Bitwise operations on Subarrays of size K
Last Updated :
11 Jul, 2022
Given an array arr[] of positive integers and a number K, the task is to find the minimum and maximum values of Bitwise operation on elements of subarray of size K.
Examples:
Input: arr[]={2, 5, 3, 6, 11, 13}, k = 3
Output:
Maximum AND = 2
Minimum AND = 0
Maximum OR = 15
Minimum OR = 7
Explanation:
Maximum AND is generated by subarray 3, 6 and 11, 3 & 6 & 11 = 2
Minimum AND is generated by subarray 2, 3 and 5, 2 & 3 & 5 = 0
Maximum OR is generated by subarray 2, 6 and 13, 2 | 6 | 13 = 15
Minimum OR is generated by subarray 2, 3 and 5, 2 | 3 | 5 = 7
Input: arr[]={5, 9, 7, 19}, k = 2
Output:
Maximum AND = 3
Minimum AND = 1
Maximum OR = 23
Minimum OR = 13
Naive Approach: The naive approach is to generate all possible subarrays of size K and check which of the above-formed subarray will give the minimum and maximum Bitwise OR and AND.
Time Complexity: O(N2)
Auxiliary Space: O(K)
Efficient Approach: The idea is to use the Sliding Window Technique to solve this problem. Below are the steps:
- Traverse the prefix array of size K and for each array, element goes through it's each bit and increases bit array (by maintaining an integer array bit of size 32) by 1 if it is set.
- Convert this bit array to a decimal number lets say ans, and move the sliding window to the next index.
- For newly added element for the next subarray of size K, Iterate through each bit of the newly added element and increase bit array by 1 if it is set.
- For removing the first element from the previous window, decrease bit array by 1 if it is set.
- Update ans with a minimum or maximum of the new decimal number generated by bit array.
- Below is the program to find the Maximum Bitwise OR subarray:
C++
// C++ program for maximum values of
// each bitwise OR operation on
// element of subarray of size K
#include <iostream>
using namespace std;
// Function to convert bit array to
// decimal number
int build_num(int bit[])
{
int ans = 0;
for (int i = 0; i < 32; i++)
if (bit[i])
ans += (1 << i);
return ans;
}
// Function to find maximum values of
// each bitwise OR operation on
// element of subarray of size K
int maximumOR(int arr[], int n, int k)
{
// Maintain an integer array bit[]
// of size 32 all initialized to 0
int bit[32] = { 0 };
// Create a sliding window of size k
for (int i = 0; i < k; i++) {
for (int j = 0; j < 32; j++) {
if (arr[i] & (1 << j))
bit[j]++;
}
}
// Function call
int max_or = build_num(bit);
for (int i = k; i < n; i++) {
// Perform operation for
// removed element
for (int j = 0; j < 32; j++) {
if (arr[i - k] & (1 << j))
bit[j]--;
}
// Perform operation for
// added_element
for (int j = 0; j < 32; j++) {
if (arr[i] & (1 << j))
bit[j]++;
}
// Taking maximum value
max_or = max(build_num(bit), max_or);
}
// Return the result
return max_or;
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 2, 5, 3, 6, 11, 13 };
// Given subarray size K
int k = 3;
int n = sizeof arr / sizeof arr[0];
// Function Call
cout << maximumOR(arr, n, k);
return 0;
}
// Java program for maximum values of
// each bitwise OR operation on
// element of subarray of size K
import java.util.*;
class GFG{
// Function to convert bit array to
// decimal number
static int build_num(int bit[])
{
int ans = 0;
for (int i = 0; i < 32; i++)
if (bit[i] > 0)
ans += (1 << i);
return ans;
}
// Function to find maximum values of
// each bitwise OR operation on
// element of subarray of size K
static int maximumOR(int arr[], int n, int k)
{
// Maintain an integer array bit[]
// of size 32 all initialized to 0
int bit[] = new int[32];
// Create a sliding window of size k
for (int i = 0; i < k; i++)
{
for (int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
}
// Function call
int max_or = build_num(bit);
for (int i = k; i < n; i++)
{
// Perform operation for
// removed element
for (int j = 0; j < 32; j++)
{
if ((arr[i - k] & (1 << j)) > 0)
bit[j]--;
}
// Perform operation for
// added_element
for (int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
// Taking maximum value
max_or = Math.max(build_num(bit), max_or);
}
// Return the result
return max_or;
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int arr[] = { 2, 5, 3, 6, 11, 13 };
// Given subarray size K
int k = 3;
int n = arr.length;
// Function Call
System.out.print(maximumOR(arr, n, k));
}
}
// This code is contributed by Rohit_ranjan
# Python3 program for maximum values of
# each bitwise OR operation on
# element of subarray of size K
# Function to convert bit array to
# decimal number
def build_num(bit):
ans = 0;
for i in range(32):
if (bit[i] > 0):
ans += (1 << i);
return ans;
# Function to find maximum values of
# each bitwise OR operation on
# element of subarray of size K
def maximumOR(arr, n, k):
# Maintain an integer array bit
# of size 32 all initialized to 0
bit = [0] * 32;
# Create a sliding window of size k
for i in range(k):
for j in range(32):
if ((arr[i] & (1 << j)) > 0):
bit[j] += 1;
# Function call
max_or = build_num(bit);
for i in range(k, n):
# Perform operation for
# removed element
for j in range(32):
if ((arr[i - k] & (1 << j)) > 0):
bit[j] -= 1;
# Perform operation for
# added_element
for j in range(32):
if ((arr[i] & (1 << j)) > 0):
bit[j] += 1;
# Taking maximum value
max_or = max(build_num(bit), max_or);
# Return the result
return max_or;
# Driver Code
if __name__ == '__main__':
# Given array arr
arr = [ 2, 5, 3, 6, 11, 13 ];
# Given subarray size K
k = 3;
n = len(arr);
# Function call
print(maximumOR(arr, n, k));
# This code is contributed by Amit Katiyar
// C# program for maximum values of
// each bitwise OR operation on
// element of subarray of size K
using System;
class GFG{
// Function to convert bit
// array to decimal number
static int build_num(int[] bit)
{
int ans = 0;
for (int i = 0; i < 32; i++)
if (bit[i] > 0)
ans += (1 << i);
return ans;
}
// Function to find maximum values of
// each bitwise OR operation on
// element of subarray of size K
static int maximumOR(int[] arr, int n, int k)
{
// Maintain an integer array bit[]
// of size 32 all initialized to 0
int[] bit = new int[32];
// Create a sliding window of size k
for (int i = 0; i < k; i++)
{
for (int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
}
// Function call
int max_or = build_num(bit);
for (int i = k; i < n; i++)
{
// Perform operation for
// removed element
for (int j = 0; j < 32; j++)
{
if ((arr[i - k] & (1 << j)) > 0)
bit[j]--;
}
// Perform operation for
// added_element
for (int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
// Taking maximum value
max_or = Math.Max(build_num(bit), max_or);
}
// Return the result
return max_or;
}
// Driver Code
public static void Main(String[] args)
{
// Given array []arr
int[] arr = {2, 5, 3, 6, 11, 13};
// Given subarray size K
int k = 3;
int n = arr.Length;
// Function Call
Console.Write(maximumOR(arr, n, k));
}
}
// This code is contributed by Rohit_ranjan
<script>
// Javascript program for maximum values of
// each bitwise OR operation on
// element of subarray of size K
// Function to convert bit array to
// decimal number
function build_num(bit)
{
let ans = 0;
for (let i = 0; i < 32; i++)
if (bit[i] > 0)
ans += (1 << i);
return ans;
}
// Function to find maximum values of
// each bitwise OR operation on
// element of subarray of size K
function maximumOR(arr, n, k)
{
// Maintain an integer array bit[]
// of size 32 all initialized to 0
let bit = new Array(32);
bit.fill(0);
// Create a sliding window of size k
for (let i = 0; i < k; i++)
{
for (let j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
}
// Function call
let max_or = build_num(bit);
for (let i = k; i < n; i++)
{
// Perform operation for
// removed element
for (let j = 0; j < 32; j++)
{
if ((arr[i - k] & (1 << j)) > 0)
bit[j]--;
}
// Perform operation for
// added_element
for (let j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
// Taking maximum value
max_or = Math.max(build_num(bit), max_or);
}
// Return the result
return max_or;
}
// Given array []arr
let arr = [2, 5, 3, 6, 11, 13];
// Given subarray size K
let k = 3;
let n = arr.length;
// Function Call
document.write(maximumOR(arr, n, k));
// This code is contributed by divyesh072019.
</script>
Time Complexity: O(n * B) where n is the size of the array and B is the integer array bit of size 32.
Auxiliary Space: O(n)
- Below is the program to find the Minimum Bitwise OR subarray:
C++
// C++ program for minimum values of
// each bitwise OR operation on
// element of subarray of size K
#include <iostream>
using namespace std;
// Function to convert bit array
// to decimal number
int build_num(int bit[])
{
int ans = 0;
for (int i = 0; i < 32; i++)
if (bit[i])
ans += (1 << i);
return ans;
}
// Function to find minimum values of
// each bitwise OR operation on
// element of subarray of size K
int minimumOR(int arr[], int n, int k)
{
// Maintain an integer array bit[]
// of size 32 all initialized to 0
int bit[32] = { 0 };
// Create a sliding window of size k
for (int i = 0; i < k; i++) {
for (int j = 0; j < 32; j++) {
if (arr[i] & (1 << j))
bit[j]++;
}
}
// Function call
int min_or = build_num(bit);
for (int i = k; i < n; i++) {
// Perform operation for
// removed element
for (int j = 0; j < 32; j++) {
if (arr[i - k] & (1 << j))
bit[j]--;
}
// Perform operation for
// added_element
for (int j = 0; j < 32; j++) {
if (arr[i] & (1 << j))
bit[j]++;
}
// Taking minimum value
min_or = min(build_num(bit),
min_or);
}
// Return the result
return min_or;
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 2, 5, 3, 6, 11, 13 };
// Given subarray size K
int k = 3;
int n = sizeof arr / sizeof arr[0];
// Function Call
cout << minimumOR(arr, n, k);
return 0;
}
// Java program for minimum values of
// each bitwise OR operation on
// element of subarray of size K
import java.util.*;
class GFG{
// Function to convert bit array
// to decimal number
static int build_num(int bit[])
{
int ans = 0;
for(int i = 0; i < 32; i++)
if (bit[i] > 0)
ans += (1 << i);
return ans;
}
// Function to find minimum values of
// each bitwise OR operation on
// element of subarray of size K
static int minimumOR(int arr[], int n, int k)
{
// Maintain an integer array bit[]
// of size 32 all initialized to 0
int bit[] = new int[32];
// Create a sliding window of size k
for(int i = 0; i < k; i++)
{
for(int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
}
// Function call
int min_or = build_num(bit);
for(int i = k; i < n; i++)
{
// Perform operation for
// removed element
for(int j = 0; j < 32; j++)
{
if ((arr[i - k] & (1 << j)) > 0)
bit[j]--;
}
// Perform operation for
// added_element
for(int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
// Taking minimum value
min_or = Math.min(build_num(bit),
min_or);
}
// Return the result
return min_or;
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int arr[] = { 2, 5, 3, 6, 11, 13 };
// Given subarray size K
int k = 3;
int n = arr.length;
// Function call
System.out.print(minimumOR(arr, n, k));
}
}
// This code is contributed by Amit Katiyar
# Python3 program for minimum values
# of each bitwise OR operation on
# element of subarray of size K
# Function to convert bit array
# to decimal number
def build_num(bit):
ans = 0;
for i in range(32):
if (bit[i] > 0):
ans += (1 << i);
return ans;
# Function to find minimum values of
# each bitwise OR operation on
# element of subarray of size K
def minimumOR(arr, n, k):
# Maintain an integer array bit
# of size 32 all initialized to 0
bit = [0] * 32;
# Create a sliding window of size k
for i in range(k):
for j in range(32):
if ((arr[i] & (1 << j)) > 0):
bit[j] += 1;
# Function call
min_or = build_num(bit);
for i in range(k, n):
# Perform operation for
# removed element
for j in range(32):
if ((arr[i - k] & (1 << j)) > 0):
bit[j] -= 1;
# Perform operation for
# added_element
for j in range(32):
if ((arr[i] & (1 << j)) > 0):
bit[j] += 1;
# Taking minimum value
min_or = min(build_num(bit), min_or);
# Return the result
return min_or;
# Driver Code
if __name__ == '__main__':
# Given array arr
arr = [ 2, 5, 3, 6, 11, 13 ];
# Given subarray size K
k = 3;
n = len(arr);
# Function call
print(minimumOR(arr, n, k));
# This code is contributed by Amit Katiyar
// C# program for minimum values of
// each bitwise OR operation on
// element of subarray of size K
using System;
class GFG{
// Function to convert bit array
// to decimal number
static int build_num(int []bit)
{
int ans = 0;
for(int i = 0; i < 32; i++)
if (bit[i] > 0)
ans += (1 << i);
return ans;
}
// Function to find minimum values of
// each bitwise OR operation on
// element of subarray of size K
static int minimumOR(int []arr, int n, int k)
{
// Maintain an integer array bit[]
// of size 32 all initialized to 0
int []bit = new int[32];
// Create a sliding window of size k
for(int i = 0; i < k; i++)
{
for(int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
}
// Function call
int min_or = build_num(bit);
for(int i = k; i < n; i++)
{
// Perform operation for
// removed element
for(int j = 0; j < 32; j++)
{
if ((arr[i - k] & (1 << j)) > 0)
bit[j]--;
}
// Perform operation for
// added_element
for(int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
// Taking minimum value
min_or = Math.Min(build_num(bit),
min_or);
}
// Return the result
return min_or;
}
// Driver Code
public static void Main(String[] args)
{
// Given array []arr
int []arr = { 2, 5, 3, 6, 11, 13 };
// Given subarray size K
int k = 3;
int n = arr.Length;
// Function call
Console.Write(minimumOR(arr, n, k));
}
}
// This code is contributed by Amit Katiyar
<script>
// Javascript program for minimum values of
// each bitwise OR operation on
// element of subarray of size K
// Function to convert bit array
// to decimal number
function build_num(bit)
{
let ans = 0;
for(let i = 0; i < 32; i++)
if (bit[i] > 0)
ans += (1 << i);
return ans;
}
// Function to find minimum values of
// each bitwise OR operation on
// element of subarray of size K
function minimumOR(arr, n, k)
{
// Maintain an integer array bit[]
// of size 32 all initialized to 0
let bit = new Array(32);
bit.fill(0);
// Create a sliding window of size k
for(let i = 0; i < k; i++)
{
for(let j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
}
// Function call
let min_or = build_num(bit);
for(let i = k; i < n; i++)
{
// Perform operation for
// removed element
for(let j = 0; j < 32; j++)
{
if ((arr[i - k] & (1 << j)) > 0)
bit[j]--;
}
// Perform operation for
// added_element
for(let j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
// Taking minimum value
min_or = Math.min(build_num(bit), min_or);
}
// Return the result
return min_or;
}
// Driver code
// Given array arr[]
let arr = [ 2, 5, 3, 6, 11, 13 ];
// Given subarray size K
let k = 3;
let n = arr.length;
// Function Call
document.write(minimumOR(arr, n, k));
// This code is contributed by rameshtravel07
</script>
Time Complexity: O(n * B) where n is the size of the array and B is the integer array bit of size 32.
Auxiliary Space: O(n)
- Below is the program to find the Maximum Bitwise AND subarray:
C++
// C++ program for maximum values of
// each bitwise AND operation on
// element of subarray of size K
#include <iostream>
using namespace std;
// Function to convert bit array
// to decimal number
int build_num(int bit[], int k)
{
int ans = 0;
for (int i = 0; i < 32; i++)
if (bit[i] == k)
ans += (1 << i);
return ans;
}
// Function to find maximum values of
// each bitwise AND operation on
// element of subarray of size K
int maximumAND(int arr[], int n, int k)
{
// Maintain an integer array bit[]
// of size 32 all initialized to 0
int bit[32] = { 0 };
// Create a sliding window of size k
for (int i = 0; i < k; i++) {
for (int j = 0; j < 32; j++) {
if (arr[i] & (1 << j))
bit[j]++;
}
}
// Function call
int max_and = build_num(bit, k);
for (int i = k; i < n; i++) {
// Perform operation for
// removed element
for (int j = 0; j < 32; j++) {
if (arr[i - k] & (1 << j))
bit[j]--;
}
// Perform operation for
// added element
for (int j = 0; j < 32; j++) {
if (arr[i] & (1 << j))
bit[j]++;
}
// Taking maximum value
max_and = max(build_num(bit, k),
max_and);
}
// Return the result
return max_and;
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 2, 5, 3, 6, 11, 13 };
// Given subarray size K
int k = 3;
int n = sizeof arr / sizeof arr[0];
// Function Call
cout << maximumAND(arr, n, k);
return 0;
}
// Java program for maximum values of
// each bitwise AND operation on
// element of subarray of size K
class GFG{
// Function to convert bit array
// to decimal number
static int build_num(int bit[], int k)
{
int ans = 0;
for(int i = 0; i < 32; i++)
if (bit[i] == k)
ans += (1 << i);
return ans;
}
// Function to find maximum values of
// each bitwise AND operation on
// element of subarray of size K
static int maximumAND(int arr[],
int n, int k)
{
// Maintain an integer array bit[]
// of size 32 all initialized to 0
int bit[] = new int[32];
// Create a sliding window of size k
for(int i = 0; i < k; i++)
{
for(int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
}
// Function call
int max_and = build_num(bit, k);
for(int i = k; i < n; i++)
{
// Perform operation for
// removed element
for(int j = 0; j < 32; j++)
{
if ((arr[i - k] & (1 << j)) > 0)
bit[j]--;
}
// Perform operation for
// added element
for(int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
// Taking maximum value
max_and = Math.max(build_num(bit, k),
max_and);
}
// Return the result
return max_and;
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int arr[] = { 2, 5, 3, 6, 11, 13 };
// Given subarray size K
int k = 3;
int n = arr.length;
// Function call
System.out.print(maximumAND(arr, n, k));
}
}
// This code is contributed by 29AjayKumar
# Python3 program for maximum values of
# each bitwise AND operation on
# element of subarray of size K
# Function to convert bit array
# to decimal number
def build_num(bit, k):
ans = 0;
for i in range(32):
if (bit[i] == k):
ans += (1 << i);
return ans;
# Function to find maximum values of
# each bitwise AND operation on
# element of subarray of size K
def maximumAND(arr, n, k):
# Maintain an integer array bit
# of size 32 all initialized to 0
bit = [0] * 32;
# Create a sliding window of size k
for i in range(k):
for j in range(32):
if ((arr[i] & (1 << j)) > 0):
bit[j] += 1;
# Function call
max_and = build_num(bit, k);
for i in range(k, n):
# Perform operation for
# removed element
for j in range(32):
if ((arr[i - k] & (1 << j)) > 0):
bit[j] -= 1;
# Perform operation for
# added element
for j in range(32):
if ((arr[i] & (1 << j)) > 0):
bit[j] += 1;
# Taking maximum value
max_and = max(build_num(bit, k),
max_and);
# Return the result
return max_and;
# Driver Code
if __name__ == '__main__':
# Given array arr
arr = [ 2, 5, 3, 6, 11, 13 ];
# Given subarray size K
k = 3;
n = len(arr);
# Function call
print(maximumAND(arr, n, k));
# This code is contributed by Amit Katiyar
// C# program for maximum values of
// each bitwise AND operation on
// element of subarray of size K
using System;
class GFG{
// Function to convert bit
// array to decimal number
static int build_num(int[] bit, int k)
{
int ans = 0;
for (int i = 0; i < 32; i++)
if (bit[i] == k)
ans += (1 << i);
return ans;
}
// Function to find maximum values of
// each bitwise AND operation on
// element of subarray of size K
static int maximumAND(int[] arr, int n, int k)
{
// Maintain an integer array bit[]
// of size 32 all initialized to 0
int[] bit = new int[32];
// Create a sliding window of size k
for (int i = 0; i < k; i++)
{
for (int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
}
// Function call
int max_and = build_num(bit, k);
for (int i = k; i < n; i++)
{
// Perform operation for
// removed element
for (int j = 0; j < 32; j++)
{
if ((arr[i - k] & (1 << j)) > 0)
bit[j]--;
}
// Perform operation for
// added element
for (int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
// Taking maximum value
max_and = Math.Max(build_num(bit, k),
max_and);
}
// Return the result
return max_and;
}
// Driver Code
public static void Main(String[] args)
{
// Given array []arr
int[] arr = {2, 5, 3, 6, 11, 13};
// Given subarray size K
int k = 3;
int n = arr.Length;
// Function call
Console.Write(maximumAND(arr, n, k));
}
}
// This code is contributed by shikhasingrajput
<script>
// Javascript program for maximum values of
// each bitwise AND operation on
// element of subarray of size K
// Function to convert bit
// array to decimal number
function build_num(bit, k)
{
let ans = 0;
for (let i = 0; i < 32; i++)
if (bit[i] == k)
ans += (1 << i);
return ans;
}
// Function to find maximum values of
// each bitwise AND operation on
// element of subarray of size K
function maximumAND(arr, n, k)
{
// Maintain an integer array bit[]
// of size 32 all initialized to 0
let bit = new Array(32);
bit.fill(0);
// Create a sliding window of size k
for (let i = 0; i < k; i++)
{
for (let j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
}
// Function call
let max_and = build_num(bit, k);
for (let i = k; i < n; i++)
{
// Perform operation for
// removed element
for (let j = 0; j < 32; j++)
{
if ((arr[i - k] & (1 << j)) > 0)
bit[j]--;
}
// Perform operation for
// added element
for (let j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
// Taking maximum value
max_and = Math.max(build_num(bit, k),
max_and);
}
// Return the result
return max_and;
}
// Given array []arr
let arr = [2, 5, 3, 6, 11, 13];
// Given subarray size K
let k = 3;
let n = arr.length;
// Function call
document.write(maximumAND(arr, n, k));
// This code is contributed by mukesh07.
</script>
Time Complexity: O(n * B) where n is the size of the array and B is the integer array bit of size 32.
Auxiliary Space: O(n)
C++
// C++ program for minimum values of
// each bitwise AND operation on
// elements of subarray of size K
#include <iostream>
using namespace std;
// Function to convert bit array
// to decimal number
int build_num(int bit[], int k)
{
int ans = 0;
for (int i = 0; i < 32; i++)
if (bit[i] == k)
ans += (1 << i);
return ans;
}
// Function to find minimum values of
// each bitwise AND operation on
// element of subarray of size K
int minimumAND(int arr[], int n, int k)
{
// Maintain an integer array bit[]
// of size 32 all initialized to 0
int bit[32] = { 0 };
// Create a sliding window of size k
for (int i = 0; i < k; i++) {
for (int j = 0; j < 32; j++) {
if (arr[i] & (1 << j))
bit[j]++;
}
}
// Function call
int min_and = build_num(bit, k);
for (int i = k; i < n; i++) {
// Perform operation to removed
// element
for (int j = 0; j < 32; j++) {
if (arr[i - k] & (1 << j))
bit[j]--;
}
// Perform operation to add
// element
for (int j = 0; j < 32; j++) {
if (arr[i] & (1 << j))
bit[j]++;
}
// Taking minimum value
min_and = min(build_num(bit, k),
min_and);
}
// Return the result
return min_and;
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 2, 5, 3, 6, 11, 13 };
// Given subarray size K
int k = 3;
int n = sizeof arr / sizeof arr[0];
// Function Call
cout << minimumAND(arr, n, k);
return 0;
}
// Java program for minimum values of
// each bitwise AND operation on
// elements of subarray of size K
class GFG{
// Function to convert bit array
// to decimal number
static int build_num(int bit[], int k)
{
int ans = 0;
for(int i = 0; i < 32; i++)
if (bit[i] == k)
ans += (1 << i);
return ans;
}
// Function to find minimum values of
// each bitwise AND operation on
// element of subarray of size K
static int minimumAND(int arr[], int n, int k)
{
// Maintain an integer array bit[]
// of size 32 all initialized to 0
int bit[] = new int[32];
// Create a sliding window of size k
for(int i = 0; i < k; i++)
{
for(int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
}
// Function call
int min_and = build_num(bit, k);
for(int i = k; i < n; i++)
{
// Perform operation to removed
// element
for(int j = 0; j < 32; j++)
{
if ((arr[i - k] & (1 << j)) > 0)
bit[j]--;
}
// Perform operation to add
// element
for(int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
// Taking minimum value
min_and = Math.min(build_num(bit, k),
min_and);
}
// Return the result
return min_and;
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int arr[] = { 2, 5, 3, 6, 11, 13 };
// Given subarray size K
int k = 3;
int n = arr.length;
// Function call
System.out.print(minimumAND(arr, n, k));
}
}
// This code is contributed by 29AjayKumar
# Python program for minimum values of
# each bitwise AND operation on
# elements of subarray of size K
# Function to convert bit array
# to decimal number
def build_num(bit, k):
ans = 0;
for i in range(32):
if (bit[i] == k):
ans += (1 << i);
return ans;
# Function to find minimum values of
# each bitwise AND operation on
# element of subarray of size K
def minimumAND(arr, n, k):
# Maintain an integer array bit
# of size 32 all initialized to 0
bit = [0] * 32;
# Create a sliding window of size k
for i in range(k):
for j in range(32):
if ((arr[i] & (1 << j)) > 0):
bit[j] += 1;
# Function call
min_and = build_num(bit, k);
for i in range(k, n):
# Perform operation to removed
# element
for j in range(32):
if ((arr[i - k] & (1 << j)) > 0):
bit[j] -=1;
# Perform operation to add
# element
for j in range(32):
if ((arr[i] & (1 << j)) > 0):
bit[j] += 1;
# Taking minimum value
min_and = min(build_num(bit, k), min_and);
# Return the result
return min_and;
# Driver Code
if __name__ == '__main__':
# Given array arr
arr = [2, 5, 3, 6, 11, 13];
# Given subarray size K
k = 3;
n = len(arr);
# Function call
print(minimumAND(arr, n, k));
# This code contributed by Rajput-Ji
// C# program for minimum values of
// each bitwise AND operation on
// elements of subarray of size K
using System;
class GFG{
// Function to convert bit array
// to decimal number
static int build_num(int []bit, int k)
{
int ans = 0;
for(int i = 0; i < 32; i++)
if (bit[i] == k)
ans += (1 << i);
return ans;
}
// Function to find minimum values of
// each bitwise AND operation on
// element of subarray of size K
static int minimumAND(int []arr, int n, int k)
{
// Maintain an integer array bit[]
// of size 32 all initialized to 0
int []bit = new int[32];
// Create a sliding window of size k
for(int i = 0; i < k; i++)
{
for(int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
}
// Function call
int min_and = build_num(bit, k);
for(int i = k; i < n; i++)
{
// Perform operation to removed
// element
for(int j = 0; j < 32; j++)
{
if ((arr[i - k] & (1 << j)) > 0)
bit[j]--;
}
// Perform operation to add
// element
for(int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
// Taking minimum value
min_and = Math.Min(build_num(bit, k),
min_and);
}
// Return the result
return min_and;
}
// Driver Code
public static void Main(String[] args)
{
// Given array []arr
int []arr = { 2, 5, 3, 6, 11, 13 };
// Given subarray size K
int k = 3;
int n = arr.Length;
// Function call
Console.Write(minimumAND(arr, n, k));
}
}
// This code is contributed by 29AjayKumar
<script>
// Javascript program for minimum values of
// each bitwise AND operation on
// elements of subarray of size K
// Function to convert bit array
// to decimal number
function build_num(bit, k)
{
let ans = 0;
for(let i = 0; i < 32; i++)
if (bit[i] == k)
ans += (1 << i);
return ans;
}
// Function to find minimum values of
// each bitwise AND operation on
// element of subarray of size K
function minimumAND(arr, n, k)
{
// Maintain an integer array bit[]
// of size 32 all initialized to 0
let bit = new Array(32);
bit.fill(0);
// Create a sliding window of size k
for(let i = 0; i < k; i++)
{
for(let j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
}
// Function call
let min_and = build_num(bit, k);
for(let i = k; i < n; i++)
{
// Perform operation to removed
// element
for(let j = 0; j < 32; j++)
{
if ((arr[i - k] & (1 << j)) > 0)
bit[j]--;
}
// Perform operation to add
// element
for(let j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
// Taking minimum value
min_and = Math.min(build_num(bit, k), min_and);
}
// Return the result
return min_and;
}
// Given array []arr
let arr = [ 2, 5, 3, 6, 11, 13 ];
// Given subarray size K
let k = 3;
let n = arr.length;
// Function call
document.write(minimumAND(arr, n, k));
</script>
Time Complexity: O(n * B) where n is the size of the array and B is the integer array bit of size 32.
Auxiliary Space: O(n)
C++
// C++ program to find the subarray
/// with minimum XOR
#include <bits/stdc++.h>
using namespace std;
// Function to find the minimum XOR
// of the subarray of size K
void findMinXORSubarray(int arr[],
int n, int k)
{
// K must be smaller than
// or equal to n
if (n < k)
return;
// Initialize the beginning
// index of result
int res_index = 0;
// Compute XOR sum of first
// subarray of size K
int curr_xor = 0;
for (int i = 0; i < k; i++)
curr_xor ^= arr[i];
// Initialize minimum XOR
// sum as current xor
int min_xor = curr_xor;
// Traverse from (k+1)'th
// element to n'th element
for (int i = k; i < n; i++) {
// XOR with current item
// and first item of
// previous subarray
curr_xor ^= (arr[i] ^ arr[i - k]);
// Update result if needed
if (curr_xor < min_xor) {
min_xor = curr_xor;
res_index = (i - k + 1);
}
}
// Print the minimum XOR
cout << min_xor << "\n";
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 3, 7, 90, 20, 10, 50, 40 };
// Given subarray size K
int k = 3;
int n = sizeof(arr) / sizeof(arr[0]);
// Function Call
findMinXORSubarray(arr, n, k);
return 0;
}
// Java program to find the subarray
// with minimum XOR
class GFG{
// Function to find the minimum XOR
// of the subarray of size K
static void findMinXORSubarray(int arr[],
int n, int k)
{
// K must be smaller than
// or equal to n
if (n < k)
return;
// Initialize the beginning
// index of result
int res_index = 0;
// Compute XOR sum of first
// subarray of size K
int curr_xor = 0;
for(int i = 0; i < k; i++)
curr_xor ^= arr[i];
// Initialize minimum XOR
// sum as current xor
int min_xor = curr_xor;
// Traverse from (k+1)'th
// element to n'th element
for(int i = k; i < n; i++)
{
// XOR with current item
// and first item of
// previous subarray
curr_xor ^= (arr[i] ^ arr[i - k]);
// Update result if needed
if (curr_xor < min_xor)
{
min_xor = curr_xor;
res_index = (i - k + 1);
}
}
// Print the minimum XOR
System.out.println(min_xor);
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int arr[] = { 3, 7, 90, 20, 10, 50, 40 };
// Given subarray size K
int k = 3;
int n = arr.length;
// Function call
findMinXORSubarray(arr, n, k);
}
}
// This code is contributed by rock_cool
# Python3 program to find the subarray
# with minimum XOR
# Function to find the minimum XOR
# of the subarray of size K
def findMinXORSubarray(arr, n, k):
# K must be smaller than
# or equal to n
if (n < k):
return;
# Initialize the beginning
# index of result
res_index = 0;
# Compute XOR sum of first
# subarray of size K
curr_xor = 0;
for i in range(k):
curr_xor ^= arr[i];
# Initialize minimum XOR
# sum as current xor
min_xor = curr_xor;
# Traverse from (k+1)'th
# element to n'th element
for i in range(k, n):
# XOR with current item
# and first item of
# previous subarray
curr_xor ^= (arr[i] ^ arr[i - k]);
# Update result if needed
if (curr_xor < min_xor):
min_xor = curr_xor;
res_index = (i - k + 1);
# Print the minimum XOR
print(min_xor);
# Driver Code
if __name__ == '__main__':
# Given array arr
arr = [ 3, 7, 90, 20, 10, 50, 40 ];
# Given subarray size K
k = 3;
n = len(arr);
# Function call
findMinXORSubarray(arr, n, k);
# This code is contributed by Amit Katiyar
// C# program to find the subarray
// with minimum XOR
using System;
class GFG{
// Function to find the minimum XOR
// of the subarray of size K
static void findMinXORSubarray(int []arr,
int n, int k)
{
// K must be smaller than
// or equal to n
if (n < k)
return;
// Initialize the beginning
// index of result
int res_index = 0;
// Compute XOR sum of first
// subarray of size K
int curr_xor = 0;
for(int i = 0; i < k; i++)
curr_xor ^= arr[i];
// Initialize minimum XOR
// sum as current xor
int min_xor = curr_xor;
// Traverse from (k+1)'th
// element to n'th element
for(int i = k; i < n; i++)
{
// XOR with current item
// and first item of
// previous subarray
curr_xor ^= (arr[i] ^ arr[i - k]);
// Update result if needed
if (curr_xor < min_xor)
{
min_xor = curr_xor;
res_index = (i - k + 1);
}
}
// Print the minimum XOR
Console.WriteLine(min_xor);
}
// Driver Code
public static void Main(String[] args)
{
// Given array []arr
int []arr = { 3, 7, 90, 20, 10, 50, 40 };
// Given subarray size K
int k = 3;
int n = arr.Length;
// Function call
findMinXORSubarray(arr, n, k);
}
}
// This code is contributed by PrinciRaj1992
<script>
// Javascript program to find the subarray
// with minimum XOR
// Function to find the minimum XOR
// of the subarray of size K
function findMinXORSubarray(arr, n, k)
{
// K must be smaller than
// or equal to n
if (n < k)
return;
// Initialize the beginning
// index of result
let res_index = 0;
// Compute XOR sum of first
// subarray of size K
let curr_xor = 0;
for(let i = 0; i < k; i++)
curr_xor ^= arr[i];
// Initialize minimum XOR
// sum as current xor
let min_xor = curr_xor;
// Traverse from (k+1)'th
// element to n'th element
for(let i = k; i < n; i++)
{
// XOR with current item
// and first item of
// previous subarray
curr_xor ^= (arr[i] ^ arr[i - k]);
// Update result if needed
if (curr_xor < min_xor)
{
min_xor = curr_xor;
res_index = (i - k + 1);
}
}
// Print the minimum XOR
document.write(min_xor);
}
// Driver code
// Given array arr[]
let arr = [ 3, 7, 90, 20, 10, 50, 40 ];
// Given subarray size K
let k = 3;
let n = arr.length;
// Function Call
findMinXORSubarray(arr, n, k);
// This code is contributed by divyeshrabadiya07
</script>
Time Complexity: O(n * B) where n is the size of the array and B is the integer array bit of size 32.
Auxiliary Space: O(n)
Related Topic: Subarrays, Subsequences, and Subsets in Array
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