15. Delete middle element of a stack

The problem can be found at the following link: Question Link

My Approach

We can utilize an additional stack to temporarily store the elements while popping until we reach the middle element in the original stack. Here are the steps involved:

• Determine the number of elements on top of the middle element by dividing the sizeOfStack by 2, and assign the result to the variable cnt.

• Iterate cnt times and temporarily store the elements in the newStack.

• Pop the middle element from the original stack s.

• Re-push the elements stored in the newStack back into the original stack s.

Time and Auxiliary Space Complexity

• Time Complexity: O(N), where N is the number of elements in the stack.

• Auxiliary Space Complexity: O(N), as we are using an auxiliary stack to temporarily store elements.

Code (C++)

class Solution {
public:
    void deleteMid(stack<int>& s, int sizeOfStack) {
        stack<int> newStack;
        int cnt = sizeOfStack / 2; // number of elements present on the top of mid element
        
        while (cnt--) {
            newStack.push(s.top());
            s.pop();
        }
        
        s.pop(); // Mid Element
        
        while (!newStack.empty()) {
            s.push(newStack.top());
            newStack.pop();
        }
    }
};

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