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Queries for bitwise AND in the index range [L, R] of the given array
• Difficulty Level : Medium
• Last Updated : 12 May, 2021

Given an array arr[] of N and Q queries consisting of a range [L, R]. the task is to find the bit-wise AND of all the elements of in that index range.
Examples:

Input: arr[] = {1, 3, 1, 2, 3, 4}, q[] = {{0, 1}, {3, 5}}
Output:

1 AND 3 = 1
2 AND 3 AND 4 = 0
Input: arr[] = {1, 2, 3, 4, 5}, q[] = {{0, 4}, {1, 3}}
Output:

Naive approach: Iterate through the range and find bit-wise AND of all the numbers in that range. This will take O(n) time for each query.
Efficient approach: If we look at the integers as binary number, we can easily see that condition for ith bit of our answer to be set is that ith bit of all the integers in the range [L, R] should be set.
So, we will calculate prefix-count for each bit. We will use this to find the number of integers in the range with ith bit set. If it is equal to the size of the range then the ith bit of our answer will also be set.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``#define MAX 100000``#define bitscount 32``using` `namespace` `std;` `// Array to store bit-wise``// prefix count``int` `prefix_count[bitscount][MAX];` `// Function to find the prefix sum``void` `findPrefixCount(``int` `arr[], ``int` `n)``{` `    ``// Loop for each bit``    ``for` `(``int` `i = 0; i < bitscount; i++) {` `        ``// Loop to find prefix count``        ``prefix_count[i] = ((arr >> i) & 1);``        ``for` `(``int` `j = 1; j < n; j++) {``            ``prefix_count[i][j] = ((arr[j] >> i) & 1);``            ``prefix_count[i][j] += prefix_count[i][j - 1];``        ``}``    ``}``}` `// Function to answer query``int` `rangeAnd(``int` `l, ``int` `r)``{` `    ``// To store the answer``    ``int` `ans = 0;` `    ``// Loop for each bit``    ``for` `(``int` `i = 0; i < bitscount; i++) {``        ``// To store the number of variables``        ``// with ith bit set``        ``int` `x;``        ``if` `(l == 0)``            ``x = prefix_count[i][r];``        ``else``            ``x = prefix_count[i][r]``                ``- prefix_count[i][l - 1];` `        ``// Condition for ith bit``        ``// of answer to be set``        ``if` `(x == r - l + 1)``            ``ans = (ans | (1 << i));``    ``}` `    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 7, 5, 3, 5, 2, 3 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``);` `    ``findPrefixCount(arr, n);` `    ``int` `queries[] = { { 1, 3 }, { 4, 5 } };``    ``int` `q = ``sizeof``(queries) / ``sizeof``(queries);` `    ``for` `(``int` `i = 0; i < q; i++)``        ``cout << rangeAnd(queries[i],``                         ``queries[i])``             ``<< endl;` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.io.*;` `class` `GFG``{``    ` `static` `int` `MAX = ``100000``;``static` `int` `bitscount =``32``;` `// Array to store bit-wise``// prefix count``static` `int` `[][]prefix_count = ``new` `int` `[bitscount][MAX];` `// Function to find the prefix sum``static` `void` `findPrefixCount(``int` `arr[], ``int` `n)``{` `    ``// Loop for each bit``    ``for` `(``int` `i = ``0``; i < bitscount; i++)``    ``{` `        ``// Loop to find prefix count``        ``prefix_count[i][``0``] = ((arr[``0``] >> i) & ``1``);``        ``for` `(``int` `j = ``1``; j < n; j++)``        ``{``            ``prefix_count[i][j] = ((arr[j] >> i) & ``1``);``            ``prefix_count[i][j] += prefix_count[i][j - ``1``];``        ``}``    ``}``}` `// Function to answer query``static` `int` `rangeAnd(``int` `l, ``int` `r)``{` `    ``// To store the answer``    ``int` `ans = ``0``;` `    ``// Loop for each bit``    ``for` `(``int` `i = ``0``; i < bitscount; i++)``    ``{``        ``// To store the number of variables``        ``// with ith bit set``        ``int` `x;``        ``if` `(l == ``0``)``            ``x = prefix_count[i][r];``        ``else``            ``x = prefix_count[i][r]``                ``- prefix_count[i][l - ``1``];` `        ``// Condition for ith bit``        ``// of answer to be set``        ``if` `(x == r - l + ``1``)``            ``ans = (ans | (``1` `<< i));``    ``}` `    ``return` `ans;``}` `// Driver code``public` `static` `void` `main (String[] args)``{` `    ``int` `arr[] = { ``7``, ``5``, ``3``, ``5``, ``2``, ``3` `};``    ``int` `n = arr.length;``    ` `    ``findPrefixCount(arr, n);``    ` `    ``int` `queries[][] = { { ``1``, ``3` `}, { ``4``, ``5` `} };``    ``int` `q = queries.length;``    ` `    ``for` `(``int` `i = ``0``; i < q; i++)``                ``System.out.println (rangeAnd(queries[i][``0``],queries[i][``1``]));``        `  `}``}` `// This code is contributed by ajit.`

## Python3

 `# Python3 implementation of the approach` `import` `numpy as np` `MAX` `=` `100000``bitscount ``=` `32` `# Array to store bit-wise``# prefix count``prefix_count ``=` `np.zeros((bitscount,``MAX``));` `# Function to find the prefix sum``def` `findPrefixCount(arr, n) :` `    ``# Loop for each bit``    ``for` `i ``in` `range``(``0``, bitscount) :``        ` `        ``# Loop to find prefix count``        ``prefix_count[i][``0``] ``=` `((arr[``0``] >> i) & ``1``);``        ` `        ``for` `j ``in` `range``(``1``, n) :``            ``prefix_count[i][j] ``=` `((arr[j] >> i) & ``1``);``            ``prefix_count[i][j] ``+``=` `prefix_count[i][j ``-` `1``];` `# Function to answer query``def` `rangeOr(l, r) :` `    ``# To store the answer``    ``ans ``=` `0``;` `    ``# Loop for each bit``    ``for` `i ``in` `range``(bitscount) :``        ` `        ``# To store the number of variables``        ``# with ith bit set``        ``x ``=` `0``;``        ` `        ``if` `(l ``=``=` `0``) :``            ``x ``=` `prefix_count[i][r];``        ``else` `:``            ``x ``=` `prefix_count[i][r] ``-` `prefix_count[i][l ``-` `1``];``            ` `        ``# Condition for ith bit``        ``# of answer to be set``        ``if` `(x ``=``=` `r ``-` `l ``+` `1``) :``            ``ans ``=` `(ans | (``1` `<< i));``            ` `    ``return` `ans;`  `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``arr ``=` `[ ``7``, ``5``, ``3``, ``5``, ``2``, ``3` `];``    ``n ``=` `len``(arr);` `    ``findPrefixCount(arr, n);` `    ``queries ``=` `[ [ ``1``, ``3` `], [ ``4``, ``5` `] ];``    ` `    ``q ``=` `len``(queries);` `    ``for` `i ``in` `range``(q) :``        ``print``(rangeOr(queries[i][``0``], queries[i][``1``]));`  `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `static` `int` `MAX = 100000;``static` `int` `bitscount =32;` `// Array to store bit-wise``// prefix count``static` `int` `[,]prefix_count = ``new` `int` `[bitscount,MAX];` `// Function to find the prefix sum``static` `void` `findPrefixCount(``int` `[]arr, ``int` `n)``{` `    ``// Loop for each bit``    ``for` `(``int` `i = 0; i < bitscount; i++)``    ``{` `        ``// Loop to find prefix count``        ``prefix_count[i,0] = ((arr >> i) & 1);``        ``for` `(``int` `j = 1; j < n; j++)``        ``{``            ``prefix_count[i,j] = ((arr[j] >> i) & 1);``            ``prefix_count[i,j] += prefix_count[i,j - 1];``        ``}``    ``}``}` `// Function to answer query``static` `int` `rangeAnd(``int` `l, ``int` `r)``{` `    ``// To store the answer``    ``int` `ans = 0;` `    ``// Loop for each bit``    ``for` `(``int` `i = 0; i < bitscount; i++)``    ``{``        ``// To store the number of variables``        ``// with ith bit set``        ``int` `x;``        ``if` `(l == 0)``            ``x = prefix_count[i,r];``        ``else``            ``x = prefix_count[i,r]``                ``- prefix_count[i,l - 1];` `        ``// Condition for ith bit``        ``// of answer to be set``        ``if` `(x == r - l + 1)``            ``ans = (ans | (1 << i));``    ``}` `    ``return` `ans;``}` `// Driver code``public` `static` `void` `Main (String[] args)``{` `    ``int` `[]arr = { 7, 5, 3, 5, 2, 3 };``    ``int` `n = arr.Length;``    ` `    ``findPrefixCount(arr, n);``    ` `    ``int` `[,]queries = { { 1, 3 }, { 4, 5 } };``    ``int` `q = queries.GetLength(0);``    ` `    ``for` `(``int` `i = 0; i < q; i++)``            ``Console.WriteLine(rangeAnd(queries[i,0],queries[i,1]));``        ` `}``}` `// This code contributed by Rajput-Ji`

## Javascript

 ``
Output:

```1
2```

Time complexity for pre-computation is O(n) and each query can be answered in O(1)

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