 2] : 2
query[0, 3, 5] : 4
query[5, 7, 10] : 2

Time Complexity: O(Q × √N).
It takes O(Q × √N) time for MO’s algorithm and O(Q × √N) time for sqrt decomposition technique to answer the sum of freq[0]+….freq[k], therefore total time complexity is O(Q × √N + Q × √N) which is O(Q × √N).

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