 # Stack¶

• push(), pop(), isEmpty() and peek() all take O(1)
• follow LIFO/FILO order.
• real life -> stack of plates

### Applications of stack:¶

• Balancing of symbols
• Infix to Postfix / Prefix conversion
• Redo-undo
• Forward and backward feature in web browsers
• Used in many algorithms like Tower of Hanoi, tree traversals, stock span problem, histogram problem.
• Other applications can be Backtracking, Knight tour problem, rat in a maze, N queen problem and sudoku solver
• In Graph Algorithms like Topological Sorting and Strongly Connected Components

### Implementation¶

#### Array¶

Pros: Easy to implement. Memory is saved as pointers are not involved. Cons: It is not dynamic. It doesn’t grow and shrink depending on needs at runtime.

``````#define MAX 1000

class Stack {
int top;

public:
int a[MAX]; // Maximum size of Stack

Stack() { top = -1; }
bool push(int x);
int pop();
int peek();
bool isEmpty();
};

bool Stack::push(int x)
{
if (top >= (MAX - 1)) {
cout << "Stack Overflow";
return false;
}
else {
a[++top] = x;
cout << x << " pushed into stack\n";
return true;
}
}

int Stack::pop()
{
if (top < 0) {
cout << "Stack Underflow";
return 0;
}
else {
int x = a[top--];
return x;
}
}
int Stack::peek()
{
if (top < 0) {
cout << "Stack is Empty";
return 0;
}
else {
int x = a[top];
return x;
}
}

bool Stack::isEmpty()
{
return (top < 0);
}

int main()
{
class Stack s;
s.push(10);
s.push(20);
s.push(30);
cout << s.pop() << " Popped from stack\n";

return 0;
}
``````

Pros: The linked list implementation of stack can grow and shrink according to the needs at runtime. Cons: Requires extra memory due to involvement of pointers.

``````class StackNode {
public:
int data;
StackNode* next;
};

StackNode* newNode(int data)
{
StackNode* stackNode = new StackNode();
stackNode->data = data;
stackNode->next = NULL;
return stackNode;
}

int isEmpty(StackNode* root)
{
return !root;
}

void push(StackNode** root, int data)
{
StackNode* stackNode = newNode(data);
stackNode->next = *root;
*root = stackNode;
cout << data << " pushed to stack\n";
}

int pop(StackNode** root)
{
if (isEmpty(*root))
return INT_MIN;
StackNode* temp = *root;
*root = (*root)->next;
int popped = temp->data;
free(temp);

return popped;
}

int peek(StackNode* root)
{
if (isEmpty(root))
return INT_MIN;
return root->data;
}

int main()
{
StackNode* root = NULL;

push(&root, 10);
push(&root, 20);
push(&root, 30);

cout << pop(&root) << " popped from stack\n";

cout << "Top element is " << peek(root) << endl;

return 0;
}
``````

# Problems¶

• reverse a string -> push push... pop pop (better -> just swap the pointers)
• 2 stacks in an array:
• divide space into two (space inefficient)
• start from extremes
• check for balanced parenthesis -> push if open the pop if closed and check the equality
• next grater element
• two loops (o(n^2))
• using stacks (O(n))
``````Push the first element to stack.
Pick rest of the elements one by one and follow the following steps in loop.
Mark the current element as next.
If stack is not empty, compare top element of stack with next.
If next is greater than the top element,Pop element from stack. next is the next greater element for the popped element.
Keep popping from the stack while the popped element is smaller than next. next becomes the next greater element for all such popped elements
Finally, push the next in the stack.
After the loop in step 2 is over, pop all the elements from stack and print -1 as next element for them.
``````

Time Complexity: O(n). The worst case occurs when all elements are sorted in decreasing order. If elements are sorted in decreasing order, then every element is processed at most 4 times.

Initially pushed to the stack. Popped from the stack when next element is being processed. Pushed back to the stack because the next element is smaller. Popped from the stack in step 3 of algorithm.

in the same order :

``````void printNGE(int arr[], int n)
{
stack<int> s;

int arr1[n];

for (int i = n - 1; i >= 0; i--)
{
while (!s.empty() && s.top() <= arr[i])
s.pop();

if (s.empty())
arr1[i] = -1;
else
arr1[i] = s.top();

s.push(arr[i]);
}

for (int i = 0; i < n; i++)
cout << arr[i] << " ---> " << arr1[i] << endl;
}
``````