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Validate Stack Sequences

Updated: Mar 25, 2021

Given two sequences pushed and popped with distinct values, return true if and only if this could have been the result of a sequence of push and pop operations on an initially empty stack.


Example 1:

Input: pushed = [1,2,3,4,5], popped = [4,5,3,2,1]
Output: true
Explanation: We might do the following sequence: push(1), push(2), push(3), push(4), pop() -> 4, push(5), pop() -> 5, pop() -> 3, pop() -> 2, pop() -> 1 

Example 2:

Input: pushed = [1,2,3,4,5], popped = [4,3,5,1,2]
Output: false
Explanation: 1 cannot be popped before 2. 

Constraints:

  • 0 <= pushed.length == popped.length <= 1000

  • 0 <= pushed[i], popped[i] < 1000

  • pushed is a permutation of popped.

  • pushed and popped have distinct values.

Solution:

Approach 1:

Time Complexity: O(N) Space Complexity: O(N)


class Solution {
    public boolean validateStackSequences(int[] pushed, int[] popped) {
       Stack<Integer> stack = new Stack<>();
        int i=0;
        for(int x:pushed)
        {
            stack.push(x);
            while(!stack.isEmpty() && stack.peek()==popped[i])
            {
                stack.pop();
                i++;
            }
        }
        return stack.isEmpty();
    }
}


Approach 2:

Time Complexity: O(N) Space Complexity: O(1)


class Solution {
    public boolean validateStackSequences(int[] pushed, int[] popped) {
        int i=0,j=0;
        for(int x:pushed)
        {
            pushed[i++]=x;
            while(i>0 && pushed[i-1]==popped[j])
            {
                --i;
                j++;
            }
        }
        return i==0;
    }
}


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