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

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 C++ 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));
}
}
```

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 contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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