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Prashant Mishra

Posted on

# Stack and Queue

Valid parenthesis
`TC: O(N)`

``````class Solution {
public boolean isValid(String s) {
Stack<Character> stack = new Stack<>();
char ch [] = s.toCharArray();
for(char i : ch){
if(stack.size()==0) {

stack.push(i);
System.out.println("push happened "+stack.peek());
}
else if(stack.peek()=='(' && i==')' || stack.peek()=='{' &&
i=='}' || stack.peek()=='[' && i ==']') {
System.out.println("here");
stack.pop();
}
else stack.push(i);
}
if(stack.size()==0) return true;
else return false;
}
}
``````

Next greater element than stack top
`TC: in worst case O(n^2), as we will iterate over nums2, and if all the elements are in descending order except the last element in nums2 (example: 7,6,5,4,3,8), then at last index we will go to else part and while loop that will run for n-1 times hence TC will be n*(n-1)`

``````class Solution {
public int[] nextGreaterElement(int[] nums1, int[] nums2) {
HashMap<Integer,Integer> map = new HashMap<>();
Stack<Integer> stack = new Stack<>();
//we will get next greater element of all the element in the nums2,
// by that we will easily be able to get the next greater element of all the
//elements in nums1;
for(int i = 0;i<nums2.length;i++){
if(stack.isEmpty() || stack.peek() > nums2[i]){
stack.push(nums2[i]);
}
else{
// below while condition means that we have found
// greater value than stack top, now we will have to keep on removing
//stack top to make sure that we have got next greater value of stack top
while(!stack.isEmpty() && nums2[i] > stack.peek()){
map.put(stack.pop(),nums2[i]);
}
stack.push(nums2[i]);
}
}
//now we have got the next greater value of all the elements in the nums2
//we can easily get the next greater value of nums1
int result[] = new int[nums1.length];
for(int i =0;i<result.length;i++){
result[i]= map.getOrDefault(nums1[i],-1);// if next greater does not return then default value will be -1

}
return result;
}
}
``````

Implement stack using single queue (design quetion)
`You can easily guess the TC:`

``````// we can easily do it with two queue,
// we will do it with one queue
class MyStack {
Queue<Integer> q ;
public MyStack() {
}

public void push(int x) {
if(q.size()>1){
int size = q.size();
int index =0;
while(index!=size-1){//just re-push size-1 value again it will insure that the value
// that needs to be poped is at head;
index++;
}
}
}

public int pop() {
return q.remove();
}

public int top() {
return q.peek();
}

public boolean empty() {
return q.isEmpty();
}
}

/**
* Your MyStack object will be instantiated and called as such:
* MyStack obj = new MyStack();
* obj.push(x);
* int param_2 = obj.pop();
* int param_3 = obj.top();
* boolean param_4 = obj.empty();
*/

``````

Sort stack in descending order

TC: o(N)

``````import java.util.*;
public class Solution {

public static void sortStack(Stack<Integer> stack) {
Stack<Integer> stack2 = new Stack<>();
while(!stack.isEmpty()) {
int top = stack.pop();
while(!stack2.isEmpty() && stack2.peek() < top){
stack.push(stack2.pop());
}
stack2.push(top);
}
while(!stack2.isEmpty()) stack.push(stack2.pop());
}
}
``````