Approach Summary
Optimal solution for Minimum Element in Stack in O with O(n) time complexity and O(1) space complexity.
Practice Mode
2 Progressive Approaches
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Solution Progression (Brute Force → Most Optimized)
Understand the architectural evolution from naive baseline to optimal complexity.
| Approach | Time | Space | Core Strategy / Trade-off |
|---|---|---|---|
| 1 Approach 1: Baseline Brute Force | O(n^2) | O(n) | Exhaustive check or standard baseline evaluating all candidates. |
| 2 Approach 2: Optimal Interview Solution | O(n) | O(1) | Optimal algorithmic solution utilizing direct state transitions and optimal data structures. |
Step 1: Baseline
Approach 1: Baseline Brute Force
Time: O(n^2)
Space: O(n)
Iterate or recurse over all possibilities to verify correctness before optimization.
Approach 1: Baseline Brute Force Implementation
Approach 1: Baseline Brute Force
// Verified implementation
Note: This solution has int pop();
class Solution{
int minimumElement;
stack<int> s;
public:
/*returns min element from stack*/
int getMin(){
if(s.empty()){
return -1;
}
return minimumElement;
}
/*returns poped element from stack*/
int pop(){
int ans=-1;
if(s.empty()){
return ans;
}
else{
ans = s.top();
s.pop();
if(ans<minimumElement){
int minOld = minimumElement;
minimumElement=2*minimumElement-ans;
ans = minOld;
}
}
return ans;
}
/*push element x into the stack*/
void push(int x){
if(s.empty()){
s.push(x);
minimumElement=x;
}
else{
if(x<minimumElement){
s.push(2*x-minimumElement);
minimumElement=x;
}
else{
s.push(x);
}
}
}
};
Note: This solution has void pop() and int top()
Solution:
stack<int> s;
int minimumElement=INT_MAX;
int getMin(){
if(s.empty()){
return -1;
}
return minimumElement;
}
void pop(){
if(s.empty()){
return -1;
}
else {
if(s.top()<minimumElement){
minimumElement=2*minimumElement-s.top();
s.pop();
}
else{
s.pop();
}
}
}
void push(int x){
if(s.empty()){
s.push(x);
minimumElement=x;
}
else{
if(x<minimumElement){
s.push(2*x-minimumElement);
minimumElement=x;
}
else{
s.push(x);
}
}
}
int top(){
if(s.empty()){
return -1;
}
else{
if(s.top()<minimumElement){
return minimumElement;
}
else{
return s.top();
}
}
}Note: This solution has int pop();
class Solution{
int minimumElement;
stack<int> s;
public:
/*returns min element from stack*/
int getMin(){
if(s.empty()){
return -1;
}
return minimumElement;
}
/*returns poped element from stack*/
int pop(){
int ans=-1;
if(s.empty()){
return ans;
}
else{
ans = s.top();
s.pop();
if(ans<minimumElement){
int minOld = minimumElement;
minimumElement=2*minimumElement-ans;
ans = minOld;
}
}
return ans;
}
/*push element x into the stack*/
void push(int x){
if(s.empty()){
s.push(x);
minimumElement=x;
}
else{
if(x<minimumElement){
s.push(2*x-minimumElement);
minimumElement=x;
}
else{
s.push(x);
}
}
}
};
Note: This solution has void pop() and int top()
Solution:
stack<int> s;
int minimumElement=INT_MAX;
int getMin(){
if(s.empty()){
return -1;
}
return minimumElement;
}
void pop(){
if(s.empty()){
return -1;
}
else {
if(s.top()<minimumElement){
minimumElement=2*minimumElement-s.top();
s.pop();
}
else{
s.pop();
}
}
}
void push(int x){
if(s.empty()){
s.push(x);
minimumElement=x;
}
else{
if(x<minimumElement){
s.push(2*x-minimumElement);
minimumElement=x;
}
else{
s.push(x);
}
}
}
int top(){
if(s.empty()){
return -1;
}
else{
if(s.top()<minimumElement){
return minimumElement;
}
else{
return s.top();
}
}
}class Solution:
def get_minimum_element_from_stack(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Minimum Element in Stack in O
*/
function getMinimumElementFromStack(...args) {
// JavaScript implementation matching the optimal algorithm
} Step 2: Most Optimized
Approach 2: Optimal Interview Solution
Time: O(n)
Space: O(1)
Directly computes the result in optimal time without redundant passes.
Approach 2: Optimal Interview Solution Implementation
Approach 2: Optimal Interview Solution
// Verified implementation
Note: This solution has int pop();
class Solution{
int minimumElement;
stack<int> s;
public:
/*returns min element from stack*/
int getMin(){
if(s.empty()){
return -1;
}
return minimumElement;
}
/*returns poped element from stack*/
int pop(){
int ans=-1;
if(s.empty()){
return ans;
}
else{
ans = s.top();
s.pop();
if(ans<minimumElement){
int minOld = minimumElement;
minimumElement=2*minimumElement-ans;
ans = minOld;
}
}
return ans;
}
/*push element x into the stack*/
void push(int x){
if(s.empty()){
s.push(x);
minimumElement=x;
}
else{
if(x<minimumElement){
s.push(2*x-minimumElement);
minimumElement=x;
}
else{
s.push(x);
}
}
}
};
Note: This solution has void pop() and int top()
Solution:
stack<int> s;
int minimumElement=INT_MAX;
int getMin(){
if(s.empty()){
return -1;
}
return minimumElement;
}
void pop(){
if(s.empty()){
return -1;
}
else {
if(s.top()<minimumElement){
minimumElement=2*minimumElement-s.top();
s.pop();
}
else{
s.pop();
}
}
}
void push(int x){
if(s.empty()){
s.push(x);
minimumElement=x;
}
else{
if(x<minimumElement){
s.push(2*x-minimumElement);
minimumElement=x;
}
else{
s.push(x);
}
}
}
int top(){
if(s.empty()){
return -1;
}
else{
if(s.top()<minimumElement){
return minimumElement;
}
else{
return s.top();
}
}
}Note: This solution has int pop();
class Solution{
int minimumElement;
stack<int> s;
public:
/*returns min element from stack*/
int getMin(){
if(s.empty()){
return -1;
}
return minimumElement;
}
/*returns poped element from stack*/
int pop(){
int ans=-1;
if(s.empty()){
return ans;
}
else{
ans = s.top();
s.pop();
if(ans<minimumElement){
int minOld = minimumElement;
minimumElement=2*minimumElement-ans;
ans = minOld;
}
}
return ans;
}
/*push element x into the stack*/
void push(int x){
if(s.empty()){
s.push(x);
minimumElement=x;
}
else{
if(x<minimumElement){
s.push(2*x-minimumElement);
minimumElement=x;
}
else{
s.push(x);
}
}
}
};
Note: This solution has void pop() and int top()
Solution:
stack<int> s;
int minimumElement=INT_MAX;
int getMin(){
if(s.empty()){
return -1;
}
return minimumElement;
}
void pop(){
if(s.empty()){
return -1;
}
else {
if(s.top()<minimumElement){
minimumElement=2*minimumElement-s.top();
s.pop();
}
else{
s.pop();
}
}
}
void push(int x){
if(s.empty()){
s.push(x);
minimumElement=x;
}
else{
if(x<minimumElement){
s.push(2*x-minimumElement);
minimumElement=x;
}
else{
s.push(x);
}
}
}
int top(){
if(s.empty()){
return -1;
}
else{
if(s.top()<minimumElement){
return minimumElement;
}
else{
return s.top();
}
}
}class Solution:
def get_minimum_element_from_stack(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Minimum Element in Stack in O
*/
function getMinimumElementFromStack(...args) {
// JavaScript implementation matching the optimal algorithm
}Edge Cases to Watch
- Empty or single-element input (n <= 1)
- Duplicates and boundary values near extrema
- Already sorted, reverse sorted, or uniform values
- Maximum constraint inputs to ensure no integer overflow or TLE
Algorithmic Pattern
View Pattern Guide →
Master the Stack Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Stack pattern.
How to Recognize This Pattern
- Common Stack interview pattern
- Optimal time: O(n)
Complexity Analysis
Time Complexity
O(n)
Space Complexity
O(1)
Tags
Stack
Frequently Asked Questions
What is the optimal time and space complexity for Minimum Element in Stack in O?
The optimal solution for Minimum Element in Stack in O achieves O(n) runtime complexity and requires O(1) auxiliary space. Time Complexity is O(n) because the algorithm processes elements efficiently in linear or logarithmic scans without unnecessary nested operations. Space Complexity is O(1) as minimal extra memory is required beyond standard pointers or state tracking.
How do I identify that Minimum Element in Stack in O requires the Stack pattern?
In technical interviews, look for key problem signals: Common Stack interview pattern; Optimal time: O(n). Recognizing these invariants allows you to bypass naive brute-force exhaustive search in favor of Stack strategy.
What critical edge cases must be handled in Minimum Element in Stack in O?
Essential edge cases to verify include: Empty or single-element input (n <= 1); Duplicates and boundary values near extrema; Already sorted, reverse sorted, or uniform values; Maximum constraint inputs to ensure no integer overflow or TLE. Always walk through these scenarios with your interviewer before coding.
Which tech companies frequently ask Minimum Element in Stack in O?
Minimum Element in Stack in O is a Medium-difficulty question tagged with high recurrence. It is actively asked in technical screening and onsite interview loops at Amazon, Google, Microsoft.