Check whether bits are in alternate pattern in the given range | Set-2

Given a non-negative number N and two values L and R. The problem is to check whether or not N has an alternate pattern in its binary representation in the range L to R.

Here, alternate pattern means that the set and unset bits occur in alternate order. The bits are numbered from right to left, i.e., the least significant bit is considered to be at first position.

Examples:

Input : N = 18, L = 1, R = 3
Output : Yes
(18)10 = (10010)2
The bits in the range 1 to 3 in the
binary representation of 18 are in
alternate order.

Input : N = 22, L = 2, R = 4
Output : No
(22)10 = (10110)2
The bits in the range 2 to 4 in the
binary representation of 22 are not in
alternate order.

Simple Approach: It has been discussed in this post which has a worst case time complexity of O(log2n).

Efficient Approach: Following are the steps:

  1. Declare two variables num and left_shift.
  2. Check if rth bit is set or not in n. Refer this post. If set then, assign num = n and left_shift = r, Else set the (r+1)th bit in n and assign it to num. Refer this post. Also assign left_shift = r + 1.
  3. Perform num = num & ((1 << left_shift) – 1).
  4. Perform num = num >> (l – 1).
  5. Finally check whether bits are in alternate pattern in num or not. Refer this post.

The entire idea of the above approach is to create a number num in which bits are in same pattern as in the given range of n and then check whether bits are in alternate pattern in num or not.

C++

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// C++ implementation to check whether bits are in
// alternate pattern in the given range
#include <bits/stdc++.h>
  
using namespace std;
  
// function to check whether rightmost
// kth bit is set or not in 'n'
bool isKthBitSet(unsigned int n,
                 unsigned int k)
{
    if ((n >> (k - 1)) & 1)
        return true;
    return false;
}
  
// function to set the rightmost kth bit in 'n'
unsigned int setKthBit(unsigned int n,
                       unsigned int k)
{
    // kth bit of n is being set by this operation
    return ((1 << (k - 1)) | n);
}
  
// function to check if all the bits are set or not
// in the binary representation of 'n'
bool allBitsAreSet(unsigned int n)
{
    // if true, then all bits are set
    if (((n + 1) & n) == 0)
        return true;
  
    // else all bits are not set
    return false;
}
  
// function to check if a number
// has bits in alternate pattern
bool bitsAreInAltOrder(unsigned int n)
{
    unsigned int num = n ^ (n >> 1);
  
    // to check if all bits are set
    // in 'num'
    return allBitsAreSet(num);
}
  
// function to check whether bits are in
// alternate pattern in the given range
bool bitsAreInAltPatrnInGivenRange(unsigned int n,
                                   unsigned int l,
                                   unsigned int r)
{
    unsigned int num, left_shift;
  
    // preparing a number 'num' and 'left_shift'
    // which can be further used for the check
    // of alternate pattern in the given range
    if (isKthBitSet(n, r)) {
        num = n;
        left_shift = r;
    }
    else {
        num = setKthBit(n, (r + 1));
        left_shift = r + 1;
    }
  
    // unset all the bits which are left to the
    // rth bit of (r+1)th bit
    num = num & ((1 << left_shift) - 1);
  
    // right shift 'num' by (l-1) bits
    num = num >> (l - 1);
  
    return bitsAreInAltOrder(num);
}
  
// Driver program to test above
int main()
{
    unsigned int n = 18;
    unsigned int l = 1, r = 3;
    if (bitsAreInAltPatrnInGivenRange(n, l, r))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

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Java

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// Java implementation to check whether bits are in
// alternate pattern in the given range
class GFG 
{
  
// function to check whether rightmost
// kth bit is set or not in 'n'
static boolean isKthBitSet(int n,
                            int k)
{
    if ((n >> (k - 1)) == 1)
        return true;
    return false;
}
  
// function to set the rightmost kth bit in 'n'
static int setKthBit(int n,
                    int k)
{
    // kth bit of n is being set by this operation
    return ((1 << (k - 1)) | n);
}
  
// function to check if all the bits are set or not
// in the binary representation of 'n'
static boolean allBitsAreSet(int n)
{
    // if true, then all bits are set
    if (((n + 1) & n) == 0)
        return true;
  
    // else all bits are not set
    return false;
}
  
// function to check if a number
// has bits in alternate pattern
static boolean bitsAreInAltOrder(int n)
{
    int num = n ^ (n >> 1);
  
    // to check if all bits are set
    // in 'num'
    return allBitsAreSet(num);
}
  
// function to check whether bits are in
// alternate pattern in the given range
static boolean bitsAreInAltPatrnInGivenRange(int n,
                                int l, int r)
{
    int num, left_shift;
  
    // preparing a number 'num' and 'left_shift'
    // which can be further used for the check
    // of alternate pattern in the given range
    if (isKthBitSet(n, r)) 
    {
        num = n;
        left_shift = r;
    }
    else
    {
        num = setKthBit(n, (r + 1));
        left_shift = r + 1;
    }
  
    // unset all the bits which are left to the
    // rth bit of (r+1)th bit
    num = num & ((1 << left_shift) - 1);
  
    // right shift 'num' by (l-1) bits
    num = num >> (l - 1);
  
    return bitsAreInAltOrder(num);
}
  
// Driver code
public static void main(String[] args)
{
    int n = 18;
    int l = 1, r = 3;
    if (bitsAreInAltPatrnInGivenRange(n, l, r))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
  
// This code has been contributed by 29AjayKumar

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Python3

# Python 3 implementation to check
# whether bits are in alternate pattern
# in the given range

# function to check whether rightmost
# kth bit is set or not in ‘n’
def isKthBitSet(n, k):
if((n >> (k – 1)) & 1):
return True
return False

# function to set the rightmost kth bit in ‘n’
def setKthBit(n, k):

# kth bit of n is being set
# by this operation
return ((1 << (k - 1)) | n) # function to check if all the bits are set or not # in the binary representation of 'n' def allBitsAreSet(n): # if true, then all bits are set if (((n + 1) & n) == 0): return True # else all bits are not set return False # function to check if a number # has bits in alternate pattern def bitsAreInAltOrder(n): num = n ^ (n >> 1)

# to check if all bits are set
# in ‘num’
return allBitsAreSet(num)

# function to check whether bits are in
# alternate pattern in the given range
def bitsAreInAltPatrnInGivenRange(n, l, r):

# preparing a number ‘num’ and ‘left_shift’
# which can be further used for the check
# of alternate pattern in the given range
if (isKthBitSet(n, r)):
num = n
left_shift = r

else:
num = setKthBit(n, (r + 1))
left_shift = r + 1

# unset all the bits which are left to the
# rth bit of (r+1)th bit
num = num & ((1 << left_shift) - 1) # right shift 'num' by (l-1) bits num = num >> (l – 1)

return bitsAreInAltOrder(num)

# Driver Code
if __name__ == ‘__main__’:
n = 18
l = 1
r = 3
if (bitsAreInAltPatrnInGivenRange(n, l, r)):
print(“Yes”)
else:
print(“No”)

# This code is contributed by
# Surendra_Gangwar

C#

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// C# implementation to check whether bits are in 
// alternate pattern in the given range 
using System;
  
class GFG 
  
// function to check whether rightmost 
// kth bit is set or not in 'n' 
static bool isKthBitSet(int n, 
                            int k) 
    if ((n >> (k - 1)) == 1) 
        return true
    return false
  
// function to set the rightmost kth bit in 'n' 
static int setKthBit(int n, 
                    int k) 
    // kth bit of n is being set by this operation 
    return ((1 << (k - 1)) | n); 
  
// function to check if all the bits are set or not 
// in the binary representation of 'n' 
static bool allBitsAreSet(int n) 
    // if true, then all bits are set 
    if (((n + 1) & n) == 0) 
        return true
  
    // else all bits are not set 
    return false
  
// function to check if a number 
// has bits in alternate pattern 
static bool bitsAreInAltOrder(int n) 
    int num = n ^ (n >> 1); 
  
    // to check if all bits are set 
    // in 'num' 
    return allBitsAreSet(num); 
  
// function to check whether bits are in 
// alternate pattern in the given range 
static bool bitsAreInAltPatrnInGivenRange(int n, 
                                int l, int r) 
    int num, left_shift; 
  
    // preparing a number 'num' and 'left_shift' 
    // which can be further used for the check 
    // of alternate pattern in the given range 
    if (isKthBitSet(n, r)) 
    
        num = n; 
        left_shift = r; 
    
    else
    
        num = setKthBit(n, (r + 1)); 
        left_shift = r + 1; 
    
  
    // unset all the bits which are left to the 
    // rth bit of (r+1)th bit 
    num = num & ((1 << left_shift) - 1); 
  
    // right shift 'num' by (l-1) bits 
    num = num >> (l - 1); 
  
    return bitsAreInAltOrder(num); 
  
// Driver code 
public static void Main() 
    int n = 18; 
    int l = 1, r = 3; 
    if (bitsAreInAltPatrnInGivenRange(n, l, r)) 
        Console.WriteLine("Yes"); 
    else
        Console.WriteLine("No"); 
  
/* This code contributed by PrinciRaj1992 */

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Output:

Yes

Time Complexity: O(1).



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