Skip to content
Related Articles

Related Articles

Minimum integer with at most K bits set such that their bitwise AND with N is maximum
  • Last Updated : 08 Nov, 2020

Given an integer N which can be represented in 32 bit, the task is to find another integer X that has at most K bits set in its binary representation and bitwise AND of X and N is maximum.

Examples: 

Input: N = 5, K = 1 
Output: X = 2 
Explanation: 
Binary representation of 5 is 101, the possible value of X such that it maximize the AND is 2 
5 -> 101 
2 -> 100 
AND sum -> 100 (2) 
4 is the maximum AND sum we can get with N and X.

Input: N = 10, K = 2 
Output: X = 10 
Explanation: 
Binary representation of 10 is 1010, X = 10 possible integer to given maximum AND sum with N with atmost 2 bit set. 
10 -> 1010 
10 -> 1010 
AND sum -> 1010 (10) 
10 is the maximum AND sum we can get with N and X. 
 

 

Naive Approach: A naive solution is to run a loop from 1 to N and take bitwise AND with N, and also check if the number of set bits is less than or equal to K. During each iteration maintain the maximum value of bitwise AND. 
Time Complexity: O(N)

Efficient Approach: This problem can be solved efficiently by using a Greedy approach on bits. Since N can have at most 32 bits. So we will start traversing from the Most Significant bit and check if it is set(or 1) in N, if it is not set then there is no need to set this bit because AND operation will make it zero in the final answer, else we will set it in our required answer. Also, we will maintain the count of set bits in each iteration and check if it does not exceed K. 
Time Complexity: O(32)

Below is the implementation of the above efficient approach :



C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program for the
// above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find
// the integer with
// maximum bitwise with N
int find_max(int n, int k)
{
  // Store answer in the bitset
  // Initialized with 0
  bitset<32> X(0);
 
  // To maintain the count
  // of set bits that should
  // exceed k
  int cnt = 0;
 
  // Start traversing from the
  // Most significantif that bit
  // is set in n then we will set
  // in our answer i.e in X
  for (int i = 31; i >= 0 &&
           cnt != k; i--)
  {
    // Checking if the ith bit
    // is set in n or not
    if (n & (1 << i))
    {
      X[i] = 1;
      cnt++;
    }
  }
 
  // Converting into integer
  return X.to_ulong();
}
 
// Driver Code
int main()
{
  int n = 10, k = 2;
 
  // Function Call
  cout << find_max(n, k) <<
          endl;
 
  return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program for the
// above approach
import java.util.*;
import java.lang.*;
class GFG{
 
// Function to find
// the integer with
// maximum bitwise with N
static int find_max(int n,
                    int k)
{
  // Store answer in the
  // bitset Initialized
  // with 0
  int[] X = new int[32];
 
  // To maintain the count
  // of set bits that should
  // exceed k
  int cnt = 0;
 
  // Start traversing from the
  // Most significant if that bit
  // is set in n then we will set
  // in our answer i.e in X
  for (int i = 31; i >= 0 &&
           cnt != k; i--)
  {
    // Checking if the ith bit
    // is set in n or not
    if ((n & (1 << i)) != 0)
    {
      X[i] = 1;
      cnt++;
    }
  }
 
  String s = "";
  for(int i = 31; i >= 0; i--)
    s += X[i] == 0 ? '0' : '1';
 
  // Converting into integer
  return Integer.parseInt(s,2);
}
 
// Driver function
public static void main (String[] args)
{
  int n = 10, k = 2;
 
  // Function Call
  System.out.println(find_max(n, k));
}
}
 
// This code is contributed by offbeat

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program for the above approach
 
# Function to find the integer with
# maximum bitwise with N
def find_max(n, k):
     
    # Store answer in the
    # bitset Initialized
    # with 0
    X = [0] * 32
     
    # To maintain the count
    # of set bits that should
    # exceed k
    cnt = 0
     
    # Start traversing from the
    # Most significant if that bit
    # is set in n then we will set
    # in our answer i.e in X
    i = 31
     
    while (i >= 0 and cnt != k):
         
        # Checking if the ith bit
        # is set in n or not
        if ((n & (1 << i)) != 0):
            X[i] = 1
            cnt += 1
             
        i -= 1
     
    s = ""
     
    for i in range(31, -1, -1):
        if X[i] == 0:
            s += '0'
        else:
            s += '1'
     
    # Converting into integer
    return int(s, 2)
     
# Driver code
n, k = 10, 2
 
# Function Call
print(find_max(n, k))
 
# This code is contributed by divyeshrabadiya07

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program for the above approach
using System;
 
class GFG{
     
// Function to find the integer with
// maximum bitwise with N
static int find_max(int n, int k)
{
     
    // Store answer in the
    // bitset Initialized
    // with 0
    int[] X = new int[32];
     
    // To maintain the count
    // of set bits that should
    // exceed k
    int cnt = 0;
     
    // Start traversing from the
    // Most significant if that bit
    // is set in n then we will set
    // in our answer i.e in X
    for(int i = 31; i >= 0 && cnt != k; i--)
    {
         
        // Checking if the ith bit
        // is set in n or not
        if ((n & (1 << i)) != 0)
        {
            X[i] = 1;
            cnt++;
        }
    }
     
    String s = "";
     
    for(int i = 31; i >= 0; i--)
        s += X[i] == 0 ? '0' : '1';
     
    // Converting into integer
    return Convert.ToInt32(s, 2);
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 10, k = 2;
     
    // Function Call
    Console.Write(find_max(n, k));
}
}
 
// This code is contributed by offbeat

chevron_right


Output

10










Performance Analysis:

  • Time complexity: O(32)
  • Auxiliary Space: O(1

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :