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To solve the problem given an array, we need to find all possible integers that are at least the bitwise AND of a non-empty subarray, e.g. -
Input : nums[ ] = { 3, 5, 1, 2, 8 } Output : { 2, 5, 0, 3, 8, 1 } Explanation: 2 is the bitwise AND of subarray {2}, 5 is the bitwise AND of subarray {5}, 0 is the bitwise AND of subarray {1, 2}, {2, 8} and {1, 2, 8}, 3 is the bitwise AND of subarray {3}, 8 is the bitwise AND of subarray {8}, 1 is the bitwise AND of subarray {1}, {3, 5} and {3, 5, 1}. Input : nums[ ] = { 2, 6, 3, 8, 1 } Output: { 1, 8, 3, 6, 2, 0 }
The simple method that can be applied is,
Find all possible non-empty subarrays.
When traversing an array, calculate the bitwise AND of each element in the subarray.
To avoid duplicate values, store all results in a collection.
#include <bits/stdc++.h> using namespace std; int main(){ int arr[] ={ 2, 6, 3, 8, 1 }; int n = sizeof(arr) / sizeof(arr[0]); // Declaring set to store result of each AND operation. unordered_set<int> result; int val; // nested loops to traverse through all the possible non empty subarrays. for (int i = 0; i < n; ++i){ for (int j = i, val = INT_MAX; j < n; ++j){ val = val & arr[j]; // storing result of AND operation result.insert(val); } } cout << "All possible numbers are: "; // printing all the values of set. for (auto i = result.begin(); i != result.end();i++) cout << *i << " "; return 0; }
All possible numbers are: 1 8 3 6 0 2
States that set stores all AND operations result.
Initialize the "val" variable using INT_MAX because we need to set all bits to 1 for the AND operation.
The inner loop iterates through all possible subarrays in the i-th index.
Relates the elements to each other and to itself Perform an AND operation and store it in the result set.
Print All
In this tutorial, we discussed a simple way to solve this problem, That is, calculate each possible subarray of the AND operation. We also discussed a C program to solve this problem. Also, you can write this code in any other language like Java, C, Python, etc. We hope you found this tutorial helpful.
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