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# Size of the smallest subset with maximum Bitwise OR

Given an array of positive integers. The task is to find the size of the smallest subset such that the Bitwise OR of that set is Maximum possible.

Examples

Input : arr[] = {5, 1, 3, 4, 2}
Output : 2
Explanation: 7 is the maximum value possible of OR, 5|2 = 7 and 5|3 = 7

Input : arr[] = {2, 6, 2, 8, 4, 5}
Output : 3
Explanation: 15 is the maximum value of OR and set elements are 8, 6, 5

Doing bitwise OR of a number with some value does not decrease its value. It either keeps the value the same or increases. If we take a closer look at the problem we can notice that the maximum OR value that we can get is by doing bitwise OR of all array elements. But this includes all elements and here want to know the smallest subset. So we do the following.

1. Find bitwise OR of all array elements. This is the OR we are looking for.
2. Now we need to find the smallest subset with this bitwise OR. This problem is similar to the subset-sum problem, We can solve it in two ways :
1. We generate all subsets and return the smallest size with the given OR
2. We use Dynamic Programming to solve the problem. This solution is going to be very similar to Maximum size subset with given sum

The time complexity of the 1st solution is O(2n) and the time complexity of the Dynamic Programming solution is O(OR * n) where OR is OR of all array elements and n is the size of the input array.

Using Method 1 : (generating all subsets and returning the smallest size with the given OR)

Implementation:

## C++

 `// CPP Code for above approach` `#include ` `using` `namespace` `std;`   `// Compute bitwise or of all elements ` `// in array of size sz` `int` `OR(``int` `data[], ``int` `sz)` `{` `    ``int` `mOR = 0;` `      ``for` `(``int` `i = 0; i < sz; ++i) {` `        ``mOR |= data[i];` `    ``}` `  `  `      ``return` `mOR;` `}`   `// calculate the size of ` `// minimum subset with maximum or` `int` `minSubset(``int` `data[], ``int` `sz,``int` `maxOR)` `{` `  ``// store the minimum size of` `  ``// the subset with maximum OR` `      ``int` `minSZ=sz;` `  `  `      ``// generates all subsets` `    ``for``(``int` `mask=0;mask<(1<

## Java

 `// Java Program for above approach` `import` `java.io.*;` `import` `java.util.*;`   `class` `Solution ` `{` `  `  `  ``// Compute bitwise or of all elements ` `  ``// in array of size sz` `  ``private` `static` `int` `OR(``int``[] arr) ` `  ``{` `    ``int` `mOR = ``0``;` `    ``for` `(``int` `i = ``0``; i < arr.length; ++i) ` `    ``{` `      ``mOR |= arr[i];` `    ``}` `    ``return` `mOR;` `  ``}` `  `  `  ``// Recursively calculating the size of ` `  ``// minimum subset with maximum or` `  ``private` `static` `int` `maxSubset(``int``[] arr, ``int` `i, ` `                ``int` `curOr, ``int` `curSize, ``int` `maxOr) ` `  ``{` `      `  `    ``// If i is arr.length` `    ``if` `(i == arr.length) ` `    ``{` `      `  `      ``// If curOr is equal to maxOr` `      ``if` `(curOr == maxOr) ` `      ``{` `          ``return` `curSize;` `      ``} ` `      `  `      ``// Return arr.length` `      ``else` `      ``{` `          ``return` `arr.length;` `      ``}` `    ``}` `    `  `    ``// Try the current element in the subset` `    ``int` `take = maxSubset(arr, i + ``1``, curOr | ` `                          ``arr[i], curSize + ``1``, maxOr);` `    `  `    ``// Skip the current element` `    ``int` `notTake = maxSubset(arr, i + ``1``, curOr, ` `                                      ``curSize, maxOr);` `    `  `    `  `    ``// Return minimum of take and notTake` `    ``return` `Math.min(take, notTake);` `  ``}` `  `  `  ``// Driver Code` `  ``public` `static` `void` `main(String[] args) ` `  ``{` `    ``int``[] data = {``5``, ``1``, ``3``, ``4``, ``2``};` `    `  `    ``int` `maxOr = OR(data);` `    `  `    ``// Function Call` `    ``int` `maxSubsetSize = maxSubset(data, ``0``, ``0``, ``0``, maxOr);` `    ``System.out.println(maxSubsetSize);` `  ``}` `}`   `// Code contributed by Abdelaziz EROUI`

## Python3

 `# Python3 Code for above approach`     `# Compute bitwise or of all elements ` `# in array of size sz` `def` `OR(data, sz):` `    ``mOR ``=` `0` `    ``for` `i ``in` `range``(sz) :` `        ``mOR |``=` `data[i]` `    `  `  `  `    ``return` `mOR`     `# calculate the size of ` `# minimum subset with maximum or` `def` `minSubset(data, sz,maxOR):` `  ``# store the minimum size of` `  ``# the subset with maximum OR` `    ``minSZ``=``sz` `  `  `    ``# generates all subsets` `    ``for` `mask ``in` `range``(``1``<

## C#

 `// C# Program for above approach` `using` `System;` `class` `Solution ` `{` `  `  `  ``// Compute bitwise or of all elements ` `  ``// in array of size sz` `  ``private` `static` `int` `OR(``int``[] arr) ` `  ``{` `    ``int` `mOR = 0;` `    ``for` `(``int` `i = 0; i < arr.Length; ++i) ` `    ``{` `      ``mOR |= arr[i];` `    ``}` `    ``return` `mOR;` `  ``}` `  `  `  ``// Recursively calculating the size of ` `  ``// minimum subset with maximum or` `  ``private` `static` `int` `maxSubset(``int``[] arr, ``int` `i, ` `                ``int` `curOr, ``int` `curSize, ``int` `maxOr) ` `  ``{` `      `  `    ``// If i is arr.length` `    ``if` `(i == arr.Length) ` `    ``{` `      `  `      ``// If curOr is equal to maxOr` `      ``if` `(curOr == maxOr) ` `      ``{` `          ``return` `curSize;` `      ``} ` `      `  `      ``// Return arr.length` `      ``else` `      ``{` `          ``return` `arr.Length;` `      ``}` `    ``}` `    `  `    ``// Try the current element in the subset` `    ``int` `take = maxSubset(arr, i + 1, curOr | ` `                          ``arr[i], curSize + 1, maxOr);` `    `  `    ``// Skip the current element` `    ``int` `notTake = maxSubset(arr, i + 1, curOr, ` `                                      ``curSize, maxOr);` `    `  `    `  `    ``// Return minimum of take and notTake` `    ``return` `Math.Min(take, notTake);` `  ``}` `  `  `  ``// Driver Code` `  ``static` `void` `Main()` `  ``{` `    ``int``[] data = {5, 1, 3, 4, 2};` `    `  `    ``int` `maxOr = OR(data);` `    `  `    ``// Function Call` `    ``int` `maxSubsetSize = maxSubset(data, 0, 0, 0, maxOr);` `     ``Console.WriteLine(maxSubsetSize);` `  ``}` `}`   `// This code is contributed by SoumikMondal`

## Javascript

 ``

Output

```2
```

Complexity Analysis:

• Time complexity: O(2n)
• Auxiliary Space: O(n)

Using Method 2:

We first find the OR of all elements of given array.Now we need to find the smallest subset with this bitwise OR.

To do so, use the similar DP approach as given in the subset sum problem. count[i][j] denotes the minimum size subset till ith element whose OR is j.

Implementation:

## C++

 `// CPP Code for above approach` `#include ` `using` `namespace` `std;`   `// Compute bitwise or of all elements` `// in array of size sz` `int` `OR(``int` `data[], ``int` `sz)` `{` `    ``int` `mOR = 0;` `    ``for` `(``int` `i = 0; i < sz; ++i) {` `        ``mOR |= data[i];` `    ``}`   `    ``return` `mOR;` `}`   `// calculate the size of` `// minimum subset with maximum or` `int` `minSubset(``int` `data[], ``int` `sz, ``int` `maxOR)` `{` `    ``// count table where` `      ``// count[i][j] => minimum size subset till ith element` `      ``// whose OR is j` `    ``vector > count(sz + 1, vector<``int``>(maxOR + 1, 1e9));` `  `  `    ``count[0][0] = 0;`   `    ``for` `(``int` `i = 0; i < sz; i++) {` `        ``for` `(``int` `j = 0; j <= maxOR; j++) {` `            ``// Do not consider ith element.` `            ``count[i + 1][j] = min(count[i + 1][j], count[i][j]);`   `            ``// Consider the ith element.` `            ``if` `(count[i][j] != 1e9) {` `                ``count[i + 1][j | data[i]] = min(` `                    ``count[i + 1][j | data[i]], count[i][j] + 1);` `            ``}` `        ``}` `    ``}`   `    ``return` `count[sz][maxOR];` `}`   `// Driver code` `int` `main()` `{` `    ``int` `data[] = { 5, 1, 3, 4, 2 };` `    ``int` `sz = ``sizeof``(data) / ``sizeof``(0);` `    ``int` `maxOR = OR(data, sz);`   `    ``// Function Call` `    ``cout << minSubset(data, sz, maxOR) << ``'\n'``;` `}`

## Java

 `/*package whatever //do not write package name here */`   `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG {` `  ``// Java Code for above approach`   `  ``// Compute bitwise or of all elements` `  ``// in array of size sz` `  ``static` `int` `OR(``int` `data[], ``int` `sz)` `  ``{` `    ``int` `mOR = ``0``;` `    ``for` `(``int` `i = ``0``; i < sz; ++i) {` `      ``mOR |= data[i];` `    ``}`   `    ``return` `mOR;` `  ``}`   `  ``// calculate the size of` `  ``// minimum subset with maximum or` `  ``static` `int` `minSubset(``int` `data[], ``int` `sz, ``int` `maxOR)` `  ``{` `    ``// count table where` `    ``// count[i][j] => minimum size subset till ith element` `    ``// whose OR is j` `    ``int` `count[][] = ``new` `int``[sz + ``1``][maxOR + ``1``];` `    ``for``(``int` `i=``0``;i

## Python3

 `# Python3 Code for above approach`     `# Compute bitwise or of all elements` `# in array of size sz` `def` `OR(data, sz):` `    ``mOR ``=` `0` `    ``for` `i ``in` `range``(sz):` `        ``mOR |``=` `data[i]` `    ``return` `mOR`     `# calculate the size of` `# minimum subset with maximum or` `def` `minSubset(data, sz, maxOR):` `    ``# count table where` `      ``# count[i][j] => minimum size subset till ith element` `      ``# whose OR is j` `    ``count``=``[[``1e9` `for` `_ ``in` `range``(maxOR``+``1``)]``for` `_ ``in` `range``(sz``+``1``)]` `  `  `    ``count[``0``][``0``] ``=` `0`   `    ``for` `i ``in` `range``(sz) :` `        ``for` `j ``in` `range``(maxOR) :` `            ``# Do not consider ith element.` `            ``count[i ``+` `1``][j] ``=` `min``(count[i ``+` `1``][j], count[i][j])`   `            ``# Consider the ith element.` `            ``if` `(count[i][j] !``=` `1e9``) :` `                ``count[i ``+` `1``][j | data[i]] ``=` `min``(` `                    ``count[i ``+` `1``][j | data[i]], count[i][j] ``+` `1``)` `            `  `        `  `    `    `    ``return` `count[sz][maxOR]`     `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `    ``data ``=` `[``5``, ``1``, ``3``, ``4``, ``2``] ` `    ``sz ``=` `len``(data)` `    ``maxOR ``=` `OR(data, sz)`   `    ``# Function Call` `    ``print``(minSubset(data, sz, maxOR))`

## C#

 `using` `System;` `class` `GFG {` `    ``// Compute bitwise or of all elements` `    ``// in array of size sz` `    ``static` `int` `OR(``int``[] data, ``int` `sz)` `    ``{` `        ``int` `mOR = 0;` `        ``for` `(``int` `i = 0; i < sz; ++i) {` `            ``mOR |= data[i];` `        ``}`   `        ``return` `mOR;` `    ``}`   `    ``// calculate the size of` `    ``// minimum subset with maximum or` `    ``static` `int` `minSubset(``int``[] data, ``int` `sz, ``int` `maxOR)` `    ``{` `        ``// count table where` `        ``// count[i][j] => minimum size subset till ith` `        ``// element whose OR is j` `        ``int``[, ] count = ``new` `int``[sz + 1, maxOR + 1];` `        ``for` `(``int` `i = 0; i <= sz; i++) {` `            ``for` `(``int` `j = 0; j <= maxOR; j++) {` `                ``count[i, j] = (``int``)1e9;` `            ``}` `        ``}` `        ``count[0, 0] = 0;`   `        ``for` `(``int` `i = 0; i < sz; i++) {` `            ``for` `(``int` `j = 0; j <= maxOR; j++) {` `                ``// Do not consider ith element.` `                ``count[i + 1, j] = Math.Min(count[i + 1, j],` `                                           ``count[i, j]);`   `                ``// Consider the ith element.` `                ``if` `(count[i, j] != 1e9) {` `                    ``count[i + 1, j | data[i]] = Math.Min(` `                        ``count[i + 1, j | data[i]],` `                        ``count[i, j] + 1);` `                ``}` `            ``}` `        ``}`   `        ``return` `count[sz, maxOR];` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int``[] data = { 5, 1, 3, 4, 2 };` `        ``int` `sz = 5;` `        ``int` `maxOR = OR(data, sz);`   `        ``// Function Call` `        ``Console.WriteLine(minSubset(data, sz, maxOR));` `    ``}` `}`

## Javascript

 ``

Output

```2
```

Complexity Analysis:

• Time complexity: O(n*maxOR) where n is the size of the array and maxOR is the maximum or that can be obtained.
• Auxiliary Space: O(n*maxOR)

Efficient approach : Space optimization

In previous approach the current value count[i][j] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 1D array to store the computations.

Implementation steps:

• Create a 1D vector count of size maxOR+1 and initialize it with 0.
• Set a base case by initializing the values of count.
• Now iterate over subproblems by the help of nested loop and get the current value from previous computations.
• Now Create a temporary 1d vector temp used to store the current values from previous computations.
• After every iteration assign the value of temp to count for further iteration.
• At last return and print the final answer stored in count[maxOR].

Implementation:

## C++

 `// CPP Code for above approach` `#include ` `using` `namespace` `std;`   `// Compute bitwise or of all elements` `// in array of size sz` `int` `OR(``int` `data[], ``int` `sz)` `{` `    ``int` `mOR = 0;` `    ``for` `(``int` `i = 0; i < sz; ++i) {` `        ``mOR |= data[i];` `    ``}`   `    ``return` `mOR;` `}`   `// calculate the size of` `// minimum subset with maximum or` `int` `minSubset(``int` `data[], ``int` `sz, ``int` `maxOR)` `{` `    ``// count table where` `    ``// count[i][j] => minimum size subset till ith element` `    ``// whose OR is j` `    ``vector<``int``> count(maxOR + 1, 1e9);`   `    ``count[0] = 0;`   `    ``for` `(``int` `i = 0; i < sz; i++) {` `        ``vector<``int``> temp(maxOR + 1, 1e9);`   `        ``for` `(``int` `j = 0; j <= maxOR; j++) {` `            ``// Do not consider ith element.` `            ``temp[j] = min(temp[j], count[j]);`   `            ``// Consider the ith element.` `            ``if` `(count[j] != 1e9) {` `                ``temp[j | data[i]] = min(` `                    ``temp[j | data[i]], count[j] + 1);` `            ``}` `        ``}`   `        ``count = temp;` `    ``}`   `    ``return` `count[maxOR];` `}`   `// Driver code` `int` `main()` `{` `    ``int` `data[] = { 5, 1, 3, 4, 2 };` `    ``int` `sz = ``sizeof``(data) / ``sizeof``(0);` `    ``int` `maxOR = OR(data, sz);`   `    ``// Function Call` `    ``cout << minSubset(data, sz, maxOR) << ``'\n'``;` `}`

## Python

 `def` `OR(data, sz):` `    ``"""` `    ``Compute bitwise OR of all elements in an array of size sz.` `    ``"""` `    ``mOR ``=` `0` `    ``for` `i ``in` `range``(sz):` `        ``mOR |``=` `data[i]` `    ``return` `mOR`   `def` `minSubset(data, sz, maxOR):` `    ``"""` `    ``Calculate the size of the minimum subset with maximum OR.` `    ``"""` `    ``# count table where count[i][j] is the minimum size subset till the ith` `    ``# element whose OR is j` `    ``count ``=` `[``float``(``'inf'``)] ``*` `(maxOR ``+` `1``)` `    ``count[``0``] ``=` `0`   `    ``for` `i ``in` `range``(sz):` `        ``temp ``=` `[``float``(``'inf'``)] ``*` `(maxOR ``+` `1``)`   `        ``for` `j ``in` `range``(maxOR ``+` `1``):` `            ``# Do not consider the ith element.` `            ``temp[j] ``=` `min``(temp[j], count[j])`   `            ``# Consider the ith element.` `            ``if` `count[j] !``=` `float``(``'inf'``):` `                ``temp[j | data[i]] ``=` `min``(temp[j | data[i]], count[j] ``+` `1``)`   `        ``count ``=` `temp`   `    ``return` `count[maxOR]`   `# Driver code` `if` `__name__ ``=``=` `"__main__"``:` `    ``data ``=` `[``5``, ``1``, ``3``, ``4``, ``2``]` `    ``sz ``=` `len``(data)` `    ``maxOR ``=` `OR(data, sz)`   `    ``# Function call` `    ``print``(minSubset(data, sz, maxOR))`

Output

```2
```

Time complexity: O(n*maxOR) where n is the size of the array and maxOR is the maximum or that can be obtained.
Auxiliary Space: O(maxOR)

Using method 3: Backtracking Approach

Implementation steps:

• Compute the bitwise or of all elements in array of size.
• Write the recursive function to find the size of the smallest element
• such that bitwise OR of that set is maximum possible.
• If the current OR value is already equal to the maximum OR value, update the minimum subset size and return.
• If we have already considered all the elements in the array, return without updating the minimum subset size.
• Write the Backtracking step.
• remove the current element from the subset and recurse.
• include the current element in the subset and recurse.
• Calculate the size of minimum subset with maximum or.
• Recursive function to find the size of the smallest subset such that the Bitwise OR of that set is Maximum possible.
• Last return and print.

Code implementation for above approach:

## C++

 `#include ` `using` `namespace` `std;`   `// Compute bitwise or of all elements in array of size sz` `int` `OR(``int` `data[], ``int` `sz)` `{` `    ``int` `mOR = 0;` `    ``for` `(``int` `i = 0; i < sz; ++i) {` `        ``mOR |= data[i];` `    ``}`   `    ``return` `mOR;` `}`   `// Recursive function` `void` `minSubsetUtil(``int` `data[], ``int` `sz, ``int` `ORval,` `                   ``int` `currOR, ``int` `subsetSize, ``int``& minSize)` `{` `    ``if` `(currOR == ORval) {` `        ``minSize = min(minSize, subsetSize);` `        ``return``;` `    ``}`   `    ``if` `(sz == 0) {` `        ``return``;` `    ``}`   `    ``// Backtracking step:` `    ``minSubsetUtil(data + 1, sz - 1, ORval, currOR,` `                  ``subsetSize, minSize);`   `    ``minSubsetUtil(data + 1, sz - 1, ORval, currOR | data[0],` `                  ``subsetSize + 1, minSize);` `}`   `// Calculate the size of minimum subset with maximum or` `int` `minSubset(``int` `data[], ``int` `sz, ``int` `maxOR)` `{` `    ``int` `minSize = INT_MAX;` `    ``int` `currOR = 0;` `    ``int` `subsetSize = 0;`   `    ``minSubsetUtil(data, sz, maxOR, currOR, subsetSize,` `                  ``minSize);`   `    ``return` `minSize;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `data[] = { 5, 1, 3, 4, 2 };` `    ``int` `sz = ``sizeof``(data) / ``sizeof``(0);` `    ``int` `maxOR = OR(data, sz);`   `    ``// Function Call` `    ``cout << minSubset(data, sz, maxOR) << ``'\n'``;` `}`

Output

```2
```

Time Complexity: O(2^N), where N is the size of the input array.
Auxiliary Space: O(N), where N is the size of the input array.

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