Given an array arr[] and an integer K, the task is to count the number of sub-arrays having bitwise OR ? K.
Examples:
Input: arr[] = { 1, 2, 3 } K = 3
Output: 4
Bitwise OR of sub-arrays:
{ 1 } = 1
{ 1, 2 } = 3
{ 1, 2, 3 } = 3
{ 2 } = 2
{ 2, 3 } = 3
{ 3 } = 3
4 sub-arrays have bitwise OR ? KInput: arr[] = { 3, 4, 5 } K = 6
Output: 2
Naive approach: Run three nested loops. The outermost loop determines the starting of the sub-array. The middle loop determines the ending of the sub-array. The innermost loop traverses the sub-array whose bounds are determined by the outermost and middle loops. For every sub-array, calculate OR and update count = count + 1 if OR is greater than K.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the count of required sub-arraysint countSubArrays(const int* arr, int n, int K){ int count = 0; for (int i = 0; i < n; i++) { for (int j = i; j < n; j++) { int bitwise_or = 0; // Traverse sub-array [i..j] for (int k = i; k <= j; k++) { bitwise_or = bitwise_or | arr[k]; } if (bitwise_or >= K) count++; } } return count;}// Driver codeint main(){ int arr[] = { 3, 4, 5 }; int n = sizeof(arr) / sizeof(arr[0]); int k = 6; cout << countSubArrays(arr, n, k); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class solution{// Function to return the count of required sub-arraysstatic int countSubArrays(int arr[], int n, int K){ int count = 0; for (int i = 0; i < n; i++) { for (int j = i; j < n; j++) { int bitwise_or = 0; // Traverse sub-array [i..j] for (int k = i; k <= j; k++) { bitwise_or = bitwise_or | arr[k]; } if (bitwise_or >= K) count++; } } return count;}// Driver codepublic static void main(String args[]){ int arr[] = { 3, 4, 5 }; int n = arr.length; int k = 6; System.out.println(countSubArrays(arr, n, k)); }}// This code is contributed by// Surendra_Gangwar |
Python3
# Python3 implementation of the approach # Function to return the count of# required sub-arrays def countSubArrays(arr, n, K) : count = 0; for i in range(n) : for j in range(i, n) : bitwise_or = 0 # Traverse sub-array [i..j] for k in range(i, j + 1) : bitwise_or = bitwise_or | arr[k] if (bitwise_or >= K) : count += 1 return count # Driver code if __name__ == "__main__" : arr = [ 3, 4, 5 ] n = len(arr) k = 6 print(countSubArrays(arr, n, k))# This code is contributed by Ryuga |
C#
// C# implementation of the approachusing System; class GFG{// Function to return the count of // required sub-arraysstatic int countSubArrays(int []arr, int n, int K){ int count = 0; for (int i = 0; i < n; i++) { for (int j = i; j < n; j++) { int bitwise_or = 0; // Traverse sub-array [i..j] for (int k = i; k <= j; k++) { bitwise_or = bitwise_or | arr[k]; } if (bitwise_or >= K) count++; } } return count;}// Driver codepublic static void Main(){ int []arr = { 3, 4, 5 }; int n = arr.Length; int k = 6; Console.WriteLine(countSubArrays(arr, n, k));}}// This code is contributed by// Mohit kumar |
PHP
<?php// PHP implementation of the approach // Function to return the count of// required sub-arrays function countSubArrays($arr, $n, $K) { $count = 0; for ($i = 0; $i < $n; $i++) { for ($j = 0; $j < $n; $j++) { $bitwise_or = 0; // Traverse sub-array [i..j] for ($k = $i; $k < $j + 1; $k++) $bitwise_or = $bitwise_or | $arr[$k]; if ($bitwise_or >= $K) $count += 1; } } return $count; }// Driver code $arr = array( 3, 4, 5 );$n = count($arr);$k = 6;print(countSubArrays($arr, $n, $k));// This code is contributed by mits?> |
Javascript
<script> // Javascript implementation of the approach // Function to return the count of required sub-arrays function countSubArrays(arr, n, K) { let count = 0; for (let i = 0; i < n; i++) { for (let j = i; j < n; j++) { let bitwise_or = 0; // Traverse sub-array [i..j] for (let k = i; k <= j; k++) { bitwise_or = bitwise_or | arr[k]; } if (bitwise_or >= K) count++; } } return count; } // Driver code let arr = [ 3, 4, 5 ]; let n = arr.length; let k = 6; document.write(countSubArrays(arr, n, k)); // This code is contributed by suresh07.</script> |
Output
2
The time complexity of the above solution is O(n3) and Auxiliary Space is O(1).
An efficient solution uses segment trees to calculate bitwise OR of a sub-array in O(log n) time. Hence, now we directly query the segment tree instead of traversing the sub-array.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define N 100002int tree[4 * N];// Function to build the segment treevoid build(int* arr, int node, int start, int end){ if (start == end) { tree[node] = arr[start]; return; } int mid = (start + end) >> 1; build(arr, 2 * node, start, mid); build(arr, 2 * node + 1, mid + 1, end); tree[node] = tree[2 * node] | tree[2 * node + 1];}// Function to return the bitwise OR of segment [L..R]int query(int node, int start, int end, int l, int r){ if (start > end || start > r || end < l) { return 0; } if (start >= l && end <= r) { return tree[node]; } int mid = (start + end) >> 1; int q1 = query(2 * node, start, mid, l, r); int q2 = query(2 * node + 1, mid + 1, end, l, r); return q1 | q2;}// Function to return the count of required sub-arraysint countSubArrays(int arr[], int n, int K){ // Build segment tree build(arr, 1, 0, n - 1); int count = 0; for (int i = 0; i < n; i++) { for (int j = i; j < n; j++) { // Query segment tree for bitwise OR // of sub-array [i..j] int bitwise_or = query(1, 0, n - 1, i, j); if (bitwise_or >= K) count++; } } return count;}// Driver codeint main(){ int arr[] = { 3, 4, 5 }; int n = sizeof(arr) / sizeof(arr[0]); int k = 6; cout << countSubArrays(arr, n, k); return 0;} |
Java
// Java implementation of the approach public class Main{ static int N = 100002; static int tree[] = new int[4 * N]; // Function to build the segment tree static void build(int arr[], int node, int start, int end) { if (start == end) { tree[node] = arr[start]; return; } int mid = (start + end) >> 1; build(arr, 2 * node, start, mid); build(arr, 2 * node + 1, mid + 1, end); tree[node] = tree[2 * node] | tree[2 * node + 1]; } // Function to return the bitwise OR of segment [L..R] static int query(int node, int start, int end, int l, int r) { if (start > end || start > r || end < l) { return 0; } if (start >= l && end <= r) { return tree[node]; } int mid = (start + end) >> 1; int q1 = query(2 * node, start, mid, l, r); int q2 = query(2 * node + 1, mid + 1, end, l, r); return q1 | q2; } // Function to return the count of required sub-arrays static int countSubArrays(int arr[], int n, int K) { // Build segment tree build(arr, 1, 0, n - 1); int count = 0; for (int i = 0; i < n; i++) { for (int j = i; j < n; j++) { // Query segment tree for bitwise OR // of sub-array [i..j] int bitwise_or = query(1, 0, n - 1, i, j); if (bitwise_or >= K) count++; } } return count; } // Driver code public static void main(String[] args) { int arr[] = { 3, 4, 5 }; int n = arr.length; int k = 6; System.out.print(countSubArrays(arr, n, k)); }}// This code is contributed by divyesh072019 |
Python3
# Python implementation of the approachN = 100002tree = [0]*(4 * N)# Function to build the segment treedef build(arr, node, start, end): if start == end: tree[node] = arr[start] return mid = (start + end) >> 1 build(arr, 2 * node, start, mid) build(arr, 2 * node + 1, mid + 1, end) tree[node] = tree[2 * node] | tree[2 * node + 1]# Function to return the bitwise OR of segment[L..R]def query(node, start, end, l, r): if start > end or start > r or end < l: return 0 if start >= l and end <= r: return tree[node] mid = (start + end) >> 1 q1 = query(2 * node, start, mid, l, r) q2 = query(2 * node + 1, mid + 1, end, l, r) return q1 or q2# Function to return the count of required sub-arraysdef countSubArrays(arr, n, K): # Build segment tree build(arr, 1, 0, n - 1) count = 0 for i in range(n): for j in range(n): # Query segment tree for bitwise OR # of sub-array[i..j] bitwise_or = query(1, 0, n - 1, i, j) if bitwise_or >= K: count += 1 return count# Driver codearr = [3, 4, 5]n = len(arr)k = 6print(countSubArrays(arr, n, k))# This code is contributed by ankush_953 |
C#
// C# implementation of the approach using System;class GFG { static int N = 100002; static int[] tree = new int[4 * N]; // Function to build the segment tree static void build(int[] arr, int node, int start, int end) { if (start == end) { tree[node] = arr[start]; return; } int mid = (start + end) >> 1; build(arr, 2 * node, start, mid); build(arr, 2 * node + 1, mid + 1, end); tree[node] = tree[2 * node] | tree[2 * node + 1]; } // Function to return the bitwise OR of segment [L..R] static int query(int node, int start, int end, int l, int r) { if (start > end || start > r || end < l) { return 0; } if (start >= l && end <= r) { return tree[node]; } int mid = (start + end) >> 1; int q1 = query(2 * node, start, mid, l, r); int q2 = query(2 * node + 1, mid + 1, end, l, r); return q1 | q2; } // Function to return the count of required sub-arrays static int countSubArrays(int[] arr, int n, int K) { // Build segment tree build(arr, 1, 0, n - 1); int count = 0; for (int i = 0; i < n; i++) { for (int j = i; j < n; j++) { // Query segment tree for bitwise OR // of sub-array [i..j] int bitwise_or = query(1, 0, n - 1, i, j); if (bitwise_or >= K) count++; } } return count; } // Driver code static void Main() { int[] arr = { 3, 4, 5 }; int n = arr.Length; int k = 6; Console.WriteLine(countSubArrays(arr, n, k)); }}// This code is contributed by divyeshrabadiya07. |
Javascript
<script>// Javascript implementation of the approachlet N = 100002;let tree = new Array(4 * N);// Function to build the segment tree function build(arr, node, start, end){ if (start == end) { tree[node] = arr[start]; return; } let mid = (start + end) >> 1; build(arr, 2 * node, start, mid); build(arr, 2 * node + 1, mid + 1, end); tree[node] = tree[2 * node] | tree[2 * node + 1];}// Function to return the bitwise OR of segment [L..R]function query(node, start, end, l, r){ if (start > end || start > r || end < l) { return 0; } if (start >= l && end <= r) { return tree[node]; } let mid = (start + end) >> 1; let q1 = query(2 * node, start, mid, l, r); let q2 = query(2 * node + 1, mid + 1, end, l, r); return q1 | q2;}// Function to return the count of // required sub-arraysfunction countSubArrays(arr, n, K){ // Build segment tree build(arr, 1, 0, n - 1); let count = 0; for(let i = 0; i < n; i++) { for(let j = i; j < n; j++) { // Query segment tree for bitwise OR // of sub-array [i..j] let bitwise_or = query(1, 0, n - 1, i, j); if (bitwise_or >= K) count++; } } return count;}// Driver codelet arr = [ 3, 4, 5 ];let n = arr.length;let k = 6;document.write(countSubArrays(arr, n, k));// This code is contributed by rag2127</script> |
Output
2
Complexity Analysis:
- Time complexity: O(n2 log n).
- Auxiliary Space: O(n).
A further efficient solution uses binary search. Bitwise OR is a function that never decreases with the number of inputs. For example:
OR(a, b) ? OR(a, b, c)
OR(a1, a2, a3, …) ? OR(a1, a2, a3, …, b)
By this property, OR(ai, …, aj) <= OR(ai, …, aj, aj+1). Hence, if OR(ai, …, aj) is greater than K then OR(ai, …, aj, aj+1) will also be greater than K. Hence, once we find a subarray [i..j] whose OR is greater than K, we don’t need to check subarrays [i..j+1], [i..j+2], .. and so on, because their OR will also be greater than K. We can add the count of remaining subarrays to the current sum. The first subarray from a particular starting point whose OR is greater than K is found using binary search.
Below is the implementation of the above idea:
C++
// C++ program to implement the above approach#include <bits/stdc++.h>#define N 100002using namespace std;int tree[4 * N];// Function which builds the segment treevoid build(int* arr, int node, int start, int end){ if (start == end) { tree[node] = arr[start]; return; } int mid = (start + end) >> 1; build(arr, 2 * node, start, mid); build(arr, 2 * node + 1, mid + 1, end); tree[node] = tree[2 * node] | tree[2 * node + 1];}// Function that returns bitwise OR of segment [L..R]int query(int node, int start, int end, int l, int r){ if (start > end || start > r || end < l) { return 0; } if (start >= l && end <= r) { return tree[node]; } int mid = (start + end) >> 1; int q1 = query(2 * node, start, mid, l, r); int q2 = query(2 * node + 1, mid + 1, end, l, r); return q1 | q2;}// Function to count requisite number of subarraysint countSubArrays(const int* arr, int n, int K){ int count = 0; for (int i = 0; i < n; i++) { // Check for subarrays starting with index i int low = i, high = n - 1, index = INT_MAX; while (low <= high) { int mid = (low + high) >> 1; // If OR of subarray [i..mid] >= K, // then all subsequent subarrays will have OR >= K // therefore reduce high to mid - 1 // to find the minimal length subarray // [i..mid] having OR >= K if (query(1, 0, n - 1, i, mid) >= K) { index = min(index, mid); high = mid - 1; } else { low = mid + 1; } } // Increase count with number of subarrays // having OR >= K and starting with index i if (index != INT_MAX) { count += n - index; } } return count;}// Driver codeint main(){ int arr[] = { 3, 4, 5 }; int n = sizeof(arr) / sizeof(arr[0]); // Build segment tree. build(arr, 1, 0, n - 1); int k = 6; cout << countSubArrays(arr, n, k); return 0;} |
Java
// Java implementation of the above approachclass GFG { static int N = 100002; static int tree[] = new int[4 * N]; // Function which builds the segment tree static void build(int[] arr, int node, int start, int end) { if (start == end) { tree[node] = arr[start]; return; } int mid = (start + end) >> 1; build(arr, 2 * node, start, mid); build(arr, 2 * node + 1, mid + 1, end); tree[node] = tree[2 * node] | tree[2 * node + 1]; } // Function that returns bitwise // OR of segment [L..R] static int query(int node, int start, int end, int l, int r) { if (start > end || start > r || end < l) { return 0; } if (start >= l && end <= r) { return tree[node]; } int mid = (start + end) >> 1; int q1 = query(2 * node, start, mid, l, r); int q2 = query(2 * node + 1, mid + 1, end, l, r); return q1 | q2; } // Function to count requisite number of subarrays static int countSubArrays(int[] arr, int n, int K) { int count = 0; for (int i = 0; i < n; i++) { // Check for subarrays starting with index i int low = i, high = n - 1, index = Integer.MAX_VALUE; while (low <= high) { int mid = (low + high) >> 1; // If OR of subarray [i..mid] >= K, // then all subsequent subarrays will // have OR >= K therefore reduce // high to mid - 1 to find the // minimal length subarray // [i..mid] having OR >= K if (query(1, 0, n - 1, i, mid) >= K) { index = Math.min(index, mid); high = mid - 1; } else { low = mid + 1; } } // Increase count with number of subarrays // having OR >= K and starting with index i if (index != Integer.MAX_VALUE) { count += n - index; } } return count; } // Driver code public static void main(String[] args) { int arr[] = {3, 4, 5}; int n = arr.length; // Build segment tree. build(arr, 1, 0, n - 1); int k = 6; System.out.println(countSubArrays(arr, n, k)); }}// This code is contributed by Rajput-Ji |
Python3
# Python3 program to implement the above approachN = 100002tree = [0 for i in range(4 * N)];# Function which builds the segment treedef build(arr, node, start, end): if (start == end): tree[node] = arr[start]; return; mid = (start + end) >> 1; build(arr, 2 * node, start, mid); build(arr, 2 * node + 1, mid + 1, end); tree[node] = tree[2 * node] | tree[2 * node + 1];# Function that returns bitwise OR of segment [L..R]def query(node, start, end, l, r): if (start > end or start > r or end < l): return 0; if (start >= l and end <= r): return tree[node]; mid = (start + end) >> 1; q1 = query(2 * node, start, mid, l, r); q2 = query(2 * node + 1, mid + 1, end, l, r); return q1 | q2;# Function to count requisite number of subarraysdef countSubArrays(arr, n, K): count = 0; for i in range(n): # Check for subarrays starting with index i low = i high = n - 1 index = 1000000000 while (low <= high): mid = (low + high) >> 1; # If OR of subarray [i..mid] >= K, # then all subsequent subarrays will have OR >= K # therefore reduce high to mid - 1 # to find the minimal length subarray # [i..mid] having OR >= K if (query(1, 0, n - 1, i, mid) >= K): index = min(index, mid); high = mid - 1; else : low = mid + 1; # Increase count with number of subarrays # having OR >= K and starting with index i if (index != 1000000000): count += n - index; return count;# Driver codeif __name__=='__main__': arr = [ 3, 4, 5 ] n = len(arr) # Build segment tree. build(arr, 1, 0, n - 1); k = 6; print(countSubArrays(arr, n, k)) # This code is contributed by rutvik_56. |
C#
// C# implementation of the above approach using System;class GFG { static int N = 100002; static int []tree = new int[4 * N]; // Function which builds the segment tree static void build(int[] arr, int node, int start, int end) { if (start == end) { tree[node] = arr[start]; return; } int mid = (start + end) >> 1; build(arr, 2 * node, start, mid); build(arr, 2 * node + 1, mid + 1, end); tree[node] = tree[2 * node] | tree[2 * node + 1]; } // Function that returns bitwise // OR of segment [L..R] static int query(int node, int start, int end, int l, int r) { if (start > end || start > r || end < l) { return 0; } if (start >= l && end <= r) { return tree[node]; } int mid = (start + end) >> 1; int q1 = query(2 * node, start, mid, l, r); int q2 = query(2 * node + 1, mid + 1, end, l, r); return q1 | q2; } // Function to count requisite number of subarrays static int countSubArrays(int[] arr, int n, int K) { int count = 0; for (int i = 0; i < n; i++) { // Check for subarrays starting with index i int low = i, high = n - 1, index = int.MaxValue; while (low <= high) { int mid = (low + high) >> 1; // If OR of subarray [i..mid] >= K, // then all subsequent subarrays will // have OR >= K therefore reduce // high to mid - 1 to find the // minimal length subarray // [i..mid] having OR >= K if (query(1, 0, n - 1, i, mid) >= K) { index = Math.Min(index, mid); high = mid - 1; } else { low = mid + 1; } } // Increase count with number of subarrays // having OR >= K and starting with index i if (index != int.MaxValue) { count += n - index; } } return count; } // Driver code public static void Main(String[] args) { int []arr = {3, 4, 5}; int n = arr.Length; // Build segment tree. build(arr, 1, 0, n - 1); int k = 6; Console.WriteLine(countSubArrays(arr, n, k)); } } // This code is contributed by PrinciRaj1992 |
Javascript
<script>// JavaScript implementation of the above approach let N = 100002; let tree=new Array(4*N); // Function which builds the segment tree function build(arr,node,start,end) { if (start == end) { tree[node] = arr[start]; return; } let mid = (start + end) >> 1; build(arr, 2 * node, start, mid); build(arr, 2 * node + 1, mid + 1, end); tree[node] = tree[2 * node] | tree[2 * node + 1]; } // Function that returns bitwise // OR of segment [L..R] function query(node,start,end,l,r) { if (start > end || start > r || end < l) { return 0; } if (start >= l && end <= r) { return tree[node]; } let mid = (start + end) >> 1; let q1 = query(2 * node, start, mid, l, r); let q2 = query(2 * node + 1, mid + 1, end, l, r); return q1 | q2; } // Function to count requisite number of subarrays function countSubArrays(arr,n,K) { let count = 0; for (let i = 0; i < n; i++) { // Check for subarrays starting with index i let low = i, high = n - 1, index = Number.MAX_VALUE; while (low <= high) { let mid = (low + high) >> 1; // If OR of subarray [i..mid] >= K, // then all subsequent subarrays will // have OR >= K therefore reduce // high to mid - 1 to find the // minimal length subarray // [i..mid] having OR >= K if (query(1, 0, n - 1, i, mid) >= K) { index = Math.min(index, mid); high = mid - 1; } else { low = mid + 1; } } // Increase count with number of subarrays // having OR >= K and starting with index i if (index != Number.MAX_VALUE) { count += n - index; } } return count; } // Driver code let arr=[3, 4, 5]; let n = arr.length; // Build segment tree. build(arr, 1, 0, n - 1); let k = 6; document.write(countSubArrays(arr, n, k));// This code is contributed by patel2127</script> |
Output
2
Complexity Analysis:
- Time complexity: O(n log2 n).
- Auxiliary Space: O(n).
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