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