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)

**Naive approach** The naive approach for each query traverse the subarray and count the number of elements which are in the given range.

**Efficient Approach** The idea is to use-Binary Index Tree.

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

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