Bitwise operations on Subarrays of size K

Given an array arr[] of positive integers and an 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 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 Sliding Window Technique to solve this problem. Below are the steps:

Traverse the prefix array of size K and for each array element go through it’s each bit and increase 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 addded element for the next subarray of size K, Iterate through each bit of newly added element and increase bit array by 1 if it is set.
For removing first element from the previous window, decrease bit array by 1 if it is set.
Update ans with minimum or maximum of the new decimal number generated by bit array.

Below is the program to find the Maximum Bitwise OR subarray:
// 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;
}
Output:

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++ 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;
}
Output:

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++ 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;
}
Output:

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 AND subarray:
// 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;
}
Output:

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 XOR subarray:
// 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;
}
Output:

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)

#arrays #bit magic #competitive programming #bitwise-and #bitwise-or #bitwise-xor #subarray

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Bitwise operations on Subarrays of size K