155. Min Stack

Reference: LeetCode
Difficulty: Easy

Problem

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

• push(x) – Push element x onto stack.
• pop() – Removes the element on top of the stack.
• top() – Get the top element.
• getMin() – Retrieve the minimum element in the stack.

Example:

MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin();   --> Returns -3.
minStack.pop();
minStack.top();      --> Returns 0.
minStack.getMin();   --> Returns -2.

Follow up: All operations in constant time.

Analysis

Two Stacks (violates the condition)

class MinStack {
  
  private Stack<Integer> stack = new Stack<>();
  private Stack<Integer> buffer = new Stack<>();
  private int min = Integer.MAX_VALUE;

  public void push(int x) {
    stack.push(x);
    if (x < min) {
      min = x;
    }
  }

  // takes O(N) time
  public void pop() { // assume not empty
    int val = stack.pop();
    if (val == min) {
      // update min (find the new min)
      int newMin = Integer.MAX_VALUE; // consider size = 1, min should be reset
      // put aside
      while (stack.size() > 0) {
        int x = stack.pop();
        newMin = Math.min(newMin, x);
        buffer.push(x);
      }
      min = newMin;
      // put it back
      while (temp.size() > 0) {
        stack.push(temp.pop());
      }
    }
  }

  public int top() { // assume not empty
    return stack.peek();
  }

  // Required O(1)
  public int getMin() { // assume not empty
    return min;
  }
}

Time: pop() takes $O(N)$ time.
Space: $O(N)$

Backup Old Min

• When pushing x (x <= min), put old min to the stack before pushing x (the new min).
• When popping x (x == min), pop one more element to get the old min.

Note: x <= min in push is very important. Consider the case pushing [4, 5, 4] (it fails if x < min is applied).

class MinStack {
  
  private Stack<Integer> stack = new Stack<>();
  private int min = Integer.MAX_VALUE;

  public void push(int x) { // there is always a MAX_VALUE at the bottom
    if (x <= min) {
      stack.push(min);
      min = x;
    }
    stack.push(x);
  }

  public void pop() { // assume not empty
    if (stack.pop() == min) {
      min = stack.pop();
    }
  }

  public int top() { // assume not empty
    return stack.peek();
  }

  public int getMin() { // assume not empty
    return min;
  }
}

Time: $O(1)$
Space: $O(N)$