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 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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