Shortcut Tricks play a major role in solving complex problems in simple, fast and efficient way.

They can be useful and efficient for working with binary representations of integers in various programming tasks.

Here are some popular bit manipulation shortcut tricks in C++:

**1. Setting a bit using the OR operator**

To set a bit at a specific position in an integer variable

`x`

, you can use the OR operator (`|`

) with a bit mask that has a 1 in the desired position and zeroes everywhere else. Here's an example code that sets the third bit of `x`

to 1:`int x = 10; // Binary representation: 1010 int pos = 2; // Position of the bit to be set x |= (1 << pos); // Set the bit at position pos cout << x; // Output: 14 (Binary representation: 1110)`

In this code, the integer variable

`x`

is initialized with the decimal value 10, which has a binary representation of 1010.The integer variable

`pos`

specifies the position of the bit to be set, starting from zero indexing.The bit mask

`(1 << pos)`

has a value of 2^pos, which has a single 1 bit in the position corresponding to `pos`

and zeroes everywhere else. In this example, the bit mask is equal to `100`

in binary notation, which is equivalent to the decimal value 4.The OR operator between

`x`

and the bit mask `(1 << pos)`

sets the bit at position `pos`

of `x`

to 1, resulting in a binary value of `1110`

, which is equivalent to the decimal value 14.**2. Clearing a bit using the AND operator with complement**

To clear a bit at a specific position in an integer variable

`x`

, you can use the AND operator (`&`

) with a bit mask that has zeroes in the desired position and ones everywhere else. Here's an example code that clears the fourth bit of `x`

:`int x = 25; // Binary representation: 11001 int pos = 3; // Position of the bit to be cleared x &= ~(1 << pos); // Clear the bit at position pos cout << x; // Output: 17 (Binary representation: 10001)`

In this code, the integer variable

`x`

is initialized with the decimal value 25, which has a binary representation of `11001`

.The integer variable

`pos`

specifies the position of the bit to be cleared, starting from zero indexing.The bit mask

`~(1 << pos)`

has a value of bitwise NOT of 2^pos, which has a single 0 bit in the position corresponding to `pos`

and ones everywhere else. In this example, the bit mask is equal to `1110111`

in binary notation, which is equivalent to the decimal value `-9`

.The AND operator between

`x`

and the bit mask `~(1 << pos)`

clears the bit at position `pos`

of `x`

, resulting in a binary value of `10001`

, which is equivalent to the decimal value 17.**3. Toggling a bit using the XOR operator**

To toggle (i.e. flip) a bit at a specific position in an integer variable

`x`

, you can use the XOR operator (`^`

) with a bit mask that has a single 1 in the desired position and zeroes everywhere else. Here's an example code that toggles the fifth bit of `x`

:`int x = 27; // Binary representation: 11011 int pos = 4; // Position of the bit to be toggled x ^= (1 << pos); // Toggle the bit at position pos cout << x; // Output: 11 (Binary representation: 1011)`

In this code, the integer variable

`x`

is initialized with the decimal value 27, which has a binary representation of `11011`

.The integer variable

`pos`

specifies the position of the bit to be toggled, starting from zero indexing.The bit mask

`(1 << pos)`

has a value of 2^pos, which has a single 1 bit in the position corresponding to `pos`

and zeroes everywhere else. In this example, the bit mask is equal to `10000`

in binary notation, which is equivalent to the decimal value 16.The XOR operator between

`x`

and the bit mask `(1 << pos)`

toggles the bit at position `pos`

of `x`

, resulting in a binary value of `1011`

, which is equivalent to the decimal value 11.**4. Testing a bit using the AND operator**

To test whether a bit is set at a specific position in an integer variable

`x`

, you can use the AND operator (`&`

) with a bit mask that has a single 1 in the desired position and zeroes everywhere else. If the result is non-zero, the bit at that position is set; otherwise, it is not set. Here's an example code that tests whether the second bit of `x`

is set:`int x = 6; // Binary representation: 0110 int pos = 1; // Position of the bit to be tested if (x & (1 << pos)) { // Test the bit at position pos cout << "Bit " << pos << " is set"; } else { cout << "Bit " << pos << " is not set"; } // Output: Bit 1 is set`

In this code, the integer variable

`x`

is initialized with the decimal value 6, which has a binary representation of `0110`

.The integer variable

`pos`

specifies the position of the bit to be tested, starting from zero indexing.The bit mask

`(1 << pos)`

has a value of 2^pos, which has a single 1 bit in the position corresponding to `pos`

and zeroes everywhere else. In this example, the bit mask is equal to `0010`

in binary notation, which is equivalent to the decimal value 2.The AND operator between

`x`

and the bit mask `(1 << pos)`

tests the bit at position `pos`

of `x`

. Since the second bit is set in this example, the result is non-zero, and the code prints "Bit 1 is set".**5. Swapping two integers using XOR**

To swap the values of two integer variables

`a`

and `b`

without using a third variable, you can use the XOR operator (`^`

) to perform a bitwise swap. The idea is to XOR `a`

with `b`

, store the result in `a`

, then XOR the new value of `a`

with `b`

again to restore the original value of `b`

. Here's an example code that swaps the values of `a`

and `b`

:`int a = 5; // Binary representation: 0101 int b = 9; // Binary representation: 1001 a ^= b; // XOR a and b to get new value of a b ^= a; // XOR new value of a with b to get new value of b a ^= b; // XOR new value of b with new value of a to restore original value of a cout << "a = " << a << ", b = " << b; // Output: a = 9, b = 5`

In this code, the integer variables

`a`

and `b`

are initialized with the decimal values 5 and 9, which have binary representations of `0101`

and `1001`

, respectively.The first XOR operation between

`a`

and `b`

produces a new value for `a`

that contains both the bits that are set in `a`

and `b`

, which is `1100`

in binary notation, equivalent to the decimal value 12.The second XOR operation between the new value of

`a`

and `b`

produces a new value for `b`

that contains the bits that were originally set in `a`

, which is `0101`

in binary notation, equivalent to the decimal value 5.The third XOR operation between the new value of

`b`

and the original value of `a`

restores the original value of `a`

, which is `0101`

in binary notation, equivalent to the decimal value 9.After the bitwise swap, the values of

`a`

and `b`

have been swapped, as evidenced by the output `a = 9, b = 5`

.**PRACTICE PROBLEMS :**