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Queries for number of array elements in a range with Kth Bit Set

Given an array of N positive (32-bit)integers, the task is to answer Q queries of the following form: 

Query(L, R, K): Print the number of elements of the array in the 
                range L to R, which have their Kth bit as set

Note: Consider LSB to be indexed at 1.

Examples: 

Input : arr[] = { 8, 9, 1, 3 } 
Query 1: L = 1, R = 3, K = 4 
Query 2: L = 2, R = 4, K = 1 
Output

3
Explanation
For the 1st query, the range (1, 3) represents elements, {8, 9, 1}. Among these elements only 8 and 9 have their 4th bit set. Thus, the answer for this query is 2.
For the 2nd query, the range (2, 4) represents elements, {9, 1, 3}. All of these elements have their 1st bit set. Thus, the answer for this query is 3

Prerequisites: Bit Manipulation | Prefix Sum Arrays



Method 1 (Brute Force) : For each query, traverse the array from L to R, and at every index check if the array element at that index has its Kth bit as set. If it does increment the counter variable.

Below is the implementation of above approach. 

C++




/* C++ Program to find the number of elements
   in a range L to R having the Kth bit as set */
#include <bits/stdc++.h>
using namespace std;
 
// Maximum bits required in binary represention
// of an array element
#define MAX_BITS 32
 
/* Returns true if n has its kth bit as set, 
   else returns false */
bool isKthBitSet(int n, int k)
{
    if (n & (1 << (k - 1)))
        return true;
    return false;
}
 
/* Returns the answer for each query with range L
   to R querying for the number of elements with
   the Kth bit set in the range */
int answerQuery(int L, int R, int K, int arr[])
{
    // counter stores the number of element in
    // the range with the kth bit set
    int counter = 0;
    for (int i = L; i <= R; i++) {
        if (isKthBitSet(arr[i], K)) {
            counter++;
        }
    }
    return counter;
}
 
// Print the answer for all queries
void answerQueries(int queries[][3], int Q,
                   int arr[], int N)
{
    int query_L, query_R, query_K;
 
    for (int i = 0; i < Q; i++) {
        query_L = queries[i][0] - 1;
        query_R = queries[i][1] - 1;
        query_K = queries[i][2];
 
        cout << "Result for Query " << i + 1 << " = "
             << answerQuery(query_L, query_R, query_K, arr)
             << endl;
    }
}
 
// Driver Code
int main()
{
 
    int arr[] = { 8, 9, 1, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    /* queries[][] denotes the array of queries
    where each query has three integers
    query[i][0] -> Value of L for ith query
    query[i][0] -> Value of R for ith query
    query[i][0] -> Value of K for ith query */
    int queries[][3] = {
        { 1, 3, 4 },
        { 2, 4, 1 }
    };
    int Q = sizeof(queries) / sizeof(queries[0]);
 
    answerQueries(queries, Q, arr, N);
 
    return 0;
}

Java




// Java Program to find the
// number of elements in a
// range L to R having the
// Kth bit as set
import java.util.*;
import java.lang.*;
import java.io.*;
 
// Maximum bits required
// in binary represention
// of an array element
class GFG
{
static final int MAX_BITS = 32;
 
/* Returns true if n
has its kth bit as set,
else returns false */
static boolean isKthBitSet(int n,
                           int k)
{
    if ((n & (1 << (k - 1))) != 0)
        return true;
    return false;
}
 
/* Returns the answer for
each query with range L
to R querying for the number
of elements with the Kth bit
set in the range */
static int answerQuery(int L, int R,
                       int K, int arr[])
{
    // counter stores the number
    // of element in the range
    // with the kth bit set
    int counter = 0;
    for (int i = L; i <= R; i++)
    {
        if (isKthBitSet(arr[i], K))
        {
            counter++;
        }
    }
    return counter;
}
 
// Print the answer
// for all queries
static void answerQueries(int queries[][], int Q,
                          int arr[], int N)
{
    int query_L, query_R, query_K;
 
    for (int i = 0; i < Q; i++)
    {
        query_L = queries[i][0] - 1;
        query_R = queries[i][1] - 1;
        query_K = queries[i][2];
 
        System.out.println("Result for Query " +
                               (i + 1) + " = " +
                   answerQuery(query_L, query_R,
                               query_K, arr));
             
    }
}
 
// Driver Code
public static void main(String args[])
{
    int arr[] = { 8, 9, 1, 3 };
    int N = arr.length;
 
    /* queries[][] denotes the array
    of queries where each query has
    three integers
    query[i][0] -> Value of L for ith query
    query[i][0] -> Value of R for ith query
    query[i][0] -> Value of K for ith query */
    int queries[][] =
    {
        { 1, 3, 4 },
        { 2, 4, 1 }
    };
    int Q = queries.length;
 
    answerQueries(queries, Q, arr, N);
}
}
 
// This code is contributed
// by Subhadeep

Python3




# Python3 Program to find the number of elements
# in a range L to R having the Kth bit as set
 
# Maximum bits required in binary represention
# of an array element
MAX_BITS = 32
 
# Returns true if n has its kth bit as set,
# else returns false
def isKthBitSet(n, k):
 
    if (n & (1 << (k - 1))):
        return True
    return False
 
# Returns the answer for each query with range L
# to R querying for the number of elements with
# the Kth bit set in the range
def answerQuery(L, R, K, arr):
     
    # counter stores the number of element
    # in the range with the kth bit set
    counter = 0
    for i in range(L, R + 1):
        if (isKthBitSet(arr[i], K)):
            counter += 1
    return counter
 
# Print the answer for all queries
def answerQueries(queries, Q, arr, N):
 
    for i in range(Q):
        query_L = queries[i][0] - 1
        query_R = queries[i][1] - 1
        query_K = queries[i][2]
 
        print("Result for Query", i + 1, "=",
                answerQuery(query_L, query_R,
                            query_K, arr))
 
# Driver Code
if __name__ == "__main__":
 
    arr = [ 8, 9, 1, 3 ]
    N = len(arr)
 
    # queries[][] denotes the array of queries
    # where each query has three integers
    # query[i][0] -> Value of L for ith query
    # query[i][0] -> Value of R for ith query
    # query[i][0] -> Value of K for ith query
    queries = [[ 1, 3, 4 ],
               [ 2, 4, 1 ]]
    Q = len(queries)
 
    answerQueries(queries, Q, arr, N)
 
# This code is contributed by ita_c

C#




// C# Program to find the number of
// elements in a range L to R having
// the Kth bit as set
using System;
 
// Maximum bits required
// in binary represention
// of an array element
class GFG
{
static readonly int MAX_BITS = 32;
 
/* Returns true if n
has its kth bit as set,
else returns false */
static bool isKthBitSet(int n,
                        int k)
{
    if ((n & (1 << (k - 1))) != 0)
        return true;
    return false;
}
 
/* Returns the answer for each query
with range L to R querying for the
number of elements with the Kth bit
set in the range */
static int answerQuery(int L, int R,
                       int K, int []arr)
{
    // counter stores the number
    // of element in the range
    // with the kth bit set
    int counter = 0;
    for (int i = L; i <= R; i++)
    {
        if (isKthBitSet(arr[i], K))
        {
            counter++;
        }
    }
    return counter;
}
 
// Print the answer for all queries
static void answerQueries(int [,]queries, int Q,
                          int []arr, int N)
{
    int query_L, query_R, query_K;
 
    for (int i = 0; i < Q; i++)
    {
        query_L = queries[i,0] - 1;
        query_R = queries[i,1] - 1;
        query_K = queries[i,2];
 
        Console.WriteLine("Result for Query " +
                              (i + 1) + " = " +
                answerQuery(query_L, query_R,
                            query_K, arr));
 
    }
}
 
// Driver Code
public static void Main()
{
    int []arr = { 8, 9, 1, 3 };
    int N = arr.Length;
 
    /* queries[][] denotes the array
    of queries where each query has
    three integers
    query[i][0] -> Value of L for ith query
    query[i][0] -> Value of R for ith query
    query[i][0] -> Value of K for ith query */
    int [,]queries = { { 1, 3, 4 },
                       { 2, 4, 1 } };
    int Q = queries.GetLength(0);
 
    answerQueries(queries, Q, arr, N);
}
}
 
// This code is contributed
// by 29AjayKumar
Output: 
Result for Query 1 = 2
Result for Query 2 = 3

 

Time Complexity : O(N) for each query.

Method 2 (Efficient) : Assuming that every integer in the array has at max 32 bits in its Binary Representation. A 2D prefix sum array can be built to solve the problem. Here the 2nd dimension of the prefix array is of size equal to the maximum number of bits required to represent a integer of the array in binary.
Let the Prefix Sum Array be P[][]. Now, P[i][j] denotes the number of Elements from 0 to i, which have their jth bit as set. This prefix sum array is built before answering the queries. If a query from L to R is encountered, querying for elements in this range having their Kth bit as set, then the answer for that query is P[R][K] – P[L – 1][K].

Below is the implementation of above approach. 

C++




/* C++ Program to find the number of elements
   in a range L to R having the Kth bit as set */
#include <bits/stdc++.h>
using namespace std;
 
// Maximum bits required in binary represention
// of an array element
#define MAX_BITS 32
 
/* Returns true if n has its kth bit as set,
   else returns false */
bool isKthBitSet(int n, int k)
{
    if (n & (1 << (k - 1)))
        return true;
    return false;
}
 
// Return pointer to the prefix sum array
int** buildPrefixArray(int N, int arr[])
{
    // Build a prefix sum array P[][]
    // where P[i][j] represents the number of
    // elements from 0 to i having the jth bit as set
    int** P = new int*[N + 1];
    for (int i = 0; i <= N; ++i) {
        P[i] = new int[MAX_BITS + 1];
    }
 
    for (int i = 0; i <= MAX_BITS; i++) {
        P[0][i] = 0;
    }
 
    for (int i = 0; i < N; i++) {
 
        for (int j = 1; j <= MAX_BITS; j++) {
            // prefix sum from 0 to i for each bit
            // position jhas the value of sum from 0
            // to i-1 for each j
            if (i)
                P[i][j] = P[i - 1][j];
 
            // if jth bit set then increment P[i][j] by 1
            bool isJthBitSet = isKthBitSet(arr[i], j);
            if (isJthBitSet) {
                P[i][j]++;
            }
        }
    }
 
    return P;
}
 
/* Returns the answer for each query with range
   L to R querying for the number of elements with
   the Kth bit set in the range */
int answerQuery(int L, int R, int K, int** P)
{
 
    /* Number of elements in range L to R with Kth
       bit set = (Number of elements from 0 to R with
       kth bit set) - (Number of elements from 0 to L-1
       with kth bit set) */
    if (L)
        return P[R][K] - P[L - 1][K];
    else
        return P[R][K];
}
 
// Print the answer for all queries
void answerQueries(int queries[][3], int Q,
                   int arr[], int N)
{
 
    // Build Prefix Array to answer queries efficiently
    int** P = buildPrefixArray(N, arr);
 
    int query_L, query_R, query_K;
 
    for (int i = 0; i < Q; i++) {
        query_L = queries[i][0] - 1;
        query_R = queries[i][1] - 1;
        query_K = queries[i][2];
 
        cout << "Result for Query " << i + 1 << " = "
             << answerQuery(query_L, query_R, query_K, P)
             << endl;
    }
}
 
// Driver Code
int main()
{
 
    int arr[] = { 8, 9, 1, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    /* queries[][] denotes the array of queries
    where each query has three integers
    query[i][0] -> Value of L for ith query
    query[i][0] -> Value of R for ith query
    query[i][0] -> Value of K for ith query */
    int queries[][3] = {
        { 1, 3, 4 },
        { 2, 4, 1 }
    };
    int Q = sizeof(queries) / sizeof(queries[0]);
 
    answerQueries(queries, Q, arr, N);
 
    return 0;
}

Java




/* Java Program to find the number of elements
   in a range L to R having the Kth bit as set */
import java.io.*;
 
class GFG
{
 
  // Maximum bits required in binary represention
  // of an array element
  static int MAX_BITS = 32;
 
  /* Returns true if n has its kth bit as set,
    else returns false */
  static boolean isKthBitSet(int n, int k)
  {
    if((n & (1 << (k - 1))) != 0)
    {
      return true;
    }
    return false;
  }
 
  // Return pointer to the prefix sum array
  static int[][] buildPrefixArray(int N, int[] arr)
  {
 
    // Build a prefix sum array P[][]
    // where P[i][j] represents the number of
    // elements from 0 to i having the jth bit as set
    int[][] P = new int[N + 1][MAX_BITS + 1];
    for(int i = 0; i <= MAX_BITS; i++)
    {
      P[0][i] = 0;
    }
    for(int i = 0; i < N; i++)
    {
      for(int j = 1; j <= MAX_BITS; j++)
      {
 
        // prefix sum from 0 to i for each bit
        // position jhas the value of sum from 0
        // to i-1 for each j
        if(i != 0)
        {
          P[i][j] = P[i - 1][j];
        }
 
        // if jth bit set then increment P[i][j] by 1
        boolean isJthBitSet = isKthBitSet(arr[i], j);
        if(isJthBitSet)
        {
          P[i][j]++;
        }
      }
    }
    return P;
  }
 
  /* Returns the answer for each query with range
    L to R querying for the number of elements with
    the Kth bit set in the range */
  static int answerQuery(int L, int R, int K, int[][] P)
  {
 
    /* Number of elements in range L to R with Kth
        bit set = (Number of elements from 0 to R with
        kth bit set) - (Number of elements from 0 to L-1
        with kth bit set) */
    if(L != 0)
    {
      return P[R][K] - P[L - 1][K];
    }
    else
    {
      return P[R][K];
    }
  }
 
  // Print the answer for all queries
  static void answerQueries(int[][] queries,int Q,
                            int[] arr, int N)
  {
 
    // Build Prefix Array to answer queries efficiently
    int[][] P = buildPrefixArray(N, arr);
    int query_L, query_R, query_K;
    for(int i = 0; i < Q; i++)
    {
      query_L = queries[i][0] - 1;
      query_R = queries[i][1] - 1;
      query_K = queries[i][2];
      System.out.println("Result for Query " + (i + 1) + " = " + answerQuery(query_L, query_R, query_K, P));
    }
  }
 
  // Driver Code
  public static void main (String[] args)
  {
    int[] arr = {8, 9, 1, 3};
    int N = arr.length;
 
    /* queries[][] denotes the array of queries
        where each query has three integers
        query[i][0] -> Value of L for ith query
        query[i][0] -> Value of R for ith query
        query[i][0] -> Value of K for ith query */
    int[][] queries = {{1, 3, 4},{2, 4, 1}};
    int Q = queries.length;
    answerQueries(queries, Q, arr, N);
  }
}
 
// This code is contributed by avanitrachhadiya2155

Python3




# Python3 program to find the number
# of elements in a range L to R having
# the Kth bit as set
 
# Maximum bits required in binary
# represention of an array element
MAX_BITS = 32
 
# Returns true if n has its kth
# bit as set,else returns false
def isKthBitSet(n, k):
    if (n & (1 << (k - 1))):
        return True
         
    return False
 
# Return pointer to the prefix sum array
def buildPrefixArray(N, arr):
     
    # Build a prefix sum array P[][]
    # where P[i][j] represents the
    # number of elements from 0 to
    # i having the jth bit as set
    P = [[0 for i in range(MAX_BITS + 1)]
            for i in range(N + 1)]
 
    for i in range(N):
        for j in range(1, MAX_BITS + 1):
             
            # prefix sum from 0 to i for each bit
            # position jhas the value of sum from 0
            # to i-1 for each j
            if (i):
                P[i][j] = P[i - 1][j]
 
            # If jth bit set then increment
            # P[i][j] by 1
            isJthBitSet = isKthBitSet(arr[i], j)
             
            if (isJthBitSet):
                P[i][j] += 1
 
    return P
 
# Returns the answer for each query
# with range L to R querying for the
# number of elements with the Kth bit
# set in the range
def answerQuery(L, R, K, P):
 
    # Number of elements in range L to
    # R with Kth bit set = (Number of
    # elements from 0 to R with kth
    # bit set) - (Number of elements
    # from 0 to L-1 with kth bit set)
    if (L):
        return P[R][K] - P[L - 1][K]
    else:
        return P[R][K]
 
# Print the answer for all queries
def answerQueries(queries, Q, arr, N):
 
    # Build Prefix Array to answer
    # queries efficiently
    P = buildPrefixArray(N, arr)
 
    for i in range(Q):
        query_L = queries[i][0] - 1
        query_R = queries[i][1] - 1
        query_K = queries[i][2]
 
        print("Result for Query ", i + 1,
              " = ", answerQuery(query_L, query_R,
                                 query_K, P))
 
# Driver Code
if __name__ == '__main__':
 
    arr = [ 8, 9, 1, 3 ]
    N = len(arr)
 
    # queries[]denotes the array of queries
    # where each query has three integers
    # query[i][0] -> Value of L for ith query
    # query[i][0] -> Value of R for ith query
    # query[i][0] -> Value of K for ith query
    queries = [ [ 1, 3, 4 ],
                [ 2, 4, 1 ] ]
                 
    Q = len(queries)
 
    answerQueries(queries, Q, arr, N)
 
# This code is contributed by mohit kumar 29

C#




// C# program to find the number of elements
// in a range L to R having the Kth bit as set
using System;
 
class GFG{
     
// Maximum bits required in binary represention
// of an array element
static int MAX_BITS = 32;
 
// Returns true if n has its kth bit as set,
// else returns false
static bool isKthBitSet(int n, int k)
{
    if ((n & (1 << (k - 1))) != 0)
    {
        return true;
    }
    return false;
}
 
// Return pointer to the prefix sum array
static int[,] buildPrefixArray(int N, int[] arr)
{
     
    // Build a prefix sum array P[][]
    // where P[i][j] represents the
    // number of elements from 0 to i
    // having the jth bit as set
    int[,] P = new int[N + 1, MAX_BITS + 1];
    for(int i = 0; i <= MAX_BITS; i++)
    {
        P[0, i] = 0;
    }
     
    for(int i = 0; i < N; i++)
    {
        for(int j = 1; j <= MAX_BITS; j++)
        {
             
            // prefix sum from 0 to i for each bit
            // position jhas the value of sum from 0
            // to i-1 for each j
            if (i != 0)
            {
                P[i, j] = P[i - 1, j];
            }
             
            // If jth bit set then increment P[i][j] by 1
            bool isJthBitSet = isKthBitSet(arr[i], j);
            if (isJthBitSet)
            {
                P[i, j]++;
            }
        }
    }
    return P;
}
 
// Returns the answer for each query with range
// L to R querying for the number of elements with
// the Kth bit set in the range
static int answerQuery(int L, int R, int K, int[,] P)
{
     
    // Number of elements in range L to R with Kth
    // bit set = (Number of elements from 0 to R with
    // kth bit set) - (Number of elements from 0 to L-1
    // with kth bit set)
    if (L != 0)
    {
        return P[R, K] - P[L - 1, K];
    }
    else
    {
        return P[R, K];
    }
}
 
// Print the answer for all queries
static void answerQueries(int[,] queries, int Q,
                          int[] arr, int N)
{
     
    // Build Prefix Array to answer queries efficiently
    int[,] P = buildPrefixArray(N, arr);
    int query_L, query_R, query_K;
    for(int i = 0; i < Q; i++)
    {
        query_L = queries[i, 0] - 1;
        query_R = queries[i, 1] - 1;
        query_K = queries[i, 2];
        Console.WriteLine("Result for Query " +
                          (i + 1) + " = " +
                          answerQuery(query_L,
                                      query_R,
                                      query_K, P));
    }
}
 
// Driver Code
static public void Main()
{
    int[] arr = { 8, 9, 1, 3 };
    int N = arr.Length;
     
    /* queries[][] denotes the array of queries
    where each query has three integers
    query[i][0] -> Value of L for ith query
    query[i][0] -> Value of R for ith query
    query[i][0] -> Value of K for ith query */
    int[,] queries = { { 1, 3, 4 }, { 2, 4, 1 } };
    int Q = queries.GetLength(0);
     
    answerQueries(queries, Q, arr, N);
}
}
 
// This code is contributed by rag2127
Output: 
Result for Query 1 = 2
Result for Query 2 = 3

 

Time Complexity of building the Prefix array is O(N * Maximum number of Bits) and each query is answered in O(1). 
Auxiliary space : O(N * Maximum Number of Bits) is required to build the Prefix Sum Array
 

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