Bitwise operations on Subarrays of size K
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++
#include <iostream>
using namespace std;
int build_num( int bit[])
{
int ans = 0;
for ( int i = 0; i < 32; i++)
if (bit[i])
ans += (1 << i);
return ans;
}
int maximumOR( int arr[], int n, int k)
{
int bit[32] = { 0 };
for ( int i = 0; i < k; i++) {
for ( int j = 0; j < 32; j++) {
if (arr[i] & (1 << j))
bit[j]++;
}
}
int max_or = build_num(bit);
for ( int i = k; i < n; i++) {
for ( int j = 0; j < 32; j++) {
if (arr[i - k] & (1 << j))
bit[j]--;
}
for ( int j = 0; j < 32; j++) {
if (arr[i] & (1 << j))
bit[j]++;
}
max_or = max(build_num(bit), max_or);
}
return max_or;
}
int main()
{
int arr[] = { 2, 5, 3, 6, 11, 13 };
int k = 3;
int n = sizeof arr / sizeof arr[0];
cout << maximumOR(arr, n, k);
return 0;
}
|
Java
import java.util.*;
class GFG{
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;
}
static int maximumOR( int arr[], int n, int k)
{
int bit[] = new int [ 32 ];
for ( int i = 0 ; i < k; i++)
{
for ( int j = 0 ; j < 32 ; j++)
{
if ((arr[i] & ( 1 << j)) > 0 )
bit[j]++;
}
}
int max_or = build_num(bit);
for ( int i = k; i < n; i++)
{
for ( int j = 0 ; j < 32 ; j++)
{
if ((arr[i - k] & ( 1 << j)) > 0 )
bit[j]--;
}
for ( int j = 0 ; j < 32 ; j++)
{
if ((arr[i] & ( 1 << j)) > 0 )
bit[j]++;
}
max_or = Math.max(build_num(bit), max_or);
}
return max_or;
}
public static void main(String[] args)
{
int arr[] = { 2 , 5 , 3 , 6 , 11 , 13 };
int k = 3 ;
int n = arr.length;
System.out.print(maximumOR(arr, n, k));
}
}
|
Python3
def build_num(bit):
ans = 0 ;
for i in range ( 32 ):
if (bit[i] > 0 ):
ans + = ( 1 << i);
return ans;
def maximumOR(arr, n, k):
bit = [ 0 ] * 32 ;
for i in range (k):
for j in range ( 32 ):
if ((arr[i] & ( 1 << j)) > 0 ):
bit[j] + = 1 ;
max_or = build_num(bit);
for i in range (k, n):
for j in range ( 32 ):
if ((arr[i - k] & ( 1 << j)) > 0 ):
bit[j] - = 1 ;
for j in range ( 32 ):
if ((arr[i] & ( 1 << j)) > 0 ):
bit[j] + = 1 ;
max_or = max (build_num(bit), max_or);
return max_or;
if __name__ = = '__main__' :
arr = [ 2 , 5 , 3 , 6 , 11 , 13 ];
k = 3 ;
n = len (arr);
print (maximumOR(arr, n, k));
|
C#
using System;
class GFG{
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;
}
static int maximumOR( int [] arr, int n, int k)
{
int [] bit = new int [32];
for ( int i = 0; i < k; i++)
{
for ( int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
}
int max_or = build_num(bit);
for ( int i = k; i < n; i++)
{
for ( int j = 0; j < 32; j++)
{
if ((arr[i - k] & (1 << j)) > 0)
bit[j]--;
}
for ( int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
max_or = Math.Max(build_num(bit), max_or);
}
return max_or;
}
public static void Main(String[] args)
{
int [] arr = {2, 5, 3, 6, 11, 13};
int k = 3;
int n = arr.Length;
Console.Write(maximumOR(arr, n, k));
}
}
|
Javascript
<script>
function build_num(bit)
{
let ans = 0;
for (let i = 0; i < 32; i++)
if (bit[i] > 0)
ans += (1 << i);
return ans;
}
function maximumOR(arr, n, k)
{
let bit = new Array(32);
bit.fill(0);
for (let i = 0; i < k; i++)
{
for (let j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
}
let max_or = build_num(bit);
for (let i = k; i < n; i++)
{
for (let j = 0; j < 32; j++)
{
if ((arr[i - k] & (1 << j)) > 0)
bit[j]--;
}
for (let j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
max_or = Math.max(build_num(bit), max_or);
}
return max_or;
}
let arr = [2, 5, 3, 6, 11, 13];
let k = 3;
let n = arr.length;
document.write(maximumOR(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)
- Below is the program to find the Minimum Bitwise OR subarray:
C++
#include <iostream>
using namespace std;
int build_num( int bit[])
{
int ans = 0;
for ( int i = 0; i < 32; i++)
if (bit[i])
ans += (1 << i);
return ans;
}
int minimumOR( int arr[], int n, int k)
{
int bit[32] = { 0 };
for ( int i = 0; i < k; i++) {
for ( int j = 0; j < 32; j++) {
if (arr[i] & (1 << j))
bit[j]++;
}
}
int min_or = build_num(bit);
for ( int i = k; i < n; i++) {
for ( int j = 0; j < 32; j++) {
if (arr[i - k] & (1 << j))
bit[j]--;
}
for ( int j = 0; j < 32; j++) {
if (arr[i] & (1 << j))
bit[j]++;
}
min_or = min(build_num(bit),
min_or);
}
return min_or;
}
int main()
{
int arr[] = { 2, 5, 3, 6, 11, 13 };
int k = 3;
int n = sizeof arr / sizeof arr[0];
cout << minimumOR(arr, n, k);
return 0;
}
|
Java
import java.util.*;
class GFG{
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;
}
static int minimumOR( int arr[], int n, int k)
{
int bit[] = new int [ 32 ];
for ( int i = 0 ; i < k; i++)
{
for ( int j = 0 ; j < 32 ; j++)
{
if ((arr[i] & ( 1 << j)) > 0 )
bit[j]++;
}
}
int min_or = build_num(bit);
for ( int i = k; i < n; i++)
{
for ( int j = 0 ; j < 32 ; j++)
{
if ((arr[i - k] & ( 1 << j)) > 0 )
bit[j]--;
}
for ( int j = 0 ; j < 32 ; j++)
{
if ((arr[i] & ( 1 << j)) > 0 )
bit[j]++;
}
min_or = Math.min(build_num(bit),
min_or);
}
return min_or;
}
public static void main(String[] args)
{
int arr[] = { 2 , 5 , 3 , 6 , 11 , 13 };
int k = 3 ;
int n = arr.length;
System.out.print(minimumOR(arr, n, k));
}
}
|
Python3
def build_num(bit):
ans = 0 ;
for i in range ( 32 ):
if (bit[i] > 0 ):
ans + = ( 1 << i);
return ans;
def minimumOR(arr, n, k):
bit = [ 0 ] * 32 ;
for i in range (k):
for j in range ( 32 ):
if ((arr[i] & ( 1 << j)) > 0 ):
bit[j] + = 1 ;
min_or = build_num(bit);
for i in range (k, n):
for j in range ( 32 ):
if ((arr[i - k] & ( 1 << j)) > 0 ):
bit[j] - = 1 ;
for j in range ( 32 ):
if ((arr[i] & ( 1 << j)) > 0 ):
bit[j] + = 1 ;
min_or = min (build_num(bit), min_or);
return min_or;
if __name__ = = '__main__' :
arr = [ 2 , 5 , 3 , 6 , 11 , 13 ];
k = 3 ;
n = len (arr);
print (minimumOR(arr, n, k));
|
C#
using System;
class GFG{
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;
}
static int minimumOR( int []arr, int n, int k)
{
int []bit = new int [32];
for ( int i = 0; i < k; i++)
{
for ( int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
}
int min_or = build_num(bit);
for ( int i = k; i < n; i++)
{
for ( int j = 0; j < 32; j++)
{
if ((arr[i - k] & (1 << j)) > 0)
bit[j]--;
}
for ( int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
min_or = Math.Min(build_num(bit),
min_or);
}
return min_or;
}
public static void Main(String[] args)
{
int []arr = { 2, 5, 3, 6, 11, 13 };
int k = 3;
int n = arr.Length;
Console.Write(minimumOR(arr, n, k));
}
}
|
Javascript
<script>
function build_num(bit)
{
let ans = 0;
for (let i = 0; i < 32; i++)
if (bit[i] > 0)
ans += (1 << i);
return ans;
}
function minimumOR(arr, n, k)
{
let bit = new Array(32);
bit.fill(0);
for (let i = 0; i < k; i++)
{
for (let j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
}
let min_or = build_num(bit);
for (let i = k; i < n; i++)
{
for (let j = 0; j < 32; j++)
{
if ((arr[i - k] & (1 << j)) > 0)
bit[j]--;
}
for (let j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
min_or = Math.min(build_num(bit), min_or);
}
return min_or;
}
let arr = [ 2, 5, 3, 6, 11, 13 ];
let k = 3;
let n = arr.length;
document.write(minimumOR(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)
- Below is the program to find the Maximum Bitwise AND subarray:
C++
#include <iostream>
using namespace std;
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;
}
int maximumAND( int arr[], int n, int k)
{
int bit[32] = { 0 };
for ( int i = 0; i < k; i++) {
for ( int j = 0; j < 32; j++) {
if (arr[i] & (1 << j))
bit[j]++;
}
}
int max_and = build_num(bit, k);
for ( int i = k; i < n; i++) {
for ( int j = 0; j < 32; j++) {
if (arr[i - k] & (1 << j))
bit[j]--;
}
for ( int j = 0; j < 32; j++) {
if (arr[i] & (1 << j))
bit[j]++;
}
max_and = max(build_num(bit, k),
max_and);
}
return max_and;
}
int main()
{
int arr[] = { 2, 5, 3, 6, 11, 13 };
int k = 3;
int n = sizeof arr / sizeof arr[0];
cout << maximumAND(arr, n, k);
return 0;
}
|
Java
class GFG{
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;
}
static int maximumAND( int arr[],
int n, int k)
{
int bit[] = new int [ 32 ];
for ( int i = 0 ; i < k; i++)
{
for ( int j = 0 ; j < 32 ; j++)
{
if ((arr[i] & ( 1 << j)) > 0 )
bit[j]++;
}
}
int max_and = build_num(bit, k);
for ( int i = k; i < n; i++)
{
for ( int j = 0 ; j < 32 ; j++)
{
if ((arr[i - k] & ( 1 << j)) > 0 )
bit[j]--;
}
for ( int j = 0 ; j < 32 ; j++)
{
if ((arr[i] & ( 1 << j)) > 0 )
bit[j]++;
}
max_and = Math.max(build_num(bit, k),
max_and);
}
return max_and;
}
public static void main(String[] args)
{
int arr[] = { 2 , 5 , 3 , 6 , 11 , 13 };
int k = 3 ;
int n = arr.length;
System.out.print(maximumAND(arr, n, k));
}
}
|
Python3
def build_num(bit, k):
ans = 0 ;
for i in range ( 32 ):
if (bit[i] = = k):
ans + = ( 1 << i);
return ans;
def maximumAND(arr, n, k):
bit = [ 0 ] * 32 ;
for i in range (k):
for j in range ( 32 ):
if ((arr[i] & ( 1 << j)) > 0 ):
bit[j] + = 1 ;
max_and = build_num(bit, k);
for i in range (k, n):
for j in range ( 32 ):
if ((arr[i - k] & ( 1 << j)) > 0 ):
bit[j] - = 1 ;
for j in range ( 32 ):
if ((arr[i] & ( 1 << j)) > 0 ):
bit[j] + = 1 ;
max_and = max (build_num(bit, k),
max_and);
return max_and;
if __name__ = = '__main__' :
arr = [ 2 , 5 , 3 , 6 , 11 , 13 ];
k = 3 ;
n = len (arr);
print (maximumAND(arr, n, k));
|
C#
using System;
class GFG{
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;
}
static int maximumAND( int [] arr, int n, int k)
{
int [] bit = new int [32];
for ( int i = 0; i < k; i++)
{
for ( int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
}
int max_and = build_num(bit, k);
for ( int i = k; i < n; i++)
{
for ( int j = 0; j < 32; j++)
{
if ((arr[i - k] & (1 << j)) > 0)
bit[j]--;
}
for ( int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
max_and = Math.Max(build_num(bit, k),
max_and);
}
return max_and;
}
public static void Main(String[] args)
{
int [] arr = {2, 5, 3, 6, 11, 13};
int k = 3;
int n = arr.Length;
Console.Write(maximumAND(arr, n, k));
}
}
|
Javascript
<script>
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 maximumAND(arr, n, k)
{
let bit = new Array(32);
bit.fill(0);
for (let i = 0; i < k; i++)
{
for (let j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
}
let max_and = build_num(bit, k);
for (let i = k; i < n; i++)
{
for (let j = 0; j < 32; j++)
{
if ((arr[i - k] & (1 << j)) > 0)
bit[j]--;
}
for (let j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
max_and = Math.max(build_num(bit, k),
max_and);
}
return max_and;
}
let arr = [2, 5, 3, 6, 11, 13];
let k = 3;
let n = arr.length;
document.write(maximumAND(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++
#include <iostream>
using namespace std;
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;
}
int minimumAND( int arr[], int n, int k)
{
int bit[32] = { 0 };
for ( int i = 0; i < k; i++) {
for ( int j = 0; j < 32; j++) {
if (arr[i] & (1 << j))
bit[j]++;
}
}
int min_and = build_num(bit, k);
for ( int i = k; i < n; i++) {
for ( int j = 0; j < 32; j++) {
if (arr[i - k] & (1 << j))
bit[j]--;
}
for ( int j = 0; j < 32; j++) {
if (arr[i] & (1 << j))
bit[j]++;
}
min_and = min(build_num(bit, k),
min_and);
}
return min_and;
}
int main()
{
int arr[] = { 2, 5, 3, 6, 11, 13 };
int k = 3;
int n = sizeof arr / sizeof arr[0];
cout << minimumAND(arr, n, k);
return 0;
}
|
Java
class GFG{
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;
}
static int minimumAND( int arr[], int n, int k)
{
int bit[] = new int [ 32 ];
for ( int i = 0 ; i < k; i++)
{
for ( int j = 0 ; j < 32 ; j++)
{
if ((arr[i] & ( 1 << j)) > 0 )
bit[j]++;
}
}
int min_and = build_num(bit, k);
for ( int i = k; i < n; i++)
{
for ( int j = 0 ; j < 32 ; j++)
{
if ((arr[i - k] & ( 1 << j)) > 0 )
bit[j]--;
}
for ( int j = 0 ; j < 32 ; j++)
{
if ((arr[i] & ( 1 << j)) > 0 )
bit[j]++;
}
min_and = Math.min(build_num(bit, k),
min_and);
}
return min_and;
}
public static void main(String[] args)
{
int arr[] = { 2 , 5 , 3 , 6 , 11 , 13 };
int k = 3 ;
int n = arr.length;
System.out.print(minimumAND(arr, n, k));
}
}
|
Python3
def build_num(bit, k):
ans = 0 ;
for i in range ( 32 ):
if (bit[i] = = k):
ans + = ( 1 << i);
return ans;
def minimumAND(arr, n, k):
bit = [ 0 ] * 32 ;
for i in range (k):
for j in range ( 32 ):
if ((arr[i] & ( 1 << j)) > 0 ):
bit[j] + = 1 ;
min_and = build_num(bit, k);
for i in range (k, n):
for j in range ( 32 ):
if ((arr[i - k] & ( 1 << j)) > 0 ):
bit[j] - = 1 ;
for j in range ( 32 ):
if ((arr[i] & ( 1 << j)) > 0 ):
bit[j] + = 1 ;
min_and = min (build_num(bit, k), min_and);
return min_and;
if __name__ = = '__main__' :
arr = [ 2 , 5 , 3 , 6 , 11 , 13 ];
k = 3 ;
n = len (arr);
print (minimumAND(arr, n, k));
|
C#
using System;
class GFG{
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;
}
static int minimumAND( int []arr, int n, int k)
{
int []bit = new int [32];
for ( int i = 0; i < k; i++)
{
for ( int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
}
int min_and = build_num(bit, k);
for ( int i = k; i < n; i++)
{
for ( int j = 0; j < 32; j++)
{
if ((arr[i - k] & (1 << j)) > 0)
bit[j]--;
}
for ( int j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
min_and = Math.Min(build_num(bit, k),
min_and);
}
return min_and;
}
public static void Main(String[] args)
{
int []arr = { 2, 5, 3, 6, 11, 13 };
int k = 3;
int n = arr.Length;
Console.Write(minimumAND(arr, n, k));
}
}
|
Javascript
<script>
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 minimumAND(arr, n, k)
{
let bit = new Array(32);
bit.fill(0);
for (let i = 0; i < k; i++)
{
for (let j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
}
let min_and = build_num(bit, k);
for (let i = k; i < n; i++)
{
for (let j = 0; j < 32; j++)
{
if ((arr[i - k] & (1 << j)) > 0)
bit[j]--;
}
for (let j = 0; j < 32; j++)
{
if ((arr[i] & (1 << j)) > 0)
bit[j]++;
}
min_and = Math.min(build_num(bit, k), min_and);
}
return min_and;
}
let arr = [ 2, 5, 3, 6, 11, 13 ];
let k = 3;
let n = arr.length;
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++
#include <bits/stdc++.h>
using namespace std;
void findMinXORSubarray( int arr[],
int n, int k)
{
if (n < k)
return ;
int res_index = 0;
int curr_xor = 0;
for ( int i = 0; i < k; i++)
curr_xor ^= arr[i];
int min_xor = curr_xor;
for ( int i = k; i < n; i++) {
curr_xor ^= (arr[i] ^ arr[i - k]);
if (curr_xor < min_xor) {
min_xor = curr_xor;
res_index = (i - k + 1);
}
}
cout << min_xor << "\n" ;
}
int main()
{
int arr[] = { 3, 7, 90, 20, 10, 50, 40 };
int k = 3;
int n = sizeof (arr) / sizeof (arr[0]);
findMinXORSubarray(arr, n, k);
return 0;
}
|
Java
class GFG{
static void findMinXORSubarray( int arr[],
int n, int k)
{
if (n < k)
return ;
int res_index = 0 ;
int curr_xor = 0 ;
for ( int i = 0 ; i < k; i++)
curr_xor ^= arr[i];
int min_xor = curr_xor;
for ( int i = k; i < n; i++)
{
curr_xor ^= (arr[i] ^ arr[i - k]);
if (curr_xor < min_xor)
{
min_xor = curr_xor;
res_index = (i - k + 1 );
}
}
System.out.println(min_xor);
}
public static void main(String[] args)
{
int arr[] = { 3 , 7 , 90 , 20 , 10 , 50 , 40 };
int k = 3 ;
int n = arr.length;
findMinXORSubarray(arr, n, k);
}
}
|
Python3
def findMinXORSubarray(arr, n, k):
if (n < k):
return ;
res_index = 0 ;
curr_xor = 0 ;
for i in range (k):
curr_xor ^ = arr[i];
min_xor = curr_xor;
for i in range (k, n):
curr_xor ^ = (arr[i] ^ arr[i - k]);
if (curr_xor < min_xor):
min_xor = curr_xor;
res_index = (i - k + 1 );
print (min_xor);
if __name__ = = '__main__' :
arr = [ 3 , 7 , 90 , 20 , 10 , 50 , 40 ];
k = 3 ;
n = len (arr);
findMinXORSubarray(arr, n, k);
|
C#
using System;
class GFG{
static void findMinXORSubarray( int []arr,
int n, int k)
{
if (n < k)
return ;
int res_index = 0;
int curr_xor = 0;
for ( int i = 0; i < k; i++)
curr_xor ^= arr[i];
int min_xor = curr_xor;
for ( int i = k; i < n; i++)
{
curr_xor ^= (arr[i] ^ arr[i - k]);
if (curr_xor < min_xor)
{
min_xor = curr_xor;
res_index = (i - k + 1);
}
}
Console.WriteLine(min_xor);
}
public static void Main(String[] args)
{
int []arr = { 3, 7, 90, 20, 10, 50, 40 };
int k = 3;
int n = arr.Length;
findMinXORSubarray(arr, n, k);
}
}
|
Javascript
<script>
function findMinXORSubarray(arr, n, k)
{
if (n < k)
return ;
let res_index = 0;
let curr_xor = 0;
for (let i = 0; i < k; i++)
curr_xor ^= arr[i];
let min_xor = curr_xor;
for (let i = k; i < n; i++)
{
curr_xor ^= (arr[i] ^ arr[i - k]);
if (curr_xor < min_xor)
{
min_xor = curr_xor;
res_index = (i - k + 1);
}
}
document.write(min_xor);
}
let arr = [ 3, 7, 90, 20, 10, 50, 40 ];
let k = 3;
let n = arr.length;
findMinXORSubarray(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)
Related Topic: Subarrays, Subsequences, and Subsets in Array
Last Updated :
11 Jul, 2022
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