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Minimum removals in range to make bitwise AND non-zero for given range queries

  • Last Updated : 26 Apr, 2022

Given an array queries[][] of Q range queries, the task is to find the minimum removals from the range[l, r] such that the bitwise AND of the range is a non-zero value.

Examples: 

Input: queries[][] = { {1, 5}, {3, 4}, {5, 10}, {10, 15}}
Output: 2 1 3 0 
Explanation: 
Query-1: l = 1, r = 5 {1, 2, 3, 4, 5} (2, 4 ) should be removed to make the AND of the array non-zero so minimum removals is 2.
Query-2: l = 3, r = 4 {3, 4} Either 3 or 4 should be removed to make the AND of the array non-zero so minimum removals is 1.
Query-3: l = 5, r = 10 {5, 6, 7, 8, 9, 10} (5, 6, 7) or (8, 9, 10) should be removed to make the AND of range non-zero. Minimum removals 3.
Query-4: l = 10, r = 15 {10, 11, 12, 13, 14, 15} the AND of the array is non-zero initially so 0 removals.

Input:  queries[][] = { {100, 115}, {30, 40}, {101, 190} };
Output: 0 2 27 

 

Naive Approach: This can be solved by iterating through every range and checking the max count of elements in the range having the common set bit and removing that from the total count of element i.e (r – l + 1).

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function the find the minimum removals
// to make the bitwise AND of range non-zero
void solve_queries(vector<vector<int> > q)
{
    for (int i = 0; i < q.size(); i++) {
        int l = q[i][0];
        int r = q[i][1];
 
        // Initialize the max_set to check
        // what count of set bit majority of
        // numbers in the range was set
        int max_set = 0;
        for (int i = 0; i < 31; i++) {
            int cnt = 0;
            for (int j = l; j <= r; j++) {
 
                // Check if is set
                if ((1 << i) & j) {
                    cnt++;
                }
            }
            max_set = max(max_set, cnt);
        }
 
        // Total length of range - count of
        // max numbers having 1 bit set in range
        cout << (r - l + 1) - max_set << " ";
    }
}
 
// Driver Code
int main()
{
    // Initialize the queries
    vector<vector<int> > queries
        = { { 1, 5 }, { 3, 4 }, { 5, 10 }, { 10, 15 } };
 
    solve_queries(queries);
    return 0;
}

Java




// Java code to implement the above approach
import java.util.*;
public class GFG
{
 
  // Function the find the minimum removals
  // to make the bitwise AND of range non-zero
  static void solve_queries(int [][]q)
  {
    for (int i = 0; i < q.length; i++) {
      int l = q[i][0];
      int r = q[i][1];
 
      // Initialize the max_set to check
      // what count of set bit majority of
      // numbers in the range was set
      int max_set = 0;
      for (int k = 0; k < 31; k++) {
        int cnt = 0;
        for (int j = l; j <= r; j++) {
 
          // Check if is set
          if (((1 << k) & j) != 0) {
            cnt++;
          }
        }
        max_set = Math. max(max_set, cnt);
      }
 
      // Total length of range - count of
      // max numbers having 1 bit set in range
      System.out.print((r - l + 1) - max_set + " ");
    }
  }
 
  // Driver code
  public static void main(String args[])
  {
    // Initialize the queries
    int [][]queries
      = { { 1, 5 }, { 3, 4 }, { 5, 10 }, { 10, 15 } };
 
    solve_queries(queries);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python3




# Python program for the above approach
 
# Function the find the minimum removals
# to make the bitwise AND of range non-zero
def solve_queries (q):
 
    for i in range(len(q)):
        l = q[i][0]
        r = q[i][1]
 
        # Initialize the max_set to check
        # what count of set bit majority of
        # numbers in the range was set
        max_set = 0
        for i in range(31):
            cnt = 0
            for j in range(l, r + 1):
 
                # Check if is set
                if ((1 << i) & j):
                    cnt += 1
            max_set = max(max_set, cnt)
 
        # Total length of range - count of
        # max numbers having 1 bit set in range
        print(f"{(r - l + 1) - max_set} ", end=" ")
 
# Driver Code
 
# Initialize the queries
queries = [[1, 5], [3, 4], [5, 10], [10, 15]]
 
solve_queries(queries)
 
# This code is contributed by gfgking

C#




// C# code to implement the above approach
using System;
public class GFG{
 
  // Function the find the minimum removals
  // to make the bitwise AND of range non-zero
  static void solve_queries(int[,] q)
  {
    for (int i = 0; i < 4; i++) {
      int l = q[i, 0];
      int r = q[i, 1];
 
      // Initialize the max_set to check
      // what count of set bit majority of
      // numbers in the range was set
      int max_set = 0;
      for (int k = 0; k < 31; k++) {
        int cnt = 0;
        for (int j = l; j <= r; j++) {
 
          // Check if is set
          if (((1 << k) & j) != 0) {
            cnt++;
          }
        }
        max_set = Math.Max(max_set, cnt);
      }
 
      // Total length of range - count of
      // max numbers having 1 bit set in range
      Console.Write((r - l + 1) - max_set + " ");
    }
  }
 
  // Driver code
  static public void Main()
  {
     
    // Initialize the queries
    int[,] queries = new int[4,2] { { 1, 5 }, { 3, 4 }, { 5, 10 }, { 10, 15 } };
 
    solve_queries(queries);
  }
}
 
// This code is contributed by Shubham Singh

Javascript




<script>
 
// JavaScript program for the above approach
 
// Function the find the minimum removals
// to make the bitwise AND of range non-zero
const solve_queries = (q) => {
     
    for(let i = 0; i < q.length; i++)
    {
        let l = q[i][0];
        let r = q[i][1];
 
        // Initialize the max_set to check
        // what count of set bit majority of
        // numbers in the range was set
        let max_set = 0;
        for(let i = 0; i < 31; i++)
        {
            let cnt = 0;
            for(let j = l; j <= r; j++)
            {
                 
                // Check if is set
                if ((1 << i) & j)
                {
                    cnt++;
                }
            }
            max_set = Math.max(max_set, cnt);
        }
 
        // Total length of range - count of
        // max numbers having 1 bit set in range
        document.write(`${(r - l + 1) - max_set} `);
    }
}
 
// Driver Code
 
// Initialize the queries
let queries = [ [ 1, 5 ], [ 3, 4 ],
                [ 5, 10 ], [ 10, 15 ] ];
 
solve_queries(queries);
 
// This code is contributed by rakeshsahni
 
</script>
Output
2 1 3 0 

Time Complexity: O (31 * Q * n )  where n is the length of the maximum range.
Auxiliary Space: O(1)

Efficient Approach: This approach is based on Dynamic programming and prefix sum technique. This can be done by storing the count of total set bits till that range in a using dp[] array at each position of 31 bits. Here the state of dp[i][j] means that the total set bits of jth bit position till i from [1, i]. Follow these steps to solve the above problem:

  • Make  a function call to count_set_bits() to precalculate the dp[][] array using prefix sum.
  • Now iterate through the queries and assign l = q[i][0], r = q[i][1].
    • Initialize max_set = INT_MIN to store the count of the max count of elements in the range having the j-th bit set.
    • Iterate through the bits from [0, 30].
    • Count the number of elements having j-th bit set by total_set = (dp[r][j] – dp[l – 1][j]).
    • Store the max count of elements having j-th bit set by taking max(max_set, total_set).
  • Print the minimum removals required by subtracting max_set from the total length (r-l+1).

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
int dp[200005][31];
 
// Function to build the dp array
// which stores the set bits for
// each bit position till 200005.
void count_set_bits()
{
    int N = 2e5 + 5;
    for (int i = 1; i < N; i++) {
        for (int j = 0; j <= 30; j++) {
            // If j th bit is set assign
            // dp[i][j] =1 where dp[i][j] means
            // number of jth bits set till i
            if (i & (1 << j))
                dp[i][j] = 1;
 
            // Calculate the prefix sum
            dp[i][j] += dp[i - 1][j];
        }
    }
}
 
// Function to solve the queries
void solve_queries(vector<vector<int> > q)
{
    // Call the function to
    // precalculate the dp array
    count_set_bits();
    for (int i = 0; i < q.size(); i++) {
 
        int l = q[i][0];
        int r = q[i][1];
 
        // Initialize the max_set = INT_MIN to
        // check what count of set bit
        // majority of numbers in the range was set
        int max_set = INT_MIN;
 
        // For each bit check the
        // maximum numbers in the range
        // having the j th bit set
        for (int j = 0; j <= 30; j++) {
            int total_set = (dp[r][j] - dp[l - 1][j]);
 
            max_set = max(max_set, total_set);
        }
 
        // Total length of range - count of max
        // numbers having 1 bit set in the range
        cout << (r - l + 1) - max_set << " ";
    }
}
 
// Driver Code
int main()
{
 
    // Initialize the queries
    vector<vector<int> > queries
        = { { 1, 5 }, { 3, 4 }, { 5, 10 }, { 10, 15 } };
 
    solve_queries(queries);
    return 0;
}

Java




// Java code to implement the above approach
import java.io.*;
 
class GFG
{
 
  public static int dp[][] = new int[200005][31];
 
  // Function to build the dp array
  // which stores the set bits for
  // each bit position till 200005.
  public static void count_set_bits()
  {
    int N = 200005;
    for (int i = 1; i < N; i++)
    {
      for (int j = 0; j <= 30; j++)
      {
 
        // If j th bit is set assign
        // dp[i][j] =1 where dp[i][j] means
        // number of jth bits set till i
        if ((i & (1 << j))!=0)
          dp[i][j] = 1;
 
        // Calculate the prefix sum
        dp[i][j] += dp[i - 1][j];
      }
    }
  }
 
  // Function to solv e the queries
  public static void solve_queries(int[][] q)
  {
 
    // Call the function to
    // precalculate the dp array
    count_set_bits();
    for (int i = 0; i < q.length; i++) {
 
      int l = q[i][0];
      int r = q[i][1];
 
      // Initialize the max_set = INT_MIN to
      // check what count of set bit
      // majority of numbers in the range was set
      int max_set = Integer.MIN_VALUE;
 
      // For each bit check the
      // maximum numbers in the range
      // having the j th bit set
      for (int j = 0; j <= 30; j++) {
        int total_set = (dp[r][j] - dp[l - 1][j]);
 
        max_set = Math.max(max_set, total_set);
      }
 
      // Total length of range - count of max
      // numbers having 1 bit set in the range
      System.out.print((r - l + 1) - max_set + " ");
    }
  }
 
  // Driver Code
  public static void main (String[] args)
  {
 
    // Initialize the queries
    int[][] queries = new int[][]
    {
      new int[] { 1, 5 },
      new int[] { 3, 4 },
      new int[] { 5, 10 },
      new int[] { 10, 15 }
    };
 
    solve_queries(queries);
  }
}
 
// This code is contributed by Shubham Singh

Python3




# Python program for the above approach
dp = [[0 for i in range(31)] for j in range(200005)]
 
# Function to build the dp array
# which stores the set bits for
# each bit position till 200005.
def count_set_bits():
    N = 200005
    for i in range(1, N):
        for j in range(31):
           
            # If j th bit is set assign
            # dp[i][j] =1 where dp[i][j] means
            # number of jth bits set till i
            if (i & (1 << j)):
                dp[i][j] = 1
                 
            # Calculate the prefix sum
            dp[i][j] += dp[i - 1][j]
 
# Function to solve the queries
def solve_queries(q):
     
    # Call the function to
    # precalculate the dp array
    count_set_bits()
    for i in range(len(q)):
        l = q[i][0]
        r = q[i][1]
         
        # Initialize the max_set = INT_MIN to
        # check what count of set bit
        # majority of numbers in the range was set
        max_set = -2**32
         
        # For each bit check the
        # maximum numbers in the range
        # having the j th bit set
        for j in range(31):
            total_set = (dp[r][j] - dp[l - 1][j])
            max_set = max(max_set, total_set)
             
        # Total length of range - count of max
        # numbers having 1 bit set in the range
        print((r - l + 1) - max_set, end=" ")
 
# Driver Code
# Initialize the queries
queries =  [[1, 5],  [3, 4],  [5, 10],  [10, 15]]
 
solve_queries(queries)
 
# This code is contributed by Shubham Singh

C#




// C# code to implement the above approach
using System;
 
public class GFG{
 
  public static int[,] dp = new int[200005,31];
 
  // Function to build the dp array
  // which stores the set bits for
  // each bit position till 200005.
  public static void count_set_bits()
  {
    int N = 200005;
    for (int i = 1; i < N; i++)
    {
      for (int j = 0; j <= 30; j++)
      {
 
        // If j th bit is set assign
        // dp[i,j] =1 where dp[i,j] means
        // number of jth bits set till i
        if ((i & (1 << j))!=0)
          dp[i,j] = 1;
 
        // Calculate the prefix sum
        dp[i,j] += dp[i - 1,j];
      }
    }
  }
 
  // Function to solv e the queries
  public static void solve_queries(int[][] q)
  {
 
    // Call the function to
    // precalculate the dp array
    count_set_bits();
    for (int i = 0; i < q.Length; i++) {
 
      int l = q[i][0];
      int r = q[i][1];
 
      // Initialize the max_set = INT_MIN to
      // check what count of set bit
      // majority of numbers in the range was set
      int max_set = Int32.MinValue;
 
      // For each bit check the
      // maximum numbers in the range
      // having the j th bit set
      for (int j = 0; j <= 30; j++) {
        int total_set = (dp[r,j] - dp[l - 1,j]);
 
        max_set = Math.Max(max_set, total_set);
      }
 
      // Total length of range - count of max
      // numbers having 1 bit set in the range
      Console.Write((r - l + 1) - max_set + " ");
    }
  }
 
  // Driver Code
  static public void Main ()
  {
 
    // Initialize the queries
    int[][] queries = new int[][]
    {
      new int[] { 1, 5 },
      new int[] { 3, 4 },
      new int[] { 5, 10 },
      new int[] { 10, 15 }
    };
 
    solve_queries(queries);
  }
}
 
// This code is contributed by Shubham Singh

Javascript




<script>
 
// JavaScript program for the above approach
let dp = new Array(200005)
for(let i = 0; i < 200005; i++){
    dp[i] = new Array(31).fill(0)
}
 
// Function to build the dp array
// which stores the set bits for
// each bit position till 200005.
function count_set_bits(){
    let N = 200005
    for(let i = 1; i < N; i++){
        for(let j = 0; j < 31; j++){
         
            // If j th bit is set assign
            // dp[i][j] =1 where dp[i][j] means
            // number of jth bits set till i
            if (i & (1 << j))
                dp[i][j] = 1
                 
            // Calculate the prefix sum
            dp[i][j] += dp[i - 1][j]
        }
    }
}
 
// Function to solve the queries
function solve_queries(q){
     
    // Call the function to
    // precalculate the dp array
    count_set_bits()
    for(let i = 0; i < q.length; i++){
        let l = q[i][0]
        let r = q[i][1]
         
        // Initialize the max_set = INT_MIN to
        // check what count of set bit
        // majority of numbers in the range was set
        let max_set = Number.MIN_VALUE
         
        // For each bit check the
        // maximum numbers in the range
        // having the j th bit set
        for(let j = 0; j < 31; j++){
            let total_set = (dp[r][j] - dp[l - 1][j])
            max_set = Math.max(max_set, total_set)
        }
             
        // Total length of range - count of max
        // numbers having 1 bit set in the range
        document.write((r - l + 1) - max_set," ")
    }
}
 
// Driver Code
// Initialize the queries
let queries = [[1, 5], [3, 4], [5, 10], [10, 15]]
 
solve_queries(queries)
 
// This code is contributed by Shinjanpatra
 
</script>
Output
2 1 3 0 

Time Complexity: O(31 * Q) 
Auxiliary Space: O(n)

 


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