Queries for elements having values within the range A to B in the given index range using Segment Tree

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, the task is to find the number of elements in the subarray arr[L…R] which lies within the range A to B (both included).

Examples:

Input: arr[] = {7, 3, 9, 13, 5, 4}, A=4, B=7 
query = { 1, 5 } 
Output: 2 
Explanation :
Only 5 and 4 lies within 4 to 7 
in the subarray {3, 9, 13, 5, 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 1 to 5 
in the subarray {3, 4, 5}.

Prerequisite: Segment tree
Naive approach: Find the answer for each query by simply traversing the array from index L till R and keep adding 1 to the count whenever the array element lies within the range A to B. Time Complexity of this approach will be O(n * q).

Efficient approach:
Build a Segment Tree.
Representation of Segment trees 
1. Leaf Nodes are the elements of the input array. 
2. Each internal node contains the number of leaves which lies within the range A to B of all leaves under it.

Construction of Segment Tree from given array
We start with a segment arr[0 . . . n-1]. and every time we divide the current segment into two halves(if it has not yet become a segment of length 1), and then call the same procedure on both halves, and for each such segment, we store the number of elements which lies within the range A to B of all nodes under it.
Time complexity of this approach will be O(q * log(n))

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Procedure to build the segment tree
void buildTree(vector<int>& tree, int* arr,
               int index, int s, int e, int A, int B)
{

    // Reached the leaf node
    // of the segment tree
    if (s == e) {
        if (arr[s] >= A && arr[s] <= B)
            tree[index] = 1;
        else
            tree[index] = 0;
        return;
    }

    // Recursively call the buildTree
    // on both the nodes of the tree
    int mid = (s + e) / 2;
    buildTree(tree, arr, 2 * index, s, mid, A, B);
    buildTree(tree, arr, 2 * index + 1, mid + 1, e, A, B);

    tree[index] = tree[2 * index] + tree[2 * index + 1];
}

// Query procedure to get the answer
// for each query l and r are query range
int query(vector<int> tree, int index, int s,
          int e, int l, int r)
{

    // out of bound or no overlap
    if (r < s || l > e)
        return 0;

    // Complete overlap
    // Query range completely lies in
    // the segment tree node range
    if (s >= l && e <= r) {
        return tree[index];
    }

    // Partially overlap
    // Query range partially lies in
    // the segment tree node range
    int mid = (s + e) / 2;
    return (query(tree, 2 * index, s, mid, l, r)
            + query(tree, 2 * index + 1, mid + 1, e, l, r));
}

// Driver code
int main()
{
    int arr[] = { 7, 3, 9, 13, 5, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    vector<int> tree(4 * n + 1);

    int L = 1, R = 5, A = 4, B = 7;

    buildTree(tree, arr, 1, 0, n - 1, A, B);

    cout << query(tree, 1, 0, n - 1, L, R)
         << endl;
    return 0;
}

Java

// Java implementation of the approach
class GFG{

// Procedure to build the segment tree
static void buildTree(int tree[] , int arr[] ,
                      int index, int s, int e,
                      int A, int B)
{

    // Reached the leaf node
    // of the segment tree
    if (s == e)
    {
        if (arr[s] >= A && arr[s] <= B)
            tree[index] = 1;
        else
            tree[index] = 0;

        return;
    }

    // Recursively call the buildTree
    // on both the nodes of the tree
    int mid = (s + e) / 2;
    buildTree(tree, arr, 2 * index,
              s, mid, A, B);
    buildTree(tree, arr, 2 * index + 1,
              mid + 1, e, A, B);

    tree[index] = tree[2 * index] +
                  tree[2 * index + 1];
}

// Query procedure to get the answer
// for each query l and r are query range
static int query(int tree[], int index, int s,
                 int e, int l, int r)
{

    // Out of bound or no overlap
    if (r < s || l > e)
        return 0;

    // Complete overlap
    // Query range completely lies in
    // the segment tree node range
    if (s >= l && e <= r)
    {
        return tree[index];
    }

    // Partially overlap
    // Query range partially lies in
    // the segment tree node range
    int mid = (s + e) / 2;
    return (query(tree, 2 * index,
                  s, mid, l, r) +
            query(tree, 2 * index + 1,
                  mid + 1, e, l, r));
}

// Driver code
public static void main (String []args)
{
    int arr[] = { 7, 3, 9, 13, 5, 4 };
    int n = arr.length;
    int tree[] = new int [(4 * n + 1)];

    int L = 1, R = 5, A = 4, B = 7;

    buildTree(tree, arr, 1, 0, n - 1, A, B);

    System.out.print(query(tree, 1, 0, n - 1, L, R));
}
}

// This code is contributed by chitranayal

Python3

# Python3 implementation of the approach

# Procedure to build the segment tree
def buildTree(tree,arr,index, s, e, A, B):

    # Reached the leaf node
    # of the segment tree
    if (s == e):
        if (arr[s] >= A and arr[s] <= B):
            tree[index] = 1
        else:
            tree[index] = 0
        return

    # Recursively call the buildTree
    # on both the nodes of the tree
    mid = (s + e) // 2
    buildTree(tree, arr, 2 * index, s, mid, A, B)
    buildTree(tree, arr, 2 * index + 1, mid + 1, e, A, B)

    tree[index] = tree[2 * index] + tree[2 * index + 1]

# Query procedure to get the answer
# for each query l and r are query range
def query(tree, index, s, e, l, r):

    # out of bound or no overlap
    if (r < s or l > e):
        return 0

    # Complete overlap
    # Query range completely lies in
    # the segment tree node range
    if (s >= l and e <= r):
        return tree[index]

    # Partially overlap
    # Query range partially lies in
    # the segment tree node range
    mid = (s + e) // 2
    return (query(tree, 2 * index, s, mid, l, r)
            + query(tree, 2 * index + 1, mid + 1, e, l, r))

# Driver code
if __name__ == '__main__':
    arr=[7, 3, 9, 13, 5, 4]
    n = len(arr)
    tree=[0]*(4 * n + 1)

    L = 1
    R = 5
    A = 4
    B = 7

    buildTree(tree, arr, 1, 0, n - 1, A, B)

    print(query(tree, 1, 0, n - 1, L, R))

# This code is contributed by mohit kumar 29

C#

// C# implementation of the approach
using System;

class GFG{

// Procedure to build the segment tree
static void buildTree(int[] tree, int[] arr,
                      int index, int s, int e,
                      int A, int B)
{

    // Reached the leaf node
    // of the segment tree
    if (s == e)
    {
        if (arr[s] >= A && arr[s] <= B)
            tree[index] = 1;
        else
            tree[index] = 0;

        return;
    }

    // Recursively call the buildTree
    // on both the nodes of the tree
    int mid = (s + e) / 2;
    buildTree(tree, arr, 2 * index,
              s, mid, A, B);
    buildTree(tree, arr, 2 * index + 1,
              mid + 1, e, A, B);

    tree[index] = tree[2 * index] +
                  tree[2 * index + 1];
}

// Query procedure to get the answer
// for each query l and r are query range
static int query(int[] tree, int index, int s,
                 int e, int l, int r)
{

    // Out of bound or no overlap
    if (r < s || l > e)
        return 0;

    // Complete overlap
    // Query range completely lies in
    // the segment tree node range
    if (s >= l && e <= r)
    {
        return tree[index];
    }

    // Partially overlap
    // Query range partially lies in
    // the segment tree node range
    int mid = (s + e) / 2;
    return (query(tree, 2 * index,
                  s, mid, l, r) +
            query(tree, 2 * index + 1,
                  mid + 1, e, l, r));
}

// Driver code
public static void Main ()
{
    int[] arr = new int[] { 7, 3, 9, 13, 5, 4 };
    int n = arr.Length;
    int[] tree = new int [(4 * n + 1)];

    int L = 1, R = 5, A = 4, B = 7;

    buildTree(tree, arr, 1, 0, n - 1, A, B);

    Console.Write(query(tree, 1, 0, n - 1, L, R));
}
}

// This code is contributed by sanjoy_62

Javascript

<script>

// Javascript implementation of the approach

// Procedure to build the segment tree
function buildTree(tree, arr, index, s, e, A, B)
{

    // Reached the leaf node
    // of the segment tree
    if (s == e)
    {
        if (arr[s] >= A && arr[s] <= B)
            tree[index] = 1;
        else
            tree[index] = 0;

        return;
    }

    // Recursively call the buildTree
    // on both the nodes of the tree
    let mid = Math.floor((s + e) / 2);
    buildTree(tree, arr, 2 * index,
              s, mid, A, B);
    buildTree(tree, arr, 2 * index + 1,
              mid + 1, e, A, B);

    tree[index] = tree[2 * index] +
                  tree[2 * index + 1];
}

// Query procedure to get the answer
// for each query l and r are query range
function query(tree, index, s, e, l, r)
{

    // Out of bound or no overlap
    if (r < s || l > e)
        return 0;

    // Complete overlap
    // Query range completely lies in
    // the segment tree node range
    if (s >= l && e <= r)
    {
        return tree[index];
    }

    // Partially overlap
    // Query range partially lies in
    // the segment tree node range
    let mid = Math.floor((s + e) / 2);
    return (query(tree, 2 * index,
                  s, mid, l, r) +
            query(tree, 2 * index + 1,
                  mid + 1, e, l, r));
}

// Driver code
let arr = [ 7, 3, 9, 13, 5, 4 ];
let n = arr.length;
let tree = new Array(4 * n + 1);
let L = 1, R = 5, A = 4, B = 7;
buildTree(tree, arr, 1, 0, n - 1, A, B);

document.write(query(tree, 1, 0, n - 1, L, R));

// This code is contributed by avanitrachhadiya2155

</script>

Output:

2