Numbsubarrayer of elements less than or equal to a given number in a given

Last Updated : 27 Apr, 2023

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

CPP

   // C++ program to answer queries to count number
// of elements smaller than 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;
}
 
  

Java

   // Java program to answer queries to count number
// of elements smaller than or equal to x.
 
import java.util.*;
 
// structure to hold queries
class Query
{
    int l, r, x, idx;
 
    public Query(int l, int r, int x, int idx) {
        this.l = l;
        this.r = r;
        this.x = x;
        this.idx = idx;
    }
}
 
// structure to hold array
class ArrayElement implements Comparable<ArrayElement>
{
    int val, idx;
 
    public ArrayElement(int val, int idx) {
        this.val = val;
        this.idx = idx;
    }
 
    // bool function to sort array according to its value
    public int compareTo(ArrayElement other) {
        return Integer.compare(this.val, other.val);
    }
}
 
public class GFG
{
    // bool function to sort queries according to k
    static boolean cmp1(Query q1, Query q2)
    {
        return q1.x < q2.x;
    }
 
    // updating the bit array
    static void update(int bit[], int idx, int val, int n)
    {
        for (; idx<=n; idx +=idx&-idx)
            bit[idx] += val;
    }
 
    // querying the bit array
    static int query(int bit[], int idx, int n)
    {
        int sum = 0;
        for (; idx > 0; idx -= idx&-idx)
            sum += bit[idx];
        return sum;
    }
 
    static void answerQueries(int n, Query queries[], int q,
                              ArrayElement arr[])
    {
        // initialising bit array
        int bit[] = new int[n+1];
        Arrays.fill(bit, 0);
 
        // sorting the array
        Arrays.sort(arr);
 
        // sorting queries
        Arrays.sort(queries, (q1, q2) -> Integer.compare(q1.x, q2.x));
 
        // current index of array
        int curr = 0;
 
        // array to hold answer of each Query
        int ans[] = new int[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 (curr < n && arr[curr].val <= queries[i].x)
            {
                // 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++)
            System.out.println(ans[i]);
    }
 
    public static void main(String[] args) {
    // size of array
    int n = 4;
 
    // initializing array value and index
    ArrayElement[] arr = new ArrayElement[n];
    arr[0] = new ArrayElement(2, 0);
    arr[1] = new ArrayElement(3, 1);
    arr[2] = new ArrayElement(4, 2);
    arr[3] = new ArrayElement(5, 3);
 
    // number of queries
    int q = 2;
    Query[] queries = new Query[q];
    queries[0] = new Query(0, 2, 2, 0);
    queries[1] = new Query(0, 3, 5, 1);
 
    answerQueries(n, queries, q, arr);
    }
}
 
//this code is contributed by bhardwajji
 
  

Python3

   from functools import cmp_to_key
import bisect
 
# structure to hold queries
class Query:
    def __init__(self, l, r, x, idx):
        self.l = l
        self.r = r
        self.x = x
        self.idx = idx
   
# structure to hold array
class ArrayElement:
    def __init__(self, val, idx):
        self.val = val
        self.idx = idx
   
# bool function to sort queries according to k
def cmp1(q1, q2):
    if q1.x != q2.x:
        return q1.x - q2.x
    else:
        return q1.r - q2.r
   
# bool function to sort array according to its value
def cmp2(x, y):
    if x.val != y.val:
        return x.val - y.val
    else:
        return x.idx - y.idx
   
# updating the bit array
def update(bit, idx, val, n):
    while idx <= n:
        bit[idx] += val
        idx += idx & -idx
   
# querying the bit array
def query(bit, idx, n):
    sum = 0
    while idx > 0:
        sum += bit[idx]
        idx -= idx & -idx
    return sum
   
def answerQueries(n, queries, q, arr):
    # initialising bit array
    bit = [0] * (n + 1)
   
    # sorting the array
    arr = sorted(arr, key=cmp_to_key(cmp2))
   
    # sorting queries
    queries = sorted(queries, key=cmp_to_key(cmp1))
   
    # current index of array
    curr = 0
   
    # array to hold answer of each Query
    ans = [0] * q
   
    # looping through each Query
    for i in range(q):
        # traversing the array values till it
        # is less than equal to Query number
        while curr < n and arr[curr].val <= queries[i].x:
            # updating the bit array for the array index
            update(bit, arr[curr].idx+1, 1, n)
            curr += 1
   
        # 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 i in range(q):
        print(ans[i])
   
# driver function
if __name__ == '__main__':
    # size of array
    n = 4
   
    # initialising array value and index
    arr = [ArrayElement(2, 0), ArrayElement(3, 1), ArrayElement(4, 2), ArrayElement(5, 3)]
   
    # number of queries
    q = 2
    queries = [Query(0, 2, 2, 0), Query(0, 3, 5, 1)]
   
    answerQueries(n, queries, q, arr)
 
  

C#

   // C# program to answer queries to count number
// of elements smaller than or equal to x.
using System;
using System.Linq;
 
// structure to hold queries
class Query
{
    public int l, r, x, idx;
 
    public Query(int l, int r, int x, int idx) {
        this.l = l;
        this.r = r;
        this.x = x;
        this.idx = idx;
    }
}
 
// structure to hold array
class ArrayElement : IComparable<ArrayElement>
{
    public int val, idx;
     
    public ArrayElement(int val, int idx) {
        this.val = val;
        this.idx = idx;
    }
 
    // bool function to sort array according to its value
    public int CompareTo(ArrayElement other) {
        return this.val.CompareTo(other.val);
    }
}
 
public class GFG
{
    // bool function to sort queries according to k
    static bool cmp1(Query q1, Query q2)
    {
        return q1.x < q2.x;
    }
    // updating the bit array
    static void update(int[] bit, int idx, int val, int n)
    {
        for (; idx <= n; idx += idx & -idx)
            bit[idx] += val;
    }
 
    // querying the bit array
    static int query(int[] bit, int idx, int n)
    {
        int sum = 0;
        for (; idx > 0; idx -= idx & -idx)
            sum += bit[idx];
        return sum;
    }
 
    static void answerQueries(int n, Query[] queries, int q, ArrayElement[] arr)
    {
        // initialising bit array
        int[] bit = new int[n + 1];
        Array.Fill(bit, 0);
     
        // sorting the array
        Array.Sort(arr);
     
        // sorting queries
        Array.Sort(queries, (q1, q2) => q1.x.CompareTo(q2.x));
 
        // current index of array
        int curr = 0;
 
        // array to hold answer of each Query
        int[] ans = new int[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 (curr < n && arr[curr].val <= queries[i].x)
            {
                // 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++)
            Console.WriteLine(ans[i]);
    }
 
    public static void Main(string[] args)
    {
        // size of array
        int n = 4;
     
        // initializing array value and index
        ArrayElement[] arr = new ArrayElement[n];
        arr[0] = new ArrayElement(2, 0);
        arr[1] = new ArrayElement(3, 1);
        arr[2] = new ArrayElement(4, 2);
        arr[3] = new ArrayElement(5, 3);
 
        // number of queries
        int q = 2;
        Query[] queries = new Query[q];
        queries[0] = new Query(0, 2, 2, 0);
        queries[1] = new Query(0, 3, 5, 1);
 
        answerQueries(n, queries, q, arr);
    }
}
// This code is contributed by prasad264
 
  

Javascript

   // JavaScript program to answer queries to count number
// of elements smaller than or equal to x.
 
// structure to hold queries
class Query {
constructor(l, r, x, idx) {
this.l = l;
this.r = r;
this.x = x;
this.idx = idx;
}
}
 
// structure to hold array
class ArrayElement {
constructor(val, idx) {
this.val = val;
this.idx = idx;
}
 
// function to compare array elements
static compare(a, b) {
return a.val - b.val;
}
}
 
function answerQueries(n, queries, q, arr) {
// initialising bit array
let bit = new Array(n + 1).fill(0);
 
// sorting the array
arr.sort(ArrayElement.compare);
 
// sorting queries
queries.sort((q1, q2) => q1.x - q2.x);
 
// current index of array
let curr = 0;
 
// array to hold answer of each Query
let ans = new Array(q).fill(0);
 
// looping through each Query
for (let i = 0; i < q; i++) {
// traversing the array values till it
// is less than equal to Query number
while (curr < n && arr[curr].val <= queries[i].x) {
// 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 (let i = 0; i < q; i++) {
console.log(ans[i]);
}
}
 
// bool function to sort queries according to k
function cmp1(q1, q2) {
return q1.x < q2.x;
}
 
// updating the bit array
function update(bit, idx, val, n) {
for (; idx <= n; idx += idx & -idx) bit[idx] += val;
}
 
// querying the bit array
function query(bit, idx, n) {
let sum = 0;
for (; idx > 0; idx -= idx & -idx) sum += bit[idx];
return sum;
}
 
// size of array
let n = 4;
 
// initializing array value and index
let arr = [
new ArrayElement(2, 0),
new ArrayElement(3, 1),
new ArrayElement(4, 2),
new ArrayElement(5, 3),
];
 
// number of queries
let q = 2;
let queries = [
new Query(0, 2, 2, 0),
new Query(0, 3, 5, 1),
];
 
answerQueries(n, queries, q, arr);
 
  

Output:

1 4

Time Complexity: O((n+q)*log(n))

Auxiliary Space: O(n+q)

XOR of numbers that appeared even number of times in given Range

Ayush Jha

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Article Tags :

Practice Tags :

Advanced Data Structure

Numbsubarrayer of elements less than or equal to a given number in a given

CPP

Java

Python3

C#

Javascript

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