Skip to content
Related Articles

Related Articles

Improve Article

Inserting m into n such that m starts at bit j and ends at bit i.

  • Difficulty Level : Hard
  • Last Updated : 31 May, 2021

We are given two numbers n and m, and two-bit positions, i and j. Insert bits of m into n starting from j to i. We can assume that the bits j through i have enough space to fit all of m. That is, if m = 10011, you can assume that there are at least 5 bits between j and i. You would not, for example, have j = 3 and i = 2, because m could not fully fit between bit 3 and bit 2. 
Examples : 
 

Input : n = 1024
        m = 19
        i = 2
        j = 6;
Output : n = 1100
Binary representations of input numbers
m in binary is (10011)2
n in binary is (10000000000)2
Binary representations of output number
(10000000000)2

Input : n = 5
        m = 3
        i = 1
        j = 2
Output : 7

 

Algorithm : 
 

1. Clear the bits j through i in n 

2. Shift m so that it lines up with bits j through i 

3. Return Bitwise AND of m and n. 

The trickiest part is Step 1. How do we clear the bits in n? We can do this with a mask. This mask will have all 1s, except for 0s in the bits j through i. We create this mask by creating the left half of the mask first, and then the right half. 
Following is the implementation of the above approach.
 

C++




// C++ program for implementation of updateBits()
#include <bits/stdc++.h>
using namespace std;
 
// Function to updateBits M insert to N.
int updateBits(int n, int m, int i, int j)
{
    /* Create a mask to clear bits i through j
      in n. EXAMPLE: i = 2, j = 4. Result
      should be 11100011. For simplicity, we'll
      use just 8 bits for the example. */
 
    int allOnes = ~0; // will equal sequence of all ls
 
    // ls before position j, then 0s. left = 11100000
    int left= allOnes << (j + 1);
 
    // l's after position i. right = 00000011
    int right = ((1 << i) - 1);
 
    // All ls, except for 0s between i and j. mask 11100011
    int mask = left | right;
 
    /* Clear bits j through i then put min there */
    int n_cleared = n & mask; // Clear bits j through i.
    int m_shifted = m << i;   // Move m into correct position.
 
    return (n_cleared | m_shifted); // OR them, and we're done!
}
 
// Driver Code
int main()
{
    int n = 1024; // in Binary N= 10000000000
    int m = 19;   // in Binary M= 10011
    int i = 2, j = 6;
 
    cout << updateBits(n,m,i,j);
 
    return 0;
}

Java




// Java program for implementation of updateBits()
 
class UpdateBits
{
    // Function to updateBits M insert to N.
    static int updateBits(int n, int m, int i, int j)
    {
        /* Create a mask to clear bits i through j
          in n. EXAMPLE: i = 2, j = 4. Result
          should be 11100011. For simplicity, we'll
          use just 8 bits for the example. */
      
        int allOnes = ~0; // will equal sequence of all ls
      
        // ls before position j, then 0s. left = 11100000
        int left= allOnes << (j + 1);
      
        // l's after position i. right = 00000011
        int right = ((1 << i) - 1);
      
        // All ls, except for 0s between i and j. mask 11100011
        int mask = left | right;
      
        /* Clear bits j through i then put min there */
        // Clear bits j through i.
        int n_cleared = n & mask;
        // Move m into correct position.
        int m_shifted = m << i; 
         
        // OR them, and we're done!
        return (n_cleared | m_shifted);
    }
     
    public static void main (String[] args)
    {
        // in Binary N= 10000000000
        int n = 1024;
         
        // in Binary M= 10011
        int m = 19;  
         
        int i = 2, j = 6;
      
        System.out.println(updateBits(n,m,i,j));
    }
}

Python3




# Python3 program for implementation
# of updateBits()
 
# Function to updateBits M insert to N.
def updateBits(n, m, i, j):
 
    # Create a mask to clear bits i through
    # j in n. EXAMPLE: i = 2, j = 4. Result
    # should be 11100011. For simplicity,
    # we'll use just 8 bits for the example.
 
    # will equal sequence of all ls
    allOnes = ~0
 
    # ls before position j,
    # then 0s. left = 11100000
    left = allOnes << (j + 1)
 
    # l's after position i. right = 00000011
    right = ((1 << i) - 1)
 
    # All ls, except for 0s between
    # i and j. mask 11100011
    mask = left | right
 
    # Clear bits j through i then put min there
    n_cleared = n & mask
     
    # Move m into correct position.
    m_shifted = m << i
 
    return (n_cleared | m_shifted)
 
 
# Driver Code
n = 1024 # in Binary N = 10000000000
m = 19   # in Binary M = 10011
i = 2; j = 6
print(updateBits(n, m, i, j))
 
# This code is contributed by Anant Agarwal.

C#




// C# program for implementation of
// updateBits()
using System;
 
class GFG {
     
    // Function to updateBits M
    // insert to N.
    static int updateBits(int n, int m,
                           int i, int j)
    {
         
        /* Create a mask to clear bits i
          through j in n. EXAMPLE: i = 2,
          j = 4. Result should be 11100011.
          For simplicity, we'll use just 8
          bits for the example. */
       
        // will equal sequence of all ls
        int allOnes = ~0;
       
        // ls before position j, then 0s.
        // left = 11100000
        int left= allOnes << (j + 1);
       
        // l's after position i.
        // right = 00000011
        int right = ((1 << i) - 1);
       
        // All ls, except for 0s between i
        // and j. mask 11100011
        int mask = left | right;
       
        /* Clear bits j through i then put
        min there */
        // Clear bits j through i.
        int n_cleared = n & mask;
         
        // Move m into correct position.
        int m_shifted = m << i; 
          
        // OR them, and we're done!
        return (n_cleared | m_shifted);
    }
      
    public static void Main()
    {
         
        // in Binary N= 10000000000
        int n = 1024;
          
        // in Binary M= 10011
        int m = 19;  
        int i = 2, j = 6;
       
        Console.WriteLine(updateBits(n, m, i, j));
    }
}
 
//This code is contributed by Anant Agarwal.

PHP




<?php
// PHP program for implementation
// of updateBits()
 
// Function to updateBits
// M insert to N.
 
function updateBits($n, $m, $i, $j)
{
    // Create a mask to clear
    // bits i through j in n.
    // EXAMPLE: i = 2, j = 4.
    // Result should be 11100011.
    // For simplicity, we'll use
    // just 8 bits for the example.
 
    // will equal sequence of all ls
    $allOnes = ~0;
 
    // ls before position j, then
    // 0s. left = 11100000
    $left= $allOnes << ($j + 1);
 
    // l's after position i.
    // right = 00000011
    $right = ((1 << $i) - 1);
 
    // All ls, except for 0s between
    // i and j. mask 11100011
    $mask = $left | $right;
 
    // Clear bits j through i
    // then put min there
     
    // Clear bits j through i.
    $n_cleared = $n & $mask;
     
    // Move m into correct position.
    $m_shifted = $m << $i;
 
    // OR them, and we're done!
    return ($n_cleared | $m_shifted);
}
 
// Driver Code
 
// in Binary N= 10000000000
$n = 1024;
 
// in Binary M= 10011
$m = 19;
$i = 2;
$j = 6;
 
echo updateBits($n, $m, $i, $j);
 
// This code is contributed by Ajit
?>

Javascript




<script>
 
// JavaScript program for implementation of updateBits()
 
 
    // Function to updateBits M insert to N.
    function updateBits(n,m,i,j)
    {
        /* Create a mask to clear bits i through j
        in n. EXAMPLE: i = 2, j = 4. Result
        should be 11100011. For simplicity, we'll
        use just 8 bits for the example. */
     
        let allOnes = ~0; // will equal sequence of all ls
     
        // ls before position j, then 0s. left = 11100000
        let left= allOnes << (j + 1);
     
        // l's after position i. right = 00000011
        let right = ((1 << i) - 1);
     
        // All ls, except for 0s between i and j. mask 11100011
        let mask = left | right;
     
        /* Clear bits j through i then put min there */
        // Clear bits j through i.
        let n_cleared = n & mask;
        // Move m into correct position.
        let m_shifted = m << i;
         
        // OR them, and we're done!
        return (n_cleared | m_shifted);
    }
     
     
        // in Binary N= 10000000000
        let n = 1024;
         
        // in Binary M= 10011
        let m = 19;
         
        let i = 2, j = 6;
     
        document.write(updateBits(n,m,i,j));
     
 
// This code is contributed by sravan
 
</script>

Output:  



 1100  // in Binary (10001001100)2

This article is contributed by Mr. Somesh Awasthi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :