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 = 7Input: 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
// 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
# 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#
// 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 |
Javascript
<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> |
Output:
15
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
// 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
# 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#
// 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 |
Javascript
<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> |
Output:
7
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
// 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
# 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#
// 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 |
Javascript
<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> |
Output:
2
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++
// 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
// 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 |
Python3
# 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#
// 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 |
Javascript
<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> |
Output:
0
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++
// 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
// 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
# 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#
// 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 |
Javascript
<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> |
Output:
16
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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