February LeetCoding Challenge 2021
Day 26
Validate Stack Sequences
PROBLEM STATEMENT:
Given two sequences
Example 1
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.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() -> 1Example 2
strong>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 ofpopped
.pushed
andpopped
have distinct values.
Explanation
The simplest way to do it is to just have a stack and see if the given operations are feasible.We traverse through the
pushed
array and also we have a pointer i
that moves over the popped array. If the stack is empty or the top element of the stack is not equal to popped[i]
we push elements from pushed
array to the stack. Else we pop the element from the stack and increment i
until the stack is empty or the top element is not equal to popped[i].
At the end if stack is empty that means the given operations are feasible.ALGORITHM:
- Maintain a stack.
- Traverse the pushed array from left to right and have a variable i initialised to zero.
- Push the current element to the stack.
- While stack is not empty and st.top() == popped[i], pop the top element of stack and increment i.
- After traversing the entire pushed array return true if the stack is empty else return false.
- Time complexity : O(n)
- Space complexity : O(n)

code
C++
class Solution {
public:
bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {
stack<int> st;
int j=0;
for(int i=0;i<pushed.size();++i){
st.push(pushed[i]);
while(!st.empty() && st.top() == popped[j]){
st.pop();
++j;
}
}
return st.empty();
}
};
PYTHON
class Solution:
def validateStackSequences(self, pushed: List[int], popped: List[int]) -> bool:
st = []
i=0
for num in pushed:
st.append(num)
while len(st) and st[-1] == popped[i]:
st.pop()
i += 1
return not len(st)