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Solution 2: Instant Minimum Access

Approach Explanation

We maintain two stacks: one for the actual values and another for the minimum values encountered so far. When pushing a value, we push the current minimum onto the second stack.

Step-by-Step Logic

  1. push(val): Always push val to stack1. For stack2, push min(val, stack2[-1]) if stack2 is not empty, otherwise push val.
  2. pop(): Pop from both stacks.
  3. top(): Return the top of stack1.
  4. getMin(): Return the top of stack2.

Complexity

  • Time Complexity: O(1) for all operations.
  • Space Complexity: O(N).

Code

class MinStack:
    def __init__(self):
        self.stack = []
        self.min_stack = []

    def push(self, val: int) -> None:
        self.stack.append(val)
        if not self.min_stack or val <= self.min_stack[-1]:
            self.min_stack.append(val)
        else:
            self.min_stack.append(self.min_stack[-1])

    def pop(self) -> None:
        self.stack.pop()
        self.min_stack.pop()

    def top(self) -> int:
        return self.stack[-1]

    def getMin(self) -> int:
        return self.min_stack[-1]