Union basics - Time & Space Complexity

Time Complexity: Union basics

O(1)

Understanding Time Complexity

When working with unions in C, it's helpful to understand how the program's steps grow as data changes.

We want to see how the time to run code with unions changes when the input size changes.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

#include <stdio.h>

union Data {
    int i;
    float f;
    char str[20];
};

void printUnion(union Data d, int type) {
    if (type == 0) printf("%d\n", d.i);
    else if (type == 1) printf("%f\n", d.f);
    else printf("%s\n", d.str);
}

int main() {
    union Data data;
    data.i = 10;
    printUnion(data, 0);
    return 0;
}
    

This code defines a union with different types and prints one value based on a type indicator.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Single function call to print a union member.
• How many times: Called once in this example.

How Execution Grows With Input

Since the code prints one value from the union, the steps do not increase with input size.

Input Size (n)	Approx. Operations
10	About 1 print operation
100	Still about 1 print operation
1000	Still about 1 print operation

Pattern observation: The number of steps stays the same no matter the input size.

Final Time Complexity

Time Complexity: O(1)

This means the program takes the same amount of time no matter how big the input is.

Common Mistake

[X] Wrong: "Accessing different union members takes more time as the union size grows."

[OK] Correct: Accessing any member of a union is a direct operation and does not depend on the size of the union or input.

Interview Connect

Understanding how unions work and their time behavior helps you explain memory use and efficiency clearly in interviews.

Self-Check

"What if we added a loop to print multiple union values? How would the time complexity change?"

Practice

1. What is the main characteristic of a union in C?

easy

A. Unions can only store integers

B. Each member has its own separate memory

C. All members share the same memory location

D. Unions automatically initialize all members

Solution

1. Step 1: Understand union memory sharing

   A union allocates one shared memory block for all its members, so they overlap.

2. Step 2: Compare with struct memory allocation

   Unlike structs, unions do not allocate separate memory for each member.

3. Final Answer:

   All members share the same memory location -> Option C

4. Quick Check:

   Union memory sharing = All members share the same memory location [OK]

Hint: Remember: union members overlap in memory [OK]

Common Mistakes:

• Thinking each member has separate memory
• Assuming unions initialize all members
• Believing unions only store one data type

2. Which of the following is the correct way to define a union named Data with an int and a float member?

easy

A. union Data { int i; float f; };

B. struct Data { int i; float f; };

C. union Data ( int i; float f; );

D. union Data = { int i; float f; };

Solution

1. Step 1: Recall union syntax

   The correct syntax uses the keyword 'union', the name, and braces with members inside.

2. Step 2: Check each option

   union Data { int i; float f; }; uses correct syntax with braces and semicolons. Options A, B, and C have syntax errors or wrong keywords.

3. Final Answer:

   union Data { int i; float f; }; -> Option A

4. Quick Check:

   Correct union syntax = union Data { int i; float f; }; [OK]

Hint: Use 'union Name { members; };' syntax [OK]

Common Mistakes:

• Using struct instead of union
• Using parentheses instead of braces
• Assigning values in the definition

3. What will be the output of the following C code?

union Number {
  int i;
  float f;
};

int main() {
  union Number num;
  num.i = 65;
  printf("%c", num.f);
  return 0;
}

medium

A. A

B. 65

C. Compilation error

D. Undefined behavior or garbage character

Solution

1. Step 1: Understand union member sharing

   Setting num.i = 65 stores integer bits in union memory.

2. Step 2: Printing num.f as char

   Printing num.f with %c treats float bits as a char, causing unpredictable output.

3. Final Answer:

   Undefined behavior or garbage character -> Option D

4. Quick Check:

   Accessing different union member types = Undefined behavior or garbage character [OK]

Hint: Accessing different union members can cause garbage output [OK]

Common Mistakes:

• Expecting integer 65 to print as 'A'
• Assuming union members convert automatically
• Ignoring type mismatch in printf

4. Identify the error in this union usage:

union Data {
  int i;
  float f;
};

int main() {
  union Data d;
  d.i = 10;
  d.f = 3.14;
  printf("%d", d.i);
  return 0;
}

medium

A. Using %d to print a union member is invalid

B. Assigning to d.f overwrites d.i value

C. Cannot assign to multiple union members sequentially

D. Union must be initialized before use

Solution

1. Step 1: Analyze assignments to union members

   Assigning d.i then d.f means d.f overwrites the same memory used by d.i.

2. Step 2: Understand printf output

   Printing d.i after assigning d.f prints the overwritten value, not 10.

3. Final Answer:

   Assigning to d.f overwrites d.i value -> Option B

4. Quick Check:

   Union members share memory, last assignment wins = Assigning to d.f overwrites d.i value [OK]

Hint: Last assigned union member overwrites previous values [OK]

Common Mistakes:

• Thinking union members hold values independently
• Believing %d cannot print union members
• Assuming union needs initialization

5. You want to store either an int, a float, or a char in the same memory space and print the stored value correctly. Which approach using a union and an additional variable is best?

hard

A. Use a union for data and an enum to track the active member type

B. Use a union and print all members every time

C. Use separate variables for each type without union

D. Use a union and always print the int member

Solution

1. Step 1: Understand union limitation

   Union stores one member at a time, so we must know which member is active.

2. Step 2: Use enum to track active member

   An enum variable can track which type is stored, so printing uses correct format.

3. Final Answer:

   Use a union for data and an enum to track the active member type -> Option A

4. Quick Check:

   Union + enum for type tracking = Use a union for data and an enum to track the active member type [OK]

Hint: Track active union member with enum for correct printing [OK]

Common Mistakes:

• Printing all union members blindly
• Not tracking which member is active
• Using separate variables wastes memory