Given an array of integers, the task is to find the AND of all elements of each subset of the array and print the minimum AND value among all those.
Examples:
Input: arr[] = {1, 2, 3}
Output: 0
AND of all possible subsets
(1 & 2) = 0,
(1 & 3) = 1,
(2 & 3) = 2 and
(1 & 2 & 3) = 0.
Minimum among these is 0.
Input: arr[] = {7, 2}
Output: 2
Approach: The minimum AND value of any subset of the array will be the AND of all the elements of the array. So, the simplest way is to find AND of all elements of sub-array.
Below is the implementation of the above approach:
Implementation:
C++
// C++ program for the above approach #include<bits/stdc++.h>using namespace std;void minAND(int arr[], int n) { int s = arr[0]; // Find AND of whole array for (int i = 1; i < n; i++) { s = s & arr[i]; } // Print the answer cout << (s) << endl; } // Driver code int main() { int arr[] = {1, 2, 3}; int n = sizeof(arr)/sizeof(int); minAND(arr, n); } // This code has been contributed by Arnab Kundu |
Java
// Java program for the above approachclass GFG { static void minAND(int[] arr, int n) { int s = arr[0]; // Find AND of whole array for (int i = 1; i < n; i++) { s = s & arr[i]; } // Print the answer System.out.println(s); } // Driver code public static void main(String[] args) { int[] arr = {1, 2, 3}; int n = arr.length; minAND(arr, n); }}// This code has been contributed by 29AjayKumar |
Python
# Python program for the above approachdef minAND(arr, n): s = arr[0] # Find AND of whole array for i in range(1, n): s = s & arr[i] # Print the answer print(s)# Driver codearr = [1, 2, 3]n = len(arr)minAND(arr, n) |
C#
// C# program for the above approachclass GFG { static void minAND(int[] arr, int n) { int s = arr[0]; // Find AND of whole array for (int i = 1; i < n; i++) { s = s & arr[i]; } // Print the answer System.Console.WriteLine(s); } // Driver code static void Main() { int[] arr = {1, 2, 3}; int n = arr.Length; minAND(arr, n); }}// This code has been contributed by chandan_jnu |
PHP
<?php// PHP program for the above approach function minAND($arr, $n) { $s = $arr[0]; // Find AND of whole array for ($i = 1; $i < $n; $i++) { $s = $s & $arr[$i]; } // Print the answer print($s . "\n"); } // Driver code $arr = array(1, 2, 3); $n = count($arr); minAND($arr, $n); // This code is contributed // by chandan_jnu?> |
Javascript
<script>// Javascript implementation of the approachfunction minAND(arr, n) { let s = arr[0]; // Find AND of whole array for (let i = 1; i < n; i++) { s = s & arr[i]; } // Print the answer document.write((s) + "<br>");}// Driver codelet arr = [1, 2, 3];let n = arr.length;minAND(arr, n);// This code is contributed by _saurabh_jaiswal</script> |
Output:
0
Time Complexity: O(n), where n is the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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