You are given two 32-bit numbers, N and M, and two bit positions, i and j. Write a method to set all bits between i and j in N equal to M (e.g., M becomes a substring of N located at i and starting at j).

Examples:

Input : N = 1, M = 2, i = 2, j = 4 Output: 9 N = 00000001(Considering 8 bits only) M = 10 (Binary of 2) For more indexes, leading zeroes will be considered. Now set 3 bits from ith index to j in the N as in the M. Bits:- 0 0 0 (0 1 0) 0 1 = 9 Indexes:- 7 6 5 4 3 2 1 0 From index 2 to 4, bits are set according to the M.

Asked in : Adobe

A simple solution is to traverse all bits in N from 0 to 31 and set the bits equals to M in the range from i to j.

An efficient solution is to do following steps.

- Set all the bits after j in a number.
- Set all the bits before i in a number.
- Then perform Bitwise Or on both then we get the number with all the bits set except from i to j.
- Perform Bitwise And with the given N as to set the bits according to the N.
- Then shift M into the correct position i.e. in the range of i to j.
- And at the last perform Bitwise Or on (Shifted M and the N modifed in 4th step).
- The result will be N with M as substring from ith to jth bits

## C++

// C++ program for above implementation #include <iostream> using namespace std; // Function to set the bits int setBits(int n, int m, int i, int j) { // number with all 1's int allOnes = ~0; // Set all the bits in the left of j int left = allOnes << (j + 1); // Set all the bits in the right of j int right = ((1 << i) - 1); // Do Bitwsie OR to get all the bits // set except in the range from i to j int mask = left | right; // clear bits j through i int masked_n = n & mask; // move m into the correct position int m_shifted = m << i; // return the Bitwise OR of masked_n // and shifted_m return (masked_n | m_shifted); } // Drivers program int main() { int n = 2, m = 4; int i = 2, j = 4; cout << setBits(n, m, i, j); return 0; }

## Java

// Java Program public class GFG { // Function to set the bits static int setBits(int n, int m, int i, int j) { // number with all 1's int allOnes = ~0; // Set all the bits in the left of j int left = allOnes << (j + 1); // Set all the bits in the right of j int right = ((1 << i) - 1); // Do Bitwise OR to get all the bits // set except in the range from i to j int mask = left | right; // clear bits j through i int masked_n = n & mask; // move m into the correct position int m_shifted = m << i; // return the Bitwise OR of masked_n // and shifted_m return (masked_n | m_shifted); } // Driver Program to test above function public static void main(String[] args) { int n = 2, m = 4; int i = 2, j = 4; System.out.println(setBits(n, m, i, j)); } } // This code is contributed by Sumit Ghosh

## Python

# Python program for above implementation # Function to set the bits def setBits(n, m, i, j): # number with all 1's allOnes = not 0 # Set all the bits in the left of j left = allOnes << (j + 1) # Set all the bits in the right of j right = ((1 << i) - 1) # Do Bitwsie OR to get all the bits # set except in the range from i to j mask = left | right # clear bits j through i masked_n = n & mask # move m into the correct position m_shifted = m << i # return the Bitwise OR of masked_n # and shifted_m return (masked_n | m_shifted) # Drivers program n, m = 2, 4 i, j = 2, 4 print setBits(n, m, i, j) # This code is submitted by Sachin Bisht

Output:

18

Reference:

https://www.careercup.com/question?id=8863294

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