# Check whether bits are in alternate pattern in the given range

• Last Updated : 30 Sep, 2022

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 ` `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

 `> (``\$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

 ``

Output:

`Yes`

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

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