Numbers that are bitwise AND of at least one non-empty sub-array
Given an array ‘arr’, the task is to find all possible integers each of which is the bitwise AND of at least one non-empty sub-array of ‘arr’.
Examples:
Input: arr = {11, 15, 7, 19}
Output: [3, 19, 7, 11, 15]
3 = arr[2] AND arr[3]
19 = arr[3]
7 = arr[2]
11 = arr[0]
15 = arr[1]
Input: arr = {5, 2, 8, 4, 1}
Output: [0, 1, 2, 4, 5, 8]
0 = arr[3] AND arr[4]
1 = arr[4]
2 = arr[1]
4 = arr[3]
5 = arr[0]
8 = arr[2]
Naive Approach:
- An array of size ‘N’ contains N*(N+1)/2 sub-arrays. So, for small ‘N’, iterate over all possible sub-arrays and add each of their AND result to a set.
- Since, sets do not contain duplicates, it’ll store each value only once.
- Finally, print the contents of the set.
Below is the implementation of the above approach:
Java
// Java implementation of the approach import java.util.HashSet; class CP { public static void main(String[] args) { int[] A = { 11, 15, 7, 19 }; int N = A.length; // Set to store all possible AND values. HashSet<Integer> set = new HashSet<>(); int i, j, res; // Starting index of the sub-array. for (i = 0; i < N; ++i) // Ending index of the sub-array. for (j = i, res = Integer.MAX_VALUE; j < N; ++j) { res &= A[j]; // AND value is added to the set. set.add(res); } // The set contains all possible AND values. System.out.println(set); } } |
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C#
// C# implementation of the approach using System; using System.Collections.Generic; class CP { public static void Main(String[] args) { int[] A = {11, 15, 7, 19}; int N = A.Length; // Set to store all possible AND values. HashSet<int> set1 = new HashSet<int>(); int i, j, res; // Starting index of the sub-array. for (i = 0; i < N; ++i) { // Ending index of the sub-array. for (j = i, res = int.MaxValue; j < N; ++j) { res &= A[j]; // AND value is added to the set. set1.Add(res); } } // displaying the values foreach(int m in set1) { Console.Write(m + " "); } } } |
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Output:
[3, 19, 7, 11, 15]
Time Complexity: O(N^2)
Efficient Approach: This problem can be solved efficiently by divide-and-conquer approach.
- Consider each element of the array as a single segment. (‘Divide’ step)
- Add the AND value of all the subarrays to the set.
- Now, for the ‘conquer’ step, keep merging consecutive subarrays and keep adding the additional AND values obtained while merging.
- Continue step 4, until a single segment containing the entire array is obtained.
Below is the implementation of the above approach:
// Java implementation of the approach import java.util.*; public class CP { static int ar[]; static int n; // Holds all possible AND results static HashSet<Integer> allPossibleAND; // driver code public static void main(String[] args) { ar = new int[] { 11, 15, 7, 19 }; n = ar.length; allPossibleAND = new HashSet<>(); // initialization divideThenConquer(0, n - 1); System.out.println(allPossibleAND); } // recursive function which adds all // possible AND results to 'allPossibleAND' static Segment divideThenConquer(int l, int r) { // can't divide into //further segments if (l == r) { allPossibleAND.add(ar[l]); // Therefore, return a segment // containing this single element. Segment ret = new Segment(); ret.leftToRight.add(ar[l]); ret.rightToLeft.add(ar[l]); return ret; } // can be further divided into segments else { Segment left = divideThenConquer(l, (l + r) / 2); Segment right = divideThenConquer((l + r) / 2 + 1, r); // Now, add all possible AND results, // contained in these two segments /* ******************************** This step may seem to be inefficient and time consuming, but it is not. Read the 'Analysis' block below for further clarification. *********************************** */ for (int itr1 : left.rightToLeft) for (int itr2 : right.leftToRight) allPossibleAND.add(itr1 & itr2); // 'conquer' step return mergeSegments(left, right); } } // returns the resulting segment after // merging segments 'a' and 'b' // 'conquer' step static Segment mergeSegments(Segment a, Segment b) { Segment res = new Segment(); // The resulting segment will have // same prefix sequence as segment 'a' res.copyLR(a.leftToRight); // The resulting segment will have // same suffix sequence as segment 'b' res.copyRL(b.rightToLeft); Iterator<Integer> itr; itr = b.leftToRight.iterator(); while (itr.hasNext()) res.addToLR(itr.next()); itr = a.rightToLeft.iterator(); while (itr.hasNext()) res.addToRL(itr.next()); return res; } } class Segment { // These 'vectors' will always // contain atmost 30 values. ArrayList<Integer> leftToRight = new ArrayList<>(); ArrayList<Integer> rightToLeft = new ArrayList<>(); void addToLR(int value) { int lastElement = leftToRight.get(leftToRight.size() - 1); // value decreased after AND-ing with 'value' if ((lastElement & value) < lastElement) leftToRight.add(lastElement & value); } void addToRL(int value) { int lastElement = rightToLeft.get(rightToLeft.size() - 1); // value decreased after AND-ing with 'value' if ((lastElement & value) < lastElement) rightToLeft.add(lastElement & value); } // copies 'lr' to 'leftToRight' void copyLR(ArrayList<Integer> lr) { Iterator<Integer> itr = lr.iterator(); while (itr.hasNext()) leftToRight.add(itr.next()); } // copies 'rl' to 'rightToLeft' void copyRL(ArrayList<Integer> rl) { Iterator<Integer> itr = rl.iterator(); while (itr.hasNext()) rightToLeft.add(itr.next()); } } |
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Output:
[19, 3, 7, 11, 15]
Analysis:
The major optimization in this algorithm is realizing that any array element can yield a maximum of 30 distinct integers (as 30 bits are needed to hold 1e9). Confused?? Let’s proceed step by step.
Let us start a sub-array from the ith element which is A[i]. As the succeeding elements are AND-ed with Ai, the result can either decrease or remain same (because a bit will never get changed from ‘0’ to ‘1’ after an AND operation).
In the worst case, A[i] can be 2^31 – 1 (all 30 bits will be ‘1’). As the elements are AND-ed, atmost 30 distinct values can be obtained because only a single bit might get changed from ‘1’ to ‘0’ in the worst case,
i.e. 111111111111111111111111111111 => 111111111111111111111111101111
So, for each ‘merge’ operation, these distinct values merge to form another set having atmost 30 integers.
Therefore, Worst Case Time Complexity for each merge can be O(30 * 30) = O(900).
Time Complexity: O(900*N*logN).
PS: The time complexity can seem too high, but in practice, the actual complexity lies around O(K*N*logN), where, K is much less than 900. This is because, the lengths of the ‘prefix’ and ‘suffix’ arrays are much less when ‘l’ and ‘r’ are quite close.
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Improved By : Kirti_Mangal