Stack and Queue

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;
    }
}

//recusive : TLE
class Solution {
    public boolean checkValidString(String s) {
        return possible(0,s,0);
    }
    public boolean possible(int index, String s, int count){
        //base case
        if(count<0) return false;
        if(index == s.length()) return count ==0;
        char c = s.charAt(index);
        if(c =='('){
            return possible(index+1,s,count+1);
        }
        else if(c ==')'){
            return possible(index+1,s,count-1);
        }
        else{
            return possible(index+1,s,count+1) || possible(index+1,s,count) || possible(index+1,s,count-1);
        }
    }
}

////////////////////dp memorization://////////////


//note: in valid parathesis if only one type of brackets are involved like ( and ) or [ and ] or { and } the use count variable instead of using stack
//this will save space, just increment count if character is ( and decrement count if the character is ),
//one edge case is if at any point count < 0 then it is not a valid string
//example : ()) => count -1, ()())( => at index 4 count will become -1, return false here only else count will be come 0 again at the last index which is still not valid
//keep the above edge case in mind
class Solution {
    public boolean checkValidString(String s) {
        int dp[][] = new int[s.length()][s.length()+1];
        for(int d[] : dp) Arrays.fill(d,-1);
        return possible(0,s,0,dp);
    }
    public boolean possible(int index, String s, int count,int dp[][]){
        //base case
        if(count<0) return false;
        if(index == s.length()) return count ==0;
        if(dp[index][count]!=-1) return dp[index][count]==1;
        char c = s.charAt(index);
        boolean result = false;
        if(c =='('){//increment count
            result  = possible(index+1,s,count+1,dp);
        }
        else if(c ==')'){// decrement count
            result = possible(index+1,s,count-1,dp);
        }
        else{// it is * > assume it to be ( or assume it to be '' or assume it to be )
            result  = possible(index+1,s,count+1,dp) || possible(index+1,s,count,dp) || possible(index+1,s,count-1,dp);
        }
        dp[index][count] = result ? 1:0;
        return result;
    }
}

class Solution {
    //this problem can easily be converted into the longest consecutive 1's in the array, we just have to mark indexes of valid pairs () as 1
    //Example: ((()) = [0,1,1,1,1], or (() = [0,1,1]
    //then simply count the number of consecutive ones
    public int longestValidParentheses(String s) {
        int arr[] = new int[s.length()];
        Stack<Pair<Character,Integer>> stack = new Stack();// < Character,Integer> to keep track of index of character being pushed in the stack
        for(int i = 0;i<s.length();i++){
            char c = s.charAt(i);
            if(c =='(') stack.push(new Pair(c,i));
            else if(c ==')' && !stack.isEmpty() && stack.peek().getKey() =='('){ // if valid pair found , just mark their indexes as 1
                Pair<Character,Integer>  p  = stack.pop();
                arr[i]=1;// '')' index i.e current index marked as 1
                arr[p.getValue()] = 1;// corresponding index of closing parathesis '(' marked as 1
            }
            else stack.push(new Pair(c,i));
        }
        int max = 0;
        int currentMax = 0;
        //finally count the number of 1's in the array
        for(int i =0;i<s.length();i++){
            if(arr[i] ==0){
                currentMax = 0;
            }
            else currentMax++;
            max = Integer.max(max,currentMax);
        }
        return max;
    }
}

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;
    }
}

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

    public void push(int x) {
        q.add(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;
                q.add(q.remove());
                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();
 */

import java.util.*;
public class Solution {

    public static void sortStack(Stack<Integer> stack) {
        // Write your code here. 
        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());
    }
}