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feat: update solution to lcci problem: No.03.02 #2323

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Feb 7, 2024
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feat: update solution to lcci problem: No.03.02 #2323

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11 changes: 10 additions & 1 deletion lcci/03.02.Min Stack/README_EN.md
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Expand Up @@ -28,7 +28,16 @@ minStack.getMin(); --&gt; return -2.</pre>

## Solutions

### Solution 1
### Solution 1: Double Stack

We use two stacks to implement this, where `stk1` is used to store data, and `stk2` is used to store the current minimum value in the stack. Initially, `stk2` stores a very large value.

- When we push an element `x` into the stack, we push `x` into `stk1`, and push `min(x, stk2[-1])` into `stk2`.
- When we pop an element from the stack, we pop the top elements of both `stk1` and `stk2`.
- When we want to get the top element in the current stack, we just need to return the top element of `stk1`.
- When we want to get the minimum value in the current stack, we just need to return the top element of `stk2`.

For each operation, the time complexity is $O(1)$, and the space complexity is $O(n)$.

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