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 = 7Input : arr[] = {2, 6, 2, 8, 4, 5}
Output : 3
Explanation: 15 is the maximum value of OR and set elements are 8, 6, 5
Source: Sprinklr on Campus Internship
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.
- Find bitwise OR of all array elements. This is the OR we are looking for.
- 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 :
- We generate all subsets and return the smallest size with the given OR
- 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 <bits/stdc++.h>using namespace std;// Compute bitwise or of all elements // in array of size szint 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 orint 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<<sz);mask++) { // stores the size of the current subset int curSZ=0; // stores the OR of all the elements // in the current subset int curOR=0; for(int i=0;i<sz;i++) { if(mask&(1<<i)) { curSZ++; curOR|=data[i]; } } if(curOR==maxOR) minSZ=min(minSZ,curSZ); } return minSZ;}// Driver codeint 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
// Java Program for above approachimport 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 szdef OR(data, sz): mOR = 0 for i in range(sz) : mOR |= data[i] return mOR# calculate the size of # minimum subset with maximum ordef minSubset(data, sz,maxOR): # store the minimum size of # the subset with maximum OR minSZ=sz # generates all subsets for mask in range(1<<sz): # stores the size of the current subset curSZ=0 # stores the OR of all the elements # in the current subset curOR=0 for i in range(sz): if(mask&(1<<i)): curSZ+=1 curOR|=data[i] if(curOR==maxOR): minSZ=min(minSZ,curSZ) return minSZ# Driver codeif __name__ == '__main__': data =[5, 1, 3, 4, 2] sz = len(data) maxOR = OR(data, sz) # Function Call print(minSubset(data, sz,maxOR)) |
C#
// C# Program for above approachusing 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
<script>// JavaScript Program for above approach// Compute bitwise or of all elements // in array of size szfunction OR(arr){ let mOR = 0; for (let i = 0; i < arr.length; ++i) { mOR |= arr[i]; } return mOR;}// Recursively calculating the size of // minimum subset with maximum orfunction maxSubset(arr,i,curOr,curSize,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 let take = maxSubset(arr, i + 1, curOr | arr[i], curSize + 1, maxOr); // Skip the current element let notTake = maxSubset(arr, i + 1, curOr, curSize, maxOr); // Return minimum of take and notTake return Math.min(take, notTake);}// Driver Codelet data=[5, 1, 3, 4, 2];let maxOr = OR(data);// Function Calllet maxSubsetSize = maxSubset(data, 0, 0, 0, maxOr);document.write(maxSubsetSize); // This code is contributed by rag2127</script> |
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 <bits/stdc++.h>using namespace std;// Compute bitwise or of all elements// in array of size szint 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 orint minSubset(int data[], int sz, int maxOR){ // count table where // count[i][j] => minimum size subset till ith element // whose OR is j vector<vector<int> > 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 codeint 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<sz+1;i++){ Arrays.fill(count[i],1000000000); } 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] != 1000000000) { count[i + 1][j | data[i]] = Math.min( count[i + 1][j | data[i]], count[i][j] + 1); } } } return count[sz][maxOR]; } /* Driver program to test above function*/ public static void main(String args[]) { int data[] = { 5, 1, 3, 4, 2 }; int sz = data.length; int maxOR = OR(data, sz); // Function Call System.out.println(minSubset(data, sz, maxOR)); }}// This code is contributed by shinjanpatra. |
Python3
# Python3 Code for above approach# Compute bitwise or of all elements# in array of size szdef OR(data, sz): mOR = 0 for i in range(sz): mOR |= data[i] return mOR# calculate the size of# minimum subset with maximum ordef 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 codeif __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
<script>// JavaScript Code for above approach// Compute bitwise or of all elements// in array of size szfunction OR(data, sz){ let mOR = 0 for(let i=0;i<sz;i++) mOR |= data[i] return mOR}// calculate the size of// minimum subset with maximum orfunction minSubset(data, sz, maxOR){ // count table where // count[i][j] => minimum size subset till ith element // whose OR is j let count = new Array(sz+1).fill(0).map(()=>new Array(maxOR+1).fill(1e9)); count[0][0] = 0 for(let i=0;i<sz;i++){ for(let 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 codelet data = [5, 1, 3, 4, 2]let sz = data.lengthlet maxOR = OR(data, sz)// Function Calldocument.write(minSubset(data, sz, maxOR),"</br>")// This code is contributed by shinjanpatra</script> |
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 <bits/stdc++.h>using namespace std;// Compute bitwise or of all elements// in array of size szint 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 orint 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 codeint 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
import java.util.Arrays;public class MinimumSubsetMaxOR { // Compute bitwise OR of all elements in the array static int OR(int[] data) { int mOR = 0; for (int i : data) { mOR |= i; } return mOR; } // Calculate the size of the minimum subset with maximum OR static int minSubset(int[] data, int maxOR) { int sz = data.length; // Create a table where count[i][j] represents the minimum size of the subset // till the ith element with an OR equal to j int[][] count = new int[sz + 1][maxOR + 1]; for (int i = 0; i <= sz; i++) { Arrays.fill(count[i], Integer.MAX_VALUE); } count[0][0] = 0; for (int i = 1; i <= sz; i++) { for (int j = 0; j <= maxOR; j++) { // Do not consider the ith element count[i][j] = Math.min(count[i][j], count[i - 1][j]); // Consider the ith element if (count[i - 1][j] != Integer.MAX_VALUE) { count[i][j | data[i - 1]] = Math.min(count[i][j | data[i - 1]], count[i - 1][j] + 1); } } } return count[sz][maxOR]; } // Driver code public static void main(String[] args) { int[] data = { 5, 1, 3, 4, 2 }; int maxOR = OR(data); // Function Call System.out.println(minSubset(data, maxOR)); }} |
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 mORdef 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 codeif __name__ == "__main__": data = [5, 1, 3, 4, 2] sz = len(data) maxOR = OR(data, sz) # Function call print(minSubset(data, sz, maxOR)) |
Javascript
// Function to compute bitwise OR of all elements in the arrayfunction OR(data) { let mOR = 0; for (let i = 0; i < data.length; ++i) { mOR |= data[i]; } return mOR;}// Function to calculate the minimum size of subset with maximum ORfunction minSubset(data, maxOR) { // Create a vector 'count' where count[i] represents the minimum size of subset // till ith element whose OR is i. let count = new Array(maxOR + 1).fill(Infinity); count[0] = 0; for (let i = 0; i < data.length; i++) { // Create a temporary vector to hold the updated count for the current element let temp = new Array(maxOR + 1).fill(Infinity); for (let j = 0; j <= maxOR; j++) { // Do not consider the current element (data[i]) in the subset. temp[j] = Math.min(temp[j], count[j]); // Consider the current element (data[i]) in the subset and update the count. if (count[j] !== Infinity) { temp[j | data[i]] = Math.min(temp[j | data[i]], count[j] + 1); } } // Update the 'count' vector for the next iteration. count = temp; } // Return the minimum size of the subset with the maximum OR (at index maxOR). return count[maxOR];}// Driver codeconst data = [5, 1, 3, 4, 2];const maxOR = OR(data);// Function Callconsole.log(minSubset(data, 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 <bits/stdc++.h>using namespace std;// Compute bitwise or of all elements in array of size szint OR(int data[], int sz){ int mOR = 0; for (int i = 0; i < sz; ++i) { mOR |= data[i]; } return mOR;}// Recursive functionvoid 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 orint 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 codeint 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
public class MinSubsetMaxOr { // 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; } // Recursive function static void minSubsetUtil(int data[], int sz, int ORval, int currOR, int subsetSize, int[] minSize) { if (currOR == ORval) { minSize[0] = Math.min(minSize[0], subsetSize); return; } if (sz == 0) { return; } // Backtracking step: minSubsetUtil(data, sz - 1, ORval, currOR, subsetSize, minSize); minSubsetUtil(data, sz - 1, ORval, currOR | data[sz - 1], subsetSize + 1, minSize); } // Calculate the size of minimum subset with maximum OR static int minSubset(int data[], int sz, int maxOR) { int[] minSize = { Integer.MAX_VALUE }; int currOR = 0; int subsetSize = 0; minSubsetUtil(data, sz, maxOR, currOR, subsetSize, minSize); return minSize[0]; } public static void main(String[] args) { int data[] = { 5, 1, 3, 4, 2 }; int sz = data.length; int maxOR = OR(data, sz); // Function Call System.out.println(minSubset(data, sz, maxOR)); }} |
C#
using System;class Program{ // Function to compute bitwise OR of all elements in an array static int OR(int[] data) { int mOR = 0; foreach (int element in data) { mOR |= element; } return mOR; } // Recursive function to calculate the minimum subset size static void MinSubsetUtil(int[] data, int sz, int ORval, int currOR, int subsetSize, ref int minSize) { if (currOR == ORval) { minSize = Math.Min(minSize, subsetSize); return; } if (sz == 0) { return; } // Backtracking step: int[] subArray = new int[sz - 1]; Array.Copy(data, 1, subArray, 0, sz - 1); MinSubsetUtil(subArray, sz - 1, ORval, currOR, subsetSize, ref minSize); MinSubsetUtil(subArray, sz - 1, ORval, currOR | data[0], subsetSize + 1, ref minSize); } // Function to calculate the size of the minimum subset with maximum OR static int MinSubset(int[] data, int maxOR) { int minSize = int.MaxValue; int currOR = 0; int subsetSize = 0; MinSubsetUtil(data, data.Length, maxOR, currOR, subsetSize, ref minSize); return minSize; } // Driver code static void Main() { int[] data = { 5, 1, 3, 4, 2 }; int maxOR = OR(data); // Function Call Console.WriteLine(MinSubset(data, maxOR)); }} |
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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