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Check whether bits are in alternate pattern in the given range

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  • Last Updated : 30 Sep, 2022
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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.
Constraint: 1 <= l <= r <= number of bits in the binary representation of n.

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.

 

Algorithm: 
 

bitsAreInAltPatrnInGivenTRange(n, l, r)
    
    // Shift bits of given number n by 
    // (l-1).
    num = n >> (l - 1)
    
    // Find first bit of range
    prev = num & 1

    // Traverse through remaining
    // bits. For every bit, check if
    // current bit is same as previous
    // or not.
    num = num >> 1    
    for i = 1 to (r - l)
        curr = num & 1
        
        if curr == prev then
            return false
        
        prev = curr
        num = num >> 1
    
    return true;

 

C++




// 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 bits are in
// alternate pattern in the given range
bool bitsAreInAltPatrnInGivenTRange(unsigned int n,
                                    unsigned int l,
                                    unsigned int r)
{
    unsigned int num, prev, curr;
 
    // right shift n by (l - 1) bits
    num = n >> (l - 1);
 
    // get the bit at the last position in 'num'
    prev = num & 1;
 
    // right shift 'num' by 1
    num = num >> 1;
 
    // loop until there are bits in the given range
    for (int i = 1; i <= (r - l); i++) {
 
        // get the bit at the last position in 'num'
        curr = num & 1;
 
        // if true, then bits are not in alternate
        // pattern
        if (curr == prev)
            return false;
 
        // update 'prev'
        prev = curr;
 
        // right shift 'num' by 1
        num = num >> 1;
    }
 
    // bits are in alternate pattern in the
    // given range
    return true;
}
 
// Driver program to test above
int main()
{
    unsigned int n = 18;
    unsigned int l = 1, r = 3;
 
    if (bitsAreInAltPatrnInGivenTRange(n, l, r))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java




// Java implementation to check
// whether bits are in alternate
// pattern in the given range
class GFG
{
     
// function to check whether
// bits are in alternate
// pattern in the given range
static boolean bitsAreInAltPatrnInGivenTRange(int n,
                                       int l, int r)
{
    int num, prev, curr;
 
    // right shift n by (l - 1) bits
    num = n >> (l - 1);
 
    // get the bit at the
    // last position in 'num'
    prev = num & 1;
 
    // right shift 'num' by 1
    num = num >> 1;
 
    // loop until there are
    // bits in the given range
    for (int i = 1; i <= (r - l); i++)
    {
 
        // get the bit at the
        // last position in 'num'
        curr = num & 1;
 
        // if true, then bits are
        // not in alternate pattern
        if (curr == prev)
            return false;
 
        // update 'prev'
        prev = curr;
 
        // right shift 'num' by 1
        num = num >> 1;
    }
 
    // bits are in alternate
    // pattern in the given range
    return true;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 18;
    int l = 1, r = 3;
 
    if (bitsAreInAltPatrnInGivenTRange(n, l, r))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by mits

Python3




# Python3 implementation to check
# whether bits are in alternate
# pattern in the given range
 
# function to check whether
# bits are in alternate
# pattern in the given range
def bitsAreInAltPatrnInGivenTRange(n, l, r):
 
    # right shift n by (l - 1) bits
    num = n >> (l - 1);
 
    # get the bit at the
    # last position in 'num'
    prev = num & 1;
 
    # right shift 'num' by 1
    num = num >> 1;
 
    # loop until there are
    # bits in the given range
    for i in range(1,(r - l)):
 
        # get the bit at the
        # last position in 'num'
        curr = num & 1;
 
        # if true, then bits are
        # not in alternate pattern
        if (curr == prev):
            return False;
 
        # update 'prev'
        prev = curr;
 
        # right shift 'num' by 1
        num = num >> 1;
 
    # bits are in alternate
    # pattern in the given range
    return True;
 
# Driver Code
if __name__ == "__main__":
    n = 18;
    l = 1;
    r = 3;
 
    if(bitsAreInAltPatrnInGivenTRange(n, l, r)):
        print("Yes");
    else:
        print("No");
 
# This Code is contributed by mits

C#




// C# implementation to check
// whether bits are in alternate
// pattern in the given range
using System;
 
class GFG
{
     
// function to check whether
// bits are in alternate
// pattern in the given range
static bool bitsAreInAltPatrnInGivenTRange(int n,
                                    int l, int r)
{
    int num, prev, curr;
 
    // right shift n by (l - 1) bits
    num = n >> (l - 1);
 
    // get the bit at the
    // last position in 'num'
    prev = num & 1;
 
    // right shift 'num' by 1
    num = num >> 1;
 
    // loop until there are
    // bits in the given range
    for (int i = 1; i <= (r - l); i++)
    {
 
        // get the bit at the
        // last position in 'num'
        curr = num & 1;
 
        // if true, then bits are
        // not in alternate pattern
        if (curr == prev)
            return false;
 
        // update 'prev'
        prev = curr;
 
        // right shift 'num' by 1
        num = num >> 1;
    }
 
    // bits are in alternate
    // pattern in the given range
    return true;
}
 
// Driver Code
static void Main()
{
    int n = 18;
    int l = 1, r = 3;
 
    if (bitsAreInAltPatrnInGivenTRange(n, l, r))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by mits

PHP




<?php
// PHP implementation to check
// whether bits are in alternate
// pattern in the given range
 
// function to check whether
// bits are in alternate
// pattern in the given range
function bitsAreInAltPatrnInGivenTRange($n, $l, $r)
{
 
    // right shift n by (l - 1) bits
    $num = $n >> ($l - 1);
 
    // get the bit at the
    // last position in 'num'
    $prev = $num & 1;
 
    // right shift 'num' by 1
    $num = $num >> 1;
 
    // loop until there are
    // bits in the given range
    for ($i = 1; $i <= ($r - $l); $i++)
    {
 
        // get the bit at the
        // last position in 'num'
        $curr = $num & 1;
 
        // if true, then bits are
        // not in alternate pattern
        if ($curr == $prev)
            return false;
 
        // update 'prev'
        $prev = $curr;
 
        // right shift 'num' by 1
        $num = $num >> 1;
    }
 
    // bits are in alternate
    // pattern in the given range
    return true;
}
 
// Driver Code
$n = 18;
$l = 1;
$r = 3;
 
if (bitsAreInAltPatrnInGivenTRange($n, $l, $r))
    echo "Yes";
else
    echo "No";
 
// This Code is contributed by mits
?>

Javascript




<script>
 
// Javascript implementation to check whether bits are in
// alternate pattern in the given range
 
// function to check whether bits are in
// alternate pattern in the given range
function bitsAreInAltPatrnInGivenTRange(n, l, r)
{
    var num, prev, curr;
 
    // right shift n by (l - 1) bits
    num = n >> (l - 1);
 
    // get the bit at the last position in 'num'
    prev = num & 1;
 
    // right shift 'num' by 1
    num = num >> 1;
 
    // loop until there are bits in the given range
    for (var i = 1; i <= (r - l); i++) {
 
        // get the bit at the last position in 'num'
        curr = num & 1;
 
        // if true, then bits are not in alternate
        // pattern
        if (curr == prev)
            return false;
 
        // update 'prev'
        prev = curr;
 
        // right shift 'num' by 1
        num = num >> 1;
    }
 
    // bits are in alternate pattern in the
    // given range
    return true;
}
 
// Driver program to test above
var n = 18;
var l = 1, r = 3;
if (bitsAreInAltPatrnInGivenTRange(n, l, r))
    document.write( "Yes");
else
    document.write( "No");
 
// This code is contributed by rrrtnx.
</script>

Output: 
 

Yes

Time complexity: O(r-l)
Auxiliary space: O(1)


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