Pseudocode Solved PYQ Bank (23 Questions)
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04. Pseudocode Solved PYQs & Comprehensive Practice Bank

Module Focus: 20 Questions in 20 Minutes
Key Skills: Bitwise Operator Tracing, Loop Execution Limits, Pass-by-Reference/Value, Recursion Call Stack Unwinding, Array Bit Manipulation, Operator Precedence

1. Bitwise Operations PYQ Bank

PYQ 1: Bitwise XOR and AND Combination

Integer a, b, c
Set a = 5, b = 10, c = 2
If ((a ^ b ^ c) > (a & b & c))
    a = b ^ c
    b = a + c
Else
    c = a & b
    a = b + c
End If
Print a + b + c

Step-by-Step Execution Trace: 1. Binary Representation: - $a = 5 = 0101_2$ - $b = 10 = 1010_2$ - $c = 2 = 0010_2$ 2. Evaluate Condition (a ^ b ^ c) > (a & b & c): - $a \oplus b = 0101_2 \oplus 1010_2 = 1111_2 = 15$. - $15 \oplus c = 1111_2 \oplus 0010_2 = 1101_2 = 13$. Left side $= 13$. - $a \ \& \ b = 0101_2 \ \& \ 1010_2 = 0000_2 = 0$. - $0 \ \& \ c = 0$. Right side $= 0$. - Condition 13 > 0 is TRUE! 3. Execute THEN block: - a = b ^ c $\implies 10 \oplus 2 = 1010_2 \oplus 0010_2 = 1000_2 = 8$. (Now $a = 8$). - b = a + c $\implies 8 + 2 = 10$. (Now $b = 10$). 4. Compute Final Expression a + b + c: - $a = 8, b = 10, c = 2$. - $8 + 10 + 2 = 20$.

Final Output: 20

PYQ 2: Bit Shift Arithmetic Combined with Logical Negation

Integer p = 12, q = 3, r = 0
r = (p >> 2) + (q << 2)
If ((r & 1) == 0)
    p = r ^ q
Else
    q = r & p
End If
Print p + q + r

Step-by-Step Execution Trace: 1. Bit Shifts: - p >> 2 $\implies 12 \gg 2 = \lfloor 12 / 2^2 \rfloor = \lfloor 12 / 4 \rfloor = 3$. - q << 2 $\implies 3 \ll 2 = 3 \times 2^2 = 3 \times 4 = 12$. - r = 3 + 12 = 15. 2. Condition Check (r & 1) == 0: - $15 \ \& \ 1 = 1111_2 \ \& \ 0001_2 = 1$. - (1 == 0) evaluates to FALSE. 3. Execute ELSE block: - q = r & p $\implies 15 \ \& \ 12 = 1111_2 \ \& \ 1100_2 = 1100_2 = 12$. (Now $q = 12$). 4. Compute Final Sum p + q + r: - $p = 12, q = 12, r = 15$. - $12 + 12 + 15 = 39$.

Final Output: 39

PYQ 3: Complex Nested Bitwise & Subtraction Loop

Integer a = 14, b = 7, c = 0
While (a > b)
    c = c + (a & b)
    a = a - 2
    b = b + 1
End While
Print c

Step-by-Step Execution Trace Table:

Iteration Initial a Initial b Condition (a > b) a & b Calculation Updated c Updated a Updated b
Start 14 7 - - 0 14 7
Pass 1 14 7 $14 > 7$ (True) $1110_2 \ \& \ 0111_2 = 0110_2 = 6$ $0 + 6 = 6$ $14 - 2 = 12$ $7 + 1 = 8$
Pass 2 12 8 $12 > 8$ (True) $1100_2 \ \& \ 1000_2 = 1000_2 = 8$ $6 + 8 = 14$ $12 - 2 = 10$ $8 + 1 = 9$
Pass 3 10 9 $10 > 9$ (True) $1010_2 \ \& \ 1001_2 = 1000_2 = 8$ $14 + 8 = 22$ $10 - 2 = 8$ $9 + 1 = 10$
Pass 4 8 10 $8 > 10$ (FALSE) Loop Terminates 22 8 10

Final Output: 22

PYQ 4: Bitwise OR, XOR, AND with Shift Operators

Integer a = 8, b = 5, c = 3
c = (a >> 1) ^ (b << 1)
If ((c | a) > (c & b))
    a = (a ^ b) + c
    b = b & c
Else
    c = c ^ a
End If
Print a + b + c

Step-by-Step Execution Trace: 1. Bit Shift and Initial XOR Assignment: - a >> 1 $\implies 8 \gg 1 = 4 = 0100_2$. - b << 1 $\implies 5 \ll 1 = 10 = 1010_2$. - c = 4 ^ 10 $\implies 0100_2 \oplus 1010_2 = 1110_2 = 14$. 2. Evaluate Condition (c | a) > (c & b): - c | a $\implies 14 \mid 8 = 1110_2 \mid 1000_2 = 1110_2 = 14$. - c & b $\implies 14 \ \& \ 5 = 1110_2 \ \& \ 0101_2 = 0100_2 = 4$. - Condition 14 > 4 is TRUE. 3. Execute THEN block: - a = (a ^ b) + c $\implies (8 \oplus 5) + 14 = (1000_2 \oplus 0101_2) + 14 = 13 + 14 = 27$. (Now $a = 27$). - b = b & c $\implies 5 \ \& \ 14 = 0101_2 \ \& \ 1110_2 = 0100_2 = 4$. (Now $b = 4$). 4. Compute Final Expression a + b + c: - $a = 27, b = 4, c = 14$. - $27 + 4 + 14 = 45$.

Final Output: 45

PYQ 5: Bitwise Accumulation Loop with AND & OR

Integer a = 7, b = 11, c = 0
For (Integer i = 1 to 3)
    c = c + ((a & i) ^ (b | i))
End For
Print c

Step-by-Step Execution Trace Table:

Iteration i a & i Binary & Value b \| i Binary & Value (a & i) ^ (b \| i) Calculation Updated c
Start - - - 0
i = 1 $0111_2 \ \& \ 0001_2 = 1$ $1011_2 \mid 0001_2 = 1011_2 = 11$ $0001_2 \oplus 1011_2 = 1010_2 = 10$ $0 + 10 = 10$
i = 2 $0111_2 \ \& \ 0010_2 = 2$ $1011_2 \mid 0010_2 = 1011_2 = 11$ $0010_2 \oplus 1011_2 = 1001_2 = 9$ $10 + 9 = 19$
i = 3 $0111_2 \ \& \ 0011_2 = 3$ $1011_2 \mid 0011_2 = 1011_2 = 11$ $0011_2 \oplus 1011_2 = 1000_2 = 8$ $19 + 8 = 27$

Final Output: 27

2. Recursion & Stack Unwinding PYQ Bank

PYQ 6: Double Recursive Call Tree Tracing

Integer fun(Integer n)
    If (n <= 0)
        Return 0
    End If
    If (n == 1)
        Return 2
    End If
    Return fun(n - 1) + fun(n - 2) + n
End Function
What is the return value of fun(4)?

Step-by-Step Stack Unwinding Trace: 1. Base Cases: - fun(0) = 0 - fun(1) = 2 2. Evaluate fun(2): - fun(2) = fun(1) + fun(0) + 2 = 2 + 0 + 2 = 4. 3. Evaluate fun(3): - fun(3) = fun(2) + fun(1) + 3 = 4 + 2 + 3 = 9. 4. Evaluate fun(4): - fun(4) = fun(3) + fun(2) + 4 = 9 + 4 + 4 = 17.

Final Output: 17

PYQ 7: Recursive Function with Static Variable State

Integer fun(Integer n)
    Static Integer count = 0
    If (n <= 0)
        Return count
    End If
    count = count + n
    Return fun(n - 1) + fun(n - 2)
End Function
What is the value returned by fun(3)?

Step-by-Step Call Order & State Unwinding: 1. fun(3): count updated $\implies 0 + 3 = 3$. Calls fun(2) + fun(1). 2. Left Child fun(2) executes first: - count updated $\implies 3 + 2 = 5$. Calls fun(1) + fun(0). - Left Child fun(1) executes: - count updated $\implies 5 + 1 = 6$. Calls fun(0) + fun(-1). - fun(0) returns current count $= 6$. - fun(-1) returns current count $= 6$. - fun(1) unwinds $\implies 6 + 6 = 12$. - Right Child fun(0) of fun(2) executes: - Returns current count $= 6$. - fun(2) unwinds $\implies 12 + 6 = 18$. 3. Right Child fun(1) of fun(3) executes: - count updated $\implies 6 + 1 = 7$. Calls fun(0) + fun(-1). - fun(0) returns current count $= 7$. - fun(-1) returns current count $= 7$. - fun(1) unwinds $\implies 7 + 7 = 14$. 4. Root fun(3) unwinds: - fun(3) = fun(2) + fun(1) = 18 + 14 = 32.

Final Output: 32

PYQ 8: Recursive Function with Bitwise XOR & Unwinding Step

Integer foo(Integer a, Integer b)
    If (b <= 0)
        Return a
    End If
    Return foo(a ^ b, b - 1) + b
End Function
What is the output of foo(5, 3)?

Step-by-Step Call Stack Recursion Trace: - Call 1: foo(5, 3) - $b = 3 > 0$. Bitwise XOR: $a \oplus b = 5 \oplus 3 = 0101_2 \oplus 0011_2 = 0110_2 = 6$. - Returns foo(6, 2) + 3. - Call 2: foo(6, 2) - $b = 2 > 0$. Bitwise XOR: $a \oplus b = 6 \oplus 2 = 0110_2 \oplus 0010_2 = 0100_2 = 4$. - Returns foo(4, 1) + 2. - Call 3: foo(4, 1) - $b = 1 > 0$. Bitwise XOR: $a \oplus b = 4 \oplus 1 = 0100_2 \oplus 0001_2 = 0101_2 = 5$. - Returns foo(5, 0) + 1. - Call 4: foo(5, 0) - $b = 0 \le 0$. Base Case Reached! Returns $a = 5$.

Stack Unwinding: - foo(5, 0) = 5 - foo(4, 1) = 5 + 1 = 6 - foo(6, 2) = 6 + 2 = 8 - foo(5, 3) = 8 + 3 = 11

Final Output: 11

PYQ 9: Alternating Parity Recursive Function

Integer solve(Integer n, Integer k)
    If (n <= 1)
        Return k
    End If
    If (n Mod 2 == 0)
        Return solve(n / 2, k + n)
    Else
        Return solve(n - 1, k * 2)
    End If
End Function
What is the return value of solve(6, 2)?

Step-by-Step Tail-Recursive Execution: 1. solve(6, 2): $n=6$ (Even) $\implies$ Returns solve(6 / 2, 2 + 6) $=$ solve(3, 8). 2. solve(3, 8): $n=3$ (Odd) $\implies$ Returns solve(3 - 1, 8 * 2) $=$ solve(2, 16). 3. solve(2, 16): $n=2$ (Even) $\implies$ Returns solve(2 / 2, 16 + 2) $=$ solve(1, 18). 4. solve(1, 18): $n=1 \le 1$ (Base Case) $\implies$ Returns $k = 18$.

Final Output: 18

3. Loop Execution & State Table PYQ Bank

PYQ 10: REPEAT-UNTIL Off-by-One Counter Verification

Integer i = 1, sum = 0
REPEAT
    sum = sum + i * i
    i = i + 1
UNTIL (i > 4)
Print sum

Step-by-Step Execution Trace: - Pass 1: sum = 0 + (1 * 1) = 1, i = 2. Condition (2 > 4) is FALSE $\to$ Continue. - Pass 2: sum = 1 + (2 * 2) = 5, i = 3. Condition (3 > 4) is FALSE $\to$ Continue. - Pass 3: sum = 5 + (3 * 3) = 14, i = 4. Condition (4 > 4) is FALSE $\to$ Continue. - Pass 4: sum = 14 + (4 * 4) = 30, i = 5. Condition (5 > 4) is TRUE $\to$ TERMINATE!

Final Output: 30 (Computes sum of squares $1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30$).

PYQ 11: Nested For Loops with Conditional Bitwise Accumulation

Integer sum = 0
For (Integer i = 1 to 3)
    For (Integer j = i to 3)
        If ((i + j) Mod 2 == 0)
            sum = sum + (i ^ j)
        Else
            sum = sum + (i & j)
        End If
    End For
End For
Print sum

Step-by-Step Execution Trace Table:

Outer i Inner j i + j Parity Evaluated Operation Operation Result Updated sum
1 1 2 Even $1 \oplus 1 = 0001_2 \oplus 0001_2$ 0 $0 + 0 = 0$
1 2 3 Odd $1 \ \& \ 2 = 0001_2 \ \& \ 0010_2$ 0 $0 + 0 = 0$
1 3 4 Even $1 \oplus 3 = 0001_2 \oplus 0011_2$ 2 $0 + 2 = 2$
2 2 4 Even $2 \oplus 2 = 0010_2 \oplus 0010_2$ 0 $2 + 0 = 2$
2 3 5 Odd $2 \ \& \ 3 = 0010_2 \ \& \ 0011_2$ 2 $2 + 2 = 4$
3 3 6 Even $3 \oplus 3 = 0011_2 \oplus 0011_2$ 0 $4 + 0 = 4$

Final Output: 4

PYQ 12: Digit Extraction Loop with Parity Branching

Integer n = 456, sum = 0, rem = 0
While (n > 0)
    rem = n Mod 10
    If (rem Mod 2 != 0)
        sum = sum + rem * 2
    Else
        sum = sum + rem
    End If
    n = n / 10
End While
Print sum

Step-by-Step Execution Trace Table:

Iteration Initial n rem = n Mod 10 rem Mod 2 != 0 Applied Operation Updated sum n = n / 10
Pass 1 456 6 False (Even) sum + rem $0 + 6 = 6$ 45
Pass 2 45 5 True (Odd) sum + rem * 2 $6 + (5 \times 2) = 16$ 4
Pass 3 4 4 False (Even) sum + rem $16 + 4 = 20$ 0
Pass 4 0 - $0 > 0$ False Loop Terminates 20 0

Final Output: 20

PYQ 13: Bitwise Right Shift & XOR Accumulator Loop

Integer x = 15, y = 9, z = 0
While (x > 0)
    z = z + (x & 1)
    x = x >> 1
    y = y ^ x
End While
Print z, y

Step-by-Step Execution Trace Table:

Iteration Initial x x & 1 Updated z x >> 1 (New x) y ^ x Calculation Updated y
Start 15 ($1111_2$) - 0 - - 9 ($1001_2$)
Pass 1 15 1 $0 + 1 = 1$ 7 ($0111_2$) $9 \oplus 7 = 1001_2 \oplus 0111_2 = 1110_2$ 14
Pass 2 7 1 $1 + 1 = 2$ 3 ($0011_2$) $14 \oplus 3 = 1110_2 \oplus 0011_2 = 1101_2$ 13
Pass 3 3 1 $2 + 1 = 3$ 1 ($0001_2$) $13 \oplus 1 = 1101_2 \oplus 0001_2 = 1100_2$ 12
Pass 4 1 1 $3 + 1 = 4$ 0 ($0000_2$) $12 \oplus 0 = 1100_2 \oplus 0000_2 = 1100_2$ 12
Pass 5 0 - Loop Ends 0 - 12

Final Output: 4 12

4. Pass-by-Reference & Array Tracing PYQ Bank

PYQ 14: Pass-by-Reference Modifying Outer Variables

Function modify(Integer &x, Integer y)
    x = x * 2
    y = y + 5
    Return x + y
End Function

Main:
    Integer a = 5, b = 10, res
    res = modify(a, b)
    Print a, b, res
End Main

Step-by-Step Execution Trace: 1. a is passed by reference (&x), so changes to x directly mutate variable a. 2. b is passed by value (y), so y is a local copy; variable b remains unaffected. 3. Inside modify: - x = x * 2 $\implies x = 5 \times 2 = 10$. Variable a becomes 10. - y = y + 5 $\implies y = 10 + 5 = 15$. Variable b stays 10. - Returns 10 + 15 = 25. 4. Printing a, b, res: - a = 10, b = 10, res = 25.

Final Output: 10 10 25

PYQ 15: Pass-by-Reference Bitwise Swap and Sum

Function update(Integer &a, Integer b, Integer &c)
    a = a ^ b
    b = b ^ a
    c = a + b
    Return a * b
End Function

Main:
    Integer x = 6, y = 4, z = 2, ans
    ans = update(x, y, z)
    Print x, y, z, ans
End Main

Step-by-Step Parameter Mutation Trace: 1. x is passed by reference (&a) $\implies$ modifications to a directly alter outer x. 2. y is passed by value (b) $\implies$ modifications to b do NOT affect outer y. 3. z is passed by reference (&c) $\implies$ modifications to c directly alter outer z. 4. Inside update: - a = a ^ b $\implies 6 \oplus 4 = 0110_2 \oplus 0100_2 = 0010_2 = 2$. (Outer x becomes 2). - b = b ^ a $\implies 4 \oplus 2 = 0100_2 \oplus 0010_2 = 0110_2 = 6$. (Local b $= 6$; outer y stays 4). - c = a + b $\implies 2 + 6 = 8$. (Outer z becomes 8). - Return a * b $\implies 2 \times 6 = 12$. 5. In Main: - x = 2, y = 4, z = 8, ans = 12.

Final Output: 2 4 8 12

PYQ 16: Array Traversal with Bit Shift & Conditional Accumulation

Integer A[5] = {4, 7, 12, 15, 8}
Integer total = 0
For (Integer i = 0 to 4)
    If ((A[i] & 1) == 0)
        total = total + (A[i] >> 1)
    Else
        total = total + (A[i] << 1)
    End If
End For
Print total

Step-by-Step Execution Trace Table:

Index i Array Element A[i] A[i] & 1 Check Parity Branch Shift Operation & Value Updated total
0 4 $4 \ \& \ 1 = 0$ Even (== 0) $4 \gg 1 = \lfloor 4/2 \rfloor = 2$ $0 + 2 = 2$
1 7 $7 \ \& \ 1 = 1$ Odd (!= 0) $7 \ll 1 = 7 \times 2 = 14$ $2 + 14 = 16$
2 12 $12 \ \& \ 1 = 0$ Even (== 0) $12 \gg 1 = \lfloor 12/2 \rfloor = 6$ $16 + 6 = 22$
3 15 $15 \ \& \ 1 = 1$ Odd (!= 0) $15 \ll 1 = 15 \times 2 = 30$ $22 + 30 = 52$
4 8 $8 \ \& \ 1 = 0$ Even (== 0) $8 \gg 1 = \lfloor 8/2 \rfloor = 4$ $52 + 4 = 56$

Final Output: 56

5. Operator Precedence & Conditional Logic Bank

PYQ 17: Bitwise AND vs XOR Operator Precedence Tracing

Integer p = 5, q = 3, r = 8
Integer res = (p + q * 2) ^ (r >> 2) & 7
Print res

Step-by-Step Precedence & Binary Evaluation: 1. Parentheses & Arithmetic: - (p + q * 2) $\implies 5 + (3 \times 2) = 5 + 6 = 11 = 1011_2$. - (r >> 2) $\implies 8 \gg 2 = 2 = 0010_2$. - Expression reduces to: 11 ^ 2 & 7. 2. Bitwise Operator Precedence (& higher than ^): - Evaluate 2 & 7 first: $0010_2 \ \& \ 0111_2 = 0010_2 = 2$. 3. Evaluate Bitwise XOR (^): - 11 ^ 2 $\implies 1011_2 \oplus 0010_2 = 1001_2 = 9$.

Final Output: 9

PYQ 18: Ternary Operator Evaluation with Relational Precedence

Integer x = 10, y = 20, z = 15, ans
ans = (x ^ y > z) ? ((y >> 2) + x) : ((x << 1) ^ z)
Print ans

Step-by-Step Precedence & Execution: 1. Evaluate Condition (x ^ y > z): - Relational operator > has higher precedence than bitwise ^. - y > z $\implies 20 > 15 \implies 1$ (True). - Expression becomes x ^ 1 $\implies 10 \oplus 1 = 1010_2 \oplus 0001_2 = 1011_2 = 11$. - Non-zero integer 11 evaluates to TRUE in boolean condition context. 2. Execute TRUE Expression ((y >> 2) + x): - y >> 2 $\implies 20 \gg 2 = 5$. - 5 + x $\implies 5 + 10 = 15$. 3. Assign & Print: - ans = 15.

Final Output: 15

6. Web-Researched CoCubes, Sanfoundry & GFG High-Frequency PYQ Expansion

PYQ 19: CoCubes Bitwise XOR Inequality Branching Trap

Integer a = 6, b = 8, c = 15
If ((b ^ a) < a)
    b = a
End If
Print a + b + c

Step-by-Step Binary Execution Trace: 1. Binary Conversions: - $a = 6 = 0110_2$ - $b = 8 = 1000_2$ - $c = 15 = 1111_2$ 2. Evaluate Condition (b ^ a) < a: - $b \oplus a = 1000_2 \oplus 0110_2 = 1110_2 = 14$. - Evaluate 14 < 6 $\implies$ FALSE. 3. Branch Decision: - The If condition fails, so b = a is skipped. b remains 8. 4. Compute Final Output a + b + c: - $6 + 8 + 15 = 29$.

State Table Summary: | Operation | Left Side | Right Side | Condition | Action Taken | Final Value | | :---: | :---: | :---: | :---: | :---: | :---: | | b ^ a | $1000_2 \oplus 0110_2 = 14$ | $a = 6$ | 14 < 6 (FALSE) | Skip b = a | b = 8 | | a + b + c | $6 + 8 + 15$ | - | - | Print Output | 29 |

Final Output: 29

PYQ 20: CoCubes Recursive Parameter Swapper & Accumulator

Function solve(Integer a, Integer b)
    If (a < b)
        Return solve(b, a)
    Else If (b != 0)
        Return (a + solve(a, b - 1))
    Else
        Return 0
    End If
End Function

What is the returned value of solve(8, 9)?

Step-by-Step Recursive Call Stack Unwinding Table: 1. Call 1: solve(8, 9) $\implies (8 < 9)$ is TRUE. Returns solve(9, 8) (Parameter order swapped so $a \ge b$). 2. Call 2 to Call 10 Winding Phase: solve(9, 8) calls 9 + solve(9, 7) down to solve(9, 0).

Call Level Invocation Condition / Action Returned Value / Expression Accumulated Value
Level 10 solve(9, 0) Base Case b == 0 Returns 0 0
Level 9 solve(9, 1) 9 + solve(9, 0) $9 + 0$ 9
Level 8 solve(9, 2) 9 + solve(9, 1) $9 + 9$ 18
Level 7 solve(9, 3) 9 + solve(9, 2) $9 + 18$ 27
Level 6 solve(9, 4) 9 + solve(9, 3) $9 + 27$ 36
Level 5 solve(9, 5) 9 + solve(9, 4) $9 + 36$ 45
Level 4 solve(9, 6) 9 + solve(9, 5) $9 + 45$ 54
Level 3 solve(9, 7) 9 + solve(9, 6) $9 + 54$ 63
Level 2 solve(9, 8) 9 + solve(9, 7) $9 + 63$ 72
Level 1 solve(8, 9) Returns solve(9, 8) Returns Level 2 result 72

Final Output: 72

PYQ 21: CoCubes Decrement Modification Loop Trap

Integer n
For (n = 3; n != 0; n--)
    Print n
    n = n - 1
End For

Step-by-Step State Table Trace: | Iteration | Initial n | Condition n != 0 | Body Action (Print n) | Internal Mutation (n = n - 1) | Post Loop Decrement (n--) | Value for Next Test | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | 1 | 3 | 3 != 0 (TRUE) | Prints 3 | n = 3 - 1 = 2 | n = 2 - 1 = 1 | 1 | | 2 | 1 | 1 != 0 (TRUE) | Prints 1 | n = 1 - 1 = 0 | n = 0 - 1 = -1 | -1 | | 3 | -1 | -1 != 0 (TRUE) | Prints -1 | n = -1 - 1 = -2 | n = -2 - 1 = -3 | -3 | | 4 | -3 | -3 != 0 (TRUE) | Prints -3 | n = -3 - 1 = -4 | n = -4 - 1 = -5 | -5 |

  • Key Observation: Notice how n skips 0 completely! It transitions from 1 $\to$ 0 inside body $\to$ -1 after loop header n--.
  • Termination Check: n != 0 will remain TRUE for all negative odd integers.

Final Output: Infinite Loop

PYQ 22: Sanfoundry REPEAT-UNTIL Shift & Mask Counter

Integer p = 10, q = 3, count = 0
REPEAT
    count = count + 1
    p = p >> 1
    q = q + (p & 1)
UNTIL (p <= 1)
Print count, p, q

Step-by-Step Variable State Table: | Pass | Initial p | count = count + 1 | p = p >> 1 | p & 1 | q = q + (p & 1) | UNTIL Test (p <= 1) | Evaluation | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | Start | 10 | 0 | - | - | 3 | - | - | | Pass 1 | 10 | 1 | $10 \gg 1 = \mathbf{5}$ (0101) | $5 \ \& \ 1 = \mathbf{1}$ | $3 + 1 = \mathbf{4}$ | (5 <= 1) | FALSE $\to$ Continue | | Pass 2 | 5 | 2 | $5 \gg 1 = \mathbf{2}$ (0010) | $2 \ \& \ 1 = \mathbf{0}$ | $4 + 0 = \mathbf{4}$ | (2 <= 1) | FALSE $\to$ Continue | | Pass 3 | 2 | 3 | $2 \gg 1 = \mathbf{1}$ (0001) | $1 \ \& \ 1 = \mathbf{1}$ | $4 + 1 = \mathbf{5}$ | (1 <= 1) | TRUE $\to$ TERMINATE |

  • Final Values: count = 3, p = 1, q = 5.

Final Output: 3 1 5

PYQ 23: GFG Bitwise Left Shift with Two's Complement NOT Operator

Integer a = 7, b = 2, c = 0
c = ((~a) << b) + (a ^ b)
Print c

Step-by-Step Binary State Table: | Step | Operation / Sub-expression | 8-Bit Binary Computation | Decimal Result | Notes | | :---: | :--- | :--- | :---: | :--- | | 1 | a | 0000 0111 | 7 | Input Value | | 2 | ~a | 1111 1000 | -8 | Bitwise NOT identity $\sim a = -(a + 1) = -8$ | | 3 | (~a) << b | $11111000_2 \ll 2 = \mathbf{1110\ 0000}_2$ | -32 | Shift left by 2 ($(-8) \times 4 = -32$) | | 4 | a ^ b | $00000111_2 \oplus 00000010_2 = \mathbf{0000\ 0101}_2$ | 5 | Bitwise XOR of 7 and 2 | | 5 | c = (-32) + 5 | $-32 + 5$ | -27 | Addition of components |

Final Output: -27

💡 Quick Summary Checklist for Exam Mastery

  1. Bitwise Shifts:
  2. Left Shift ($x \ll n$) $= x \times 2^n$
  3. Right Shift ($x \gg n$) $= \lfloor x / 2^n \rfloor$
  4. Bitwise Parity Check:
  5. $(x \ \& \ 1) == 0 \implies x$ is Even
  6. $(x \ \& \ 1) == 1 \implies x$ is Odd
  7. Bitwise Identity Properties:
  8. $x \oplus x = 0$
  9. $x \oplus 0 = x$
  10. $x \ \& \ x = x$
  11. $x \mid x = x$
  12. Pass-by-Reference:
  13. Parameter prefixed with & modifies original calling variable.
  14. Value parameter makes local copy, leaving outer variable intact.