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 modified 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 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); } // 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 |
Python3
# 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 Bitwise 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, 4i, j = 2, 4print(setBits(n, m, i, j)) # This code is submitted by Sachin Bisht |
C#
// C# Program for above implementation using System; 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() { int n = 2, m = 4; int i = 2, j = 4; Console.WriteLine(setBits(n, m, i, j)); } } // This code is contributed by Anant Agarwal. |
PHP
<?php // PHP program for above implementation // Function to set the bits function setBits($n, $m, $i, $j) { // number with all 1's $allOnes = ~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 Bitwise 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); } // Driver Code $n = 2; $m = 4; $i = 2; $j = 4; echo setBits($n, $m, $i, $j); // This code is contributed by ajit ?> |
Javascript
<script> // Javascript Program for above implementation // Function to set the bits function setBits(n, m, i, j) { // number with all 1's let allOnes = ~0; // Set all the bits in the left of j let left = allOnes << (j + 1); // Set all the bits in the right of j let right = ((1 << i) - 1); // Do Bitwise OR to get all the bits // set except in the range from i to j let mask = left | right; // clear bits j through i let masked_n = n & mask; // move m into the correct position let m_shifted = m << i; // return the Bitwise OR of masked_n // and shifted_m return (masked_n | m_shifted); } let n = 2, m = 4; let i = 2, j = 4; document.write(setBits(n, m, i, j)); </script> |
Output :
18
Time Complexity: O(1)
Auxiliary Space: O(1)
Reference:
https://www.careercup.com/question?id=8863294
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