Number of elements less than or equal to a given number in a given subarray

Hello Everyone,

Given an array ‘a[]’ and number of queries q. Each query can be represented by l, r, x. Your task is to print the number of elements less than or equal to x in the subarray represented by l to r.

Examples:

Input : arr[] = {2, 3, 4, 5}
q = 2
0 3 5
0 2 2
Output : 4
1
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)

Note in the following steps x is the number according to which you have to find the elements and the subarray is represented by l, r.
Step 1: Sort the array in ascending order.
Step 2: Sort the queries according to x in ascending order, initialize bit array as 0.
Step 3: Start from the first query and traverse the array till the value in the array is less than equal to x. For each such element update the BIT with value equal to 1
Step 4: Query the BIT array in the range l to r

// C++ program to answer queries to count number

// of elements smaller tban or equal to x.

#include<bits/stdc++.h>

using namespace std;

// structure to hold queries

struct Query

{

int l, r, x, idx;

};

// structure to hold array

struct ArrayElement

{

int val, idx;

};

// bool function to sort queries according to k

bool cmp1(Query q1, Query q2)

{

return q1.x < q2.x;

}

// bool function to sort array according to its value

bool cmp2(ArrayElement x, ArrayElement y)

{

return x.val < y.val;

}

// updating the bit array

void update( int bit[], int idx, int val, int n)

{

for (; idx<=n; idx +=idx&-idx)

bit[idx] += val;

}

// querying the bit array

int query( int bit[], int idx, int n)

{

int sum = 0;

for (; idx > 0; idx -= idx&-idx)

sum += bit[idx];

return sum;

}

void answerQueries( int n, Query queries[], int q,

ArrayElement arr[])

{

// initialising bit array

int bit[n+1];

memset (bit, 0, sizeof (bit));

// sorting the array

sort(arr, arr+n, cmp2);

// sorting queries

sort(queries, queries+q, cmp1);

// current index of array

int curr = 0;

// array to hold answer of each Query

int ans[q];

// looping through each Query

for ( int i=0; i<q; i++)

{

// traversing the array values till it

// is less than equal to Query number

while (arr[curr].val <= queries[i].x && curr<n)

{

// updating the bit array for the array index

update(bit, arr[curr].idx+1, 1, n);

curr++;

}

// Answer for each Query will be number of

// values less than equal to x upto r minus

// number of values less than equal to x

// upto l-1

ans[queries[i].idx] = query(bit, queries[i].r+1, n) -

query(bit, queries[i].l, n);

}

// printing answer for each Query

for ( int i=0 ; i<q; i++)

cout << ans[i] << endl;

}

// driver function

int main()

{

// size of array

int n = 4;

// initialising array value and index

ArrayElement arr[n];

arr[0].val = 2;

arr[0].idx = 0;

arr[1].val = 3;

arr[1].idx = 1;

arr[2].val = 4;

arr[2].idx = 2;

arr[3].val = 5;

arr[3].idx = 3;

// number of queries

int q = 2;

Query queries[q];

queries[0].l = 0;

queries[0].r = 2;

queries[0].x = 2;

queries[0].idx = 0;

queries[1].l = 0;

queries[1].r = 3;

queries[1].x = 5;

queries[1].idx = 1;

answerQueries(n, queries, q, arr);

return 0;

}

Output:

1
4