# 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 :
2
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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 `

## Python 3

# Python 3 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)

# 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 ` `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; ` `} `

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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