Validate Stack Sequences - Leetcode 946 - Python
Key Takeaways
Solves Validate Stack Sequences using Python on Leetcode 946
Full Transcript
hey everyone welcome back and let's write some more neat code today so today let's solve the problem validate stack sequences we're given two integer arrays pushed and popped pushed is supposed to represent the sequence of elements that we are pushing onto a stack so this is the order that we would push elements onto a stack like this so like one two three four five but it's not just that simple because actually we do have some popped operations as well so in this case we are going to pop the fourth element first that's going to be our first pop operation So based on that let me ask you should this pop be executed before or after we add this Five Element probably before because we can't really pop the four if the five is in the way we only pop from the top of a stack so just running through the simulation here we're gonna add one we're gonna push two we're gonna push three we're gonna push four we pushed all four these but now we're gonna pop four so we pop that now we are looking to pop five but five isn't here it's not at the top of the stack so the only way we can pop it now is if we go ahead and push it and thankfully there is a five here available for us but let me ask you what would happen if this five did not exist popped is supposed to pop a five but clearly it doesn't exist here and there's no more elements left for us to push so that brings us to the ultimate question that we're trying to answer here are these sequence of operations valid for a stack we're told that we have to push these elements in this order and we're told that we have to pop elements in this order and after all of these operations we should be left with an empty stack well technically in the prompt they tell us the stack is initially empty but they also below the problem clarify that pushed and popped should be of equal length therefore by the end of this we should be left with an empty stack that's kind of why it's important to read the constraints at the bottom of the problem that's a little Pro tip and in this example if we did not have this 5 over here let's assume to make these two arrays equal we also didn't have an element over here this would be impossible but for the real example where we do have this five we push this 5 and then we are looking to pop it so we pop the five then we're looking to pop a three thankfully that is at the top of our stack so we pop the three we then pop the two we then pop the one and then we are left with an empty stack so in this case these are valid operations for a stack so we can return true in the other case we would have returned false one last Edge case you might be thinking of is what if we got to a point Let's ignore this but let's say we pushed a 4 onto the stack that's great we can go ahead and pop from the top of our stack the four but what if we had a sequence where we pushed then a three and then we pushed another four how do you know whether we're supposed to pop this four or we're supposed to pop this four with this operation thankfully this case will never happen because they do clarify for us that each of the values in each of these arrays is going to be distinct that's why this problem is possible if they ask you to solve it without this constraint it would not really be possible at least not without a little bit of backtracking I think but in this case we are able to solve this problem with a single data structure where the memory complexity is going to be o of n we are pushing and popping each of the elements so the time complexity is also going to be Big O of n so let's code it up now so I'm going to first initialize the stack well it's going to be empty we're going to go through every number in our pushed array we know for each number we are just gonna push it to the end of the stack in Python you can do append so of course we're going to push before before we try to pop but after each element is pushed this could be the element that we are looking to pop how do we know if it is well we look at the beginning of our popped array so probably we should keep track of what position we're at in the popped array let's use a pointer for that I'm going to go ahead and call it I so what are we looking to check we're looking to check if the element in popped at index I is equal to the last element in our stack in Python you can take negative one to check the last element or you can do the length minus one and if this is the case then we are looking to pop from this stack there's a couple edge cases here though we might not just want to pop a single element we might want to pop multiple consecutive elements in a row so instead of making this an if statement let's make it a while loop but now that we've made it a while loop and I actually forgot to mention this we also after we pop the element from this index we probably want to increment our eye pointer as well but what happens if we increment this so much that it goes out of bounds then this is going to give us an index out of bounds error to ensure that doesn't happen let's make sure to check that I is less than the length of pop and let's do that before we even check this because if this evaluates to false this won't even execute there's actually one more Edge case here can you spot it what happens if our stack becomes empty then this would throw an index out of bounds error so before we do this let's also check that our stack is non-empty you can just check that like this in Python at least now we won't get any errors and this will do exactly what we want it to do but by the end of this Loop how do we know if this was valid or not well our stack should be empty so we can actually just return not stack if our stack is empty this evaluates to false the opposite of that is going to be true so we return true if it's non-empty we will end up returning false so let's run the code to make sure that it works and as you can see yes it does and it's pretty efficient if this was helpful please like And subscribe if you're preparing for coding interviews check out neatcode.io it has a ton of free resources to help you prepare thanks for watching and hopefully I'll see you pretty soon
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3:40 - Coding Explanation
leetcode 946
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Coding Explanation
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