The problem can be found at the following link: Question Link
To reverse the stack, we can use an auxiliary queue to store the elements temporarily. Here's the step-by-step approach:
- Create a temporary queue.
- Pop each element from the given stack and push it into the temporary queue.
- Pop each element from the temporary queue and push it back into the original stack.
- Time Complexity:
O(N)
whereN
is the number of elements in the stack. - Auxiliary Space Complexity:
O(N)
whereN
is the number of elements in the stack.
class Solution {
public:
void Reverse(stack<int> &St) {
queue<int> temp;
while (!St.empty()) {
temp.push(St.top());
St.pop();
}
while (!temp.empty()) {
St.push(temp.front());
temp.pop();
}
}
};
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