Min Stack Design in DSA C - Execution Trace

Concept Flow - Min Stack Design

Push x

↓

Push x to main stack

↓

Is min stack empty or x <= top of min stack?

No→Do nothing

Yes↓

Push x to min stack

↓

Done

↓

Pop

↓

Pop from main stack

↓

Is popped value == top of min stack?

No→Do nothing

Yes↓

Pop from min stack

↓

Done

↓

GetMin

↓

Return top of min stack

Push adds to main stack and min stack if needed; Pop removes from both if top matches; GetMin returns min stack top.

Execution Sample

DSA C

push(5)
push(3)
push(7)
pop()
getMin()

Push values 5, 3, 7; pop top; get minimum value in stack.

Execution Table

Step	Operation	Main Stack	Min Stack	Pointer Changes	Visual State
1	push(5)	5	5	main_top=0, min_top=0	Main: 5 -> null; Min: 5 -> null
2	push(3)	5 -> 3	5 -> 3	main_top=1, min_top=1	Main: 5 -> 3 -> null; Min: 5 -> 3 -> null
3	push(7)	5 -> 3 -> 7	5 -> 3	main_top=2, min_top=1	Main: 5 -> 3 -> 7 -> null; Min: 5 -> 3 -> null
4	pop()	5 -> 3	5 -> 3	main_top=1, min_top=1	Main: 5 -> 3 -> null; Min: 5 -> 3 -> null
5	getMin()	5 -> 3	5 -> 3	No change	Min top is 3

💡 Execution stops after getMin returns current minimum 3.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	Final
main_stack	empty	[5]	[5,3]	[5,3,7]	[5,3]	[5,3]
min_stack	empty	[5]	[5,3]	[5,3]	[5,3]	[5,3]
main_top	-1	0	1	2	1	1
min_top	-1	0	1	1	1	1

Key Moments - 3 Insights

Why do we push to min stack only when the new value is less or equal to the current min?

Because min stack keeps track of minimums. See step 3 in execution_table: 7 is not pushed to min stack since 7 > current min 3.

What happens to min stack when we pop a value that is not the current minimum?

Min stack stays unchanged. See step 4: popped 7 from main stack but min stack remains same.

How does getMin() return the minimum value efficiently?

It returns the top of min stack instantly without scanning main stack, as shown in step 5.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what is the state of the min stack?

A [5, 3, 7]

B [5, 3]

C [3]

D [7]

💡 Hint

Check the 'Min Stack' column at step 3 in execution_table.

At which step does the main stack size decrease?

A Step 4

B Step 2

C Step 3

D Step 5

💡 Hint

Look at 'Main Stack' column changes in execution_table.

If we push(2) after step 4, what happens to min stack?

A Min stack pops top

B Min stack remains unchanged

C 2 is pushed to min stack

D Min stack resets

💡 Hint

Recall min stack pushes new value if it is <= current min (see concept_flow).

Concept Snapshot

Min Stack Design:
- Use two stacks: main and min.
- Push: add to main; if new <= min top, push to min.
- Pop: remove from main; if popped == min top, pop min.
- getMin: return top of min stack instantly.
- Keeps track of minimum in O(1) time.

Full Transcript

Min Stack uses two stacks to track values and minimums. When pushing, the value goes to main stack. If it is smaller or equal to current minimum, it also goes to min stack. When popping, if the popped value equals min stack top, pop min stack too. getMin returns the top of min stack, giving minimum in constant time. This design ensures efficient minimum retrieval while maintaining stack behavior.

Practice

1. What is the main purpose of a Min Stack data structure?

easy

A. To reverse the order of elements

B. To retrieve the minimum element in constant time

C. To store elements in a queue fashion

D. To sort elements in ascending order

Solution

1. Step 1: Understand Min Stack functionality

   A Min Stack is designed to allow retrieval of the minimum element quickly, usually in constant time.

2. Step 2: Compare with other options

   Sorting or queue behavior is not the main goal of a Min Stack; it focuses on minimum element retrieval.

3. Final Answer:

   To retrieve the minimum element in constant time -> Option B

4. Quick Check:

   Min Stack = fast minimum retrieval [OK]

Hint: Min Stack always tracks minimum element fast [OK]

Common Mistakes:

• Confusing Min Stack with sorting algorithms
• Thinking it behaves like a queue
• Assuming it reverses elements

2. Which of the following is the correct way to declare a stack node structure for a Min Stack in C?

easy

A. typedef struct Node { int value; struct Node* next; int min; } Node;

B. typedef struct Node { int value; int min; } Node;

C. typedef struct Node { int value; struct Node* prev; } Node;

D. typedef struct Node { int value; struct Node* next; } Node;

Solution

1. Step 1: Identify required fields for Min Stack node

   A Min Stack node needs to store the value, a pointer to the next node, and the minimum value at that node.

2. Step 2: Match with options

   typedef struct Node { int value; struct Node* next; int min; } Node; includes value, next pointer, and min field, which is correct for Min Stack implementation.

3. Final Answer:

   typedef struct Node { int value; struct Node* next; int min; } Node; -> Option A

4. Quick Check:

   Node stores value, next, and min [OK]

Hint: Min Stack node needs value, next pointer, and min [OK]

Common Mistakes:

• Omitting the min field in node
• Using prev pointer instead of next
• Not using struct pointer for next

3. Consider the following C code snippet for a Min Stack push operation:

void push(Node** top, int x) {
    Node* newNode = malloc(sizeof(Node));
    newNode->value = x;
    if (*top == NULL) {
        newNode->min = x;
    } else {
        newNode->min = (x < (*top)->min) ? x : (*top)->min;
    }
    newNode->next = *top;
    *top = newNode;
}

What will be the minimum value after pushing elements 3, 5, 2 in this order on an empty stack?

medium

A. 2

B. 3

C. 5

D. Error

Solution

1. Step 1: Push 3 on empty stack

   Stack top is NULL, so min = 3.

2. Step 2: Push 5

   Compare 5 with current min 3; min remains 3.

3. Step 3: Push 2

   Compare 2 with current min 3; min updates to 2.

4. Final Answer:

   2 -> Option A

5. Quick Check:

   Minimum after pushes = 2 [OK]

Hint: Min updates only if new value is smaller [OK]

Common Mistakes:

• Not updating min on push
• Confusing min with last pushed value
• Assuming min resets on each push

4. The following C function is intended to pop an element from a Min Stack. Identify the error:

int pop(Node** top) {
    if (*top == NULL) return -1;
    int val = (*top)->value;
    free(*top);
    *top = (*top)->next;
    return val;
}

medium

A. Not checking if stack is empty

B. Returning value before freeing node

C. Accessing (*top)->next after freeing *top

D. Missing return statement

Solution

1. Step 1: Analyze order of operations

   The code frees *top, then accesses (*top)->next, which is invalid after free.

2. Step 2: Correct order

   Should save next pointer before freeing current node to avoid invalid access.

3. Final Answer:

   Accessing (*top)->next after freeing *top -> Option C

4. Quick Check:

   Free before next access causes error [OK]

Hint: Save next pointer before freeing node [OK]

Common Mistakes:

• Freeing node before accessing next
• Ignoring NULL check
• Returning value after free without saving

5. You want to design a Min Stack that supports push, pop, and getMin all in O(1) time using two stacks. Which approach correctly implements getMin?

hard

A. Use a queue to track minimum values

B. Sort the main stack on every push to keep minimum on top

C. Traverse the main stack on every getMin call to find minimum

D. Maintain a second stack that stores the minimum values; push new min or duplicate current min on push

Solution

1. Step 1: Understand two-stack Min Stack design

   One stack stores all values; the second stack keeps track of minimums for O(1) retrieval.

2. Step 2: Evaluate options

   Maintain a second stack that stores the minimum values; push new min or duplicate current min on push correctly describes pushing new minimum or current minimum duplicate onto second stack.

3. Final Answer:

   Maintain a second stack that stores the minimum values; push new min or duplicate current min on push -> Option D

4. Quick Check:

   Two stacks: one for values, one for mins [OK]

Hint: Use second stack to track mins for O(1) getMin [OK]

Common Mistakes:

• Sorting stack on every push (slow)
• Searching stack on getMin (not O(1))
• Using queue instead of stack