no need to reverse the stack ,we have to return an array as answer , so we can get the stack size , create an array of that size and pop and directly start inserting into the array from backward direction(last index).
00:04 Solving the problem of asteroid collisions in the given array. 02:20 Illustrates asteroid collisions and elimination process. 04:29 Using stack data structure to track element traversal 06:35 Asteroid collisions simulation using stack data structure 08:42 Using stack or list in asteroid collisions 10:47 Demonstration of asteroid collisions in a stack 12:59 Handling asteroid collisions using stack and queue 15:29 Explaining time and space complexity
solved by my own in around 33 minutes. Was confused at the starting like how are the collision happening then read the description carefully and i got it
This is only for my understanding pls ignore. Each asteroid is travelling at same speed. We are traversing from left to right in an array so if asteroid travelling from left to right it means it would not collide at that particular of timeframe which is kind of equivalent to the index of an array and if asteroid is coming from right to left it would collide because we are traversing from left to right (if there ) So we need to take negative number into consideration at that time and see if any asteroid is coming from left to right and if it is coming then it collides but once the asteroid collides (which was coming from right to left) it exits our timeframe. eg :- -2, -1, 1, 2 -2 at index 0 comes from right to left it should collide but there is no asteroid coming from left so nothing collides -> -3 -1 at index 1 comes from right to left it should collide but there is no asteroid coming from left so no collision -> -2 1 at index 2 going from left to right it shouldn't collide because we are travelling from left to right -> 1 -> +1 2 at index 3 doesn't collide -> +2
@@SoulFrmTitanic i dont remember much, just thought to keep removing elements from stack until the current asteroid is destroyed (if the asteroid is weaker than the previous coz otherwise just remove the current asteroid and break out)
def sign(num): return num//abs(num) def asteroidCollision(arr): n = len(arr) i = 0 lst = [] while(i < n): # will process asteroid i if len(lst) > 0 and lst[-1] > 0 and arr[i] < 0: # collision will happen if abs(lst[-1]) == abs(arr[i]): # tie lst.pop() i += 1 elif abs(lst[-1]) > abs(arr[i]): # last wins i += 1 else: # opp wins lst.pop() else: # no collision lst.append(arr[i]) i += 1 return lst print(asteroidCollision([-2,-1,1,2]))
class Solution { public int[] asteroidCollision(int[] asteroids) { int n = asteroids.length; Stack st = new Stack(); for(int i=0;i0) { st.push(asteroids[i]); }else{ while(!st.isEmpty() && st.peek()>0 && st.peek()
Java Solution TC : O(2N), O(N) for traversing and another O(N) for pushing and popping at max 'N' elements onto the stack. SC : O(2N), O(N) is for using external list data structure and another O(N) for converting the list into array to return the answer. class Solution { public int[] asteroidCollision(int[] asteroids) { // List to store the resulting asteroids after collisions List list = new ArrayList(); // Loop through each asteroid in the array for (int i = 0; i < asteroids.length; i++) { // If the current asteroid is moving to the right (positive direction) if (asteroids[i] > 0) { // Add it directly to the list (no collision with left-moving asteroids) list.add(asteroids[i]); } // If the current asteroid is moving to the left (negative direction) else { // Check for collisions with right-moving asteroids in the list while (!list.isEmpty() && list.get(list.size() - 1) > 0 && list.get(list.size() - 1) < Math.abs(asteroids[i])) { // Remove the smaller right-moving asteroid since it collides and explodes list.remove(list.size() - 1); } // If the list is empty or the last asteroid in the list is also moving to the left, // or there are no more right-moving asteroids to collide with if (list.isEmpty() || list.get(list.size() - 1) < 0) { // Add the current left-moving asteroid to the list list.add(asteroids[i]); } // If the last asteroid in the list is the same size but moving in the opposite direction else if (list.get(list.size() - 1) == Math.abs(asteroids[i])) { // Both asteroids destroy each other (equal in magnitude), so remove the last one list.remove(list.size() - 1); } // If the current left-moving asteroid is smaller, it is destroyed by the larger right-moving asteroid, // and we do not add it to the list (handled implicitly by not adding it to the list). } } // Convert the List of remaining asteroids to an array to return as the result int[] result = new int[list.size()]; for (int i = 0; i < list.size(); i++) { result[i] = list.get(i); } return result; } }
koi check karke batao na kya error hai isme. test cases pass nahi ho rahe class Solution { public: vector asteroidCollision(vector& asteroids) { vector st; int n = asteroids.size(); for(int i = 0; i0) st.push_back(asteroids[i]); else{ while(!st.empty() && st.back()>0 && st.back()
Solved by myself without watching video : class Solution { public: vector asteroidCollision(vector& nums) { int n = nums.size(); stack st; vector ans; for (int i=0 ; i 0){ st.push(nums[i]); } if (i != n &&(st.empty() && nums[i] < 0 || (!st.empty() && st.top() < 0)) && f){ st.push(nums[i]); } } if (st.empty()){ return {}; } while (!st.empty()){ ans.push_back(st.top()); st.pop(); } reverse(ans.begin() , ans.end()); return ans; } }; TBH Got cooked by a ton of edge cases !
the first question that i made by myself without any help, getting more confidence day by day!!!!
08:43 is very important to understand for an assumption that we need to insert everytime there is a positive element in array.
thank you so much i literally was confused for straight up hours
thank you!
@@Rohan-cn4ji same :)
no need to reverse the stack ,we have to return an array as answer , so we can get the stack size , create an array of that size and pop and directly start inserting into the array from backward direction(last index).
Damn! Good observation
that will take O(stack size) time, just use vector as stack.
@@randomshorts5200 but is it a good practice?
thats what we are trying to avoid.
Self solved, thanks again Striver for building up the intuitions in prev lectures🔥
00:04 Solving the problem of asteroid collisions in the given array.
02:20 Illustrates asteroid collisions and elimination process.
04:29 Using stack data structure to track element traversal
06:35 Asteroid collisions simulation using stack data structure
08:42 Using stack or list in asteroid collisions
10:47 Demonstration of asteroid collisions in a stack
12:59 Handling asteroid collisions using stack and queue
15:29 Explaining time and space complexity
solved by my own in around 33 minutes. Was confused at the starting like how are the collision happening then read the description carefully and i got it
Same
Then provide code
@@avengergirl_0464 vector asteroidCollision(int N, vector &arr) {
// code here
stackst;
int n=N;
for(int i=0;i=0 ||
(st.top()0)) st.push(arr[i]);
else{
while(!st.empty() && st.top()>0 && arr[i]
@@avengergirl_0464
class Solution {
public:
vector asteroidCollision(vector& asteroids) {
stack st;
for(int i=0;i0 and asteroids[i]=top and top>0 and !st.empty()){
if(top==absval){
st.pop();
break;
}
st.pop();
if(!st.empty())
top = st.top();
}
if(top!=absval and (asteroids[i]top))
st.push(asteroids[i]);
}
else
st.push(asteroids[i]);
}
}
vector res(st.size());
for(int i=res.size()-1;i>=0;i--){
res[i] = st.top();
st.pop();
}
return res;
}
};
@@avengergirl_0464
class Solution {
public:
vector asteroidCollision(vector& asteroids) {
vector ans;
int n=asteroids.size();
stack stk;
stk.push(asteroids[0]);
for(int i=1;i0) stk.push(asteroids[i]);
else
{
while (!stk.empty() && stk.top() > 0 && stk.top() < abs(asteroids[i]))
{
stk.pop();
}
if(!stk.empty() && asteroids[i]
these type of question the bruteforce solution is the difficult one
THANK YOU STRIVER !
i figured out the solution in just 5 minutes, pretty easy question if you could figure out that you have to use a stack.
We have to add one more condition in the last els if, and that is when the
list.back()
this condition is written in else block therefore arr[i] is already < 0 why are you checking it again? so (st.empty || st.back() < 0) will suffice.
Yep this will require
great explanation bhaiya
Thank you
Thank you so much !
Understood 👍
This is only for my understanding pls ignore.
Each asteroid is travelling at same speed.
We are traversing from left to right in an array so if asteroid travelling from left to right it means it would not collide at that particular of timeframe which is kind of equivalent to the index of an array and if asteroid is coming from right to left it would collide because we are traversing from left to right (if there )
So we need to take negative number into consideration at that time and see if any asteroid is coming from left to right and if it is coming then it collides but once the asteroid collides (which was coming from right to left) it exits our timeframe.
eg :- -2, -1, 1, 2
-2 at index 0 comes from right to left it should collide but there is no asteroid coming from left so nothing collides -> -3
-1 at index 1 comes from right to left it should collide but there is no asteroid coming from left so no collision -> -2
1 at index 2 going from left to right it shouldn't collide because we are travelling from left to right -> 1 -> +1
2 at index 3 doesn't collide -> +2
Most of the time i got the idea whats happening and solve it through brute force but unable to optimize it may be 3-4 out of 10 time able to do so.
class Solution {
public:
vector asteroidCollision(vector& asteroids) {
stack st;
for(int i = 0; i < asteroids.size(); i++) {
bool flag = false;
while(!st.empty() && asteroids[i] < 0 && st.top() > 0) {
if (abs(asteroids[i]) > abs(st.top())) {
st.pop();
}
else if (abs(asteroids[i]) == abs(st.top())) {
st.pop();
flag = true;
break;
}
else {
flag = true;
break;
}
}
if (!flag) {
st.push(asteroids[i]);
}
}
vector ans;
while (!st.empty()) {
ans.insert(ans.begin(), st.top());
st.pop();
}
return ans;
}
};
I have doubt with input
-2 -1 1 2, what would be the output
in leetcode expected output is the same as input
Yes, as 1st two move in left (stack is empty so push) and then rest of the element are moving right, opposite direction, therefore no collision
@@AbhishekGupta-zf2sw yup this!
everything will get added in the stack
thanks bhaiya
CPP Code I came up with:
class Solution {
public:
vector asteroidCollision(vector& asteroids) {
stack st;
for(const auto& it: asteroids){
bool exploded = false;
while(!(st.empty()) && it 0){
if(abs(it) > abs(st.top())){
st.pop();
}else if(abs(it) == abs(st.top())){
st.pop();
exploded = true;
break;
}else{
exploded = true;
break;
}
}
if(!exploded){
st.push(it);
}
}
vector ans(st.size());
for(int i=st.size()-1;i>=0;i--){
ans[i] = st.top();
st.pop();
}
return ans;
}
};
bro thats a great one!! Can you just tell me what was the intuition behind this beautiful approach??
Would be very helpful for me!! 🤝
@@SoulFrmTitanic i dont remember much, just thought to keep removing elements from stack until the current asteroid is destroyed (if the asteroid is weaker than the previous coz otherwise just remove the current asteroid and break out)
@@shreyxnsh.14 achaa , ok bhai
def sign(num):
return num//abs(num)
def asteroidCollision(arr):
n = len(arr)
i = 0
lst = []
while(i < n):
# will process asteroid i
if len(lst) > 0 and lst[-1] > 0 and arr[i] < 0:
# collision will happen
if abs(lst[-1]) == abs(arr[i]): # tie
lst.pop()
i += 1
elif abs(lst[-1]) > abs(arr[i]): # last wins
i += 1
else: # opp wins
lst.pop()
else:
# no collision
lst.append(arr[i])
i += 1
return lst
print(asteroidCollision([-2,-1,1,2]))
Thanks
UNDERSTOOD;
C++ solution with stack :
vector asteroidCollision(int N, vector &arr) {
// code here
stackst;
int n=N;
for(int i=0;i=0 ||
(st.top()0)) st.push(arr[i]);
else{
while(!st.empty() && st.top()>0 && arr[i]
The implementation is hard for this problem
understood
Understood
class Solution {
public int[] asteroidCollision(int[] asteroids) {
Stack stack = new Stack();
for(int i=0;i0){
stack.push(asteroids[i]);
} else {
while(!stack.isEmpty()){
int top = stack.peek();
if(top=0;i--){
ansArray[i] = stack.pop();
}
return ansArray;
}
}
//
solved this without watching the video in 15 mins in o(n) w
You are so talented bro
chaalak bro
nice
This is simple , TC - O(4N)~N , SC - O(2N)
vector asteroidCollision(vector& asteroids) {
int n=asteroids.size();
stack st;
vector nums;
for(int i=0;i0 && valabs(top))
val=val;
else if(abs(val)
😊
we could just use a vector instead of a list
sir notes upload krdo site par
class Solution {
public int[] asteroidCollision(int[] asteroids) {
int n = asteroids.length;
Stack st = new Stack();
for(int i=0;i0)
{
st.push(asteroids[i]);
}else{
while(!st.isEmpty() && st.peek()>0 && st.peek()
Java Solution
TC : O(2N), O(N) for traversing and another O(N) for pushing and popping at max 'N'
elements onto the stack.
SC : O(2N), O(N) is for using external list data structure and another O(N) for converting the
list into array to return the answer.
class Solution {
public int[] asteroidCollision(int[] asteroids) {
// List to store the resulting asteroids after collisions
List list = new ArrayList();
// Loop through each asteroid in the array
for (int i = 0; i < asteroids.length; i++) {
// If the current asteroid is moving to the right (positive direction)
if (asteroids[i] > 0) {
// Add it directly to the list (no collision with left-moving asteroids)
list.add(asteroids[i]);
}
// If the current asteroid is moving to the left (negative direction)
else {
// Check for collisions with right-moving asteroids in the list
while (!list.isEmpty() && list.get(list.size() - 1) > 0 &&
list.get(list.size() - 1) < Math.abs(asteroids[i])) {
// Remove the smaller right-moving asteroid since it collides and explodes
list.remove(list.size() - 1);
}
// If the list is empty or the last asteroid in the list is also moving to the left,
// or there are no more right-moving asteroids to collide with
if (list.isEmpty() || list.get(list.size() - 1) < 0) {
// Add the current left-moving asteroid to the list
list.add(asteroids[i]);
}
// If the last asteroid in the list is the same size but moving in the opposite direction
else if (list.get(list.size() - 1) == Math.abs(asteroids[i])) {
// Both asteroids destroy each other (equal in magnitude), so remove the last one
list.remove(list.size() - 1);
}
// If the current left-moving asteroid is smaller, it is destroyed by the larger right-moving asteroid,
// and we do not add it to the list (handled implicitly by not adding it to the list).
}
}
// Convert the List of remaining asteroids to an array to return as the result
int[] result = new int[list.size()];
for (int i = 0; i < list.size(); i++) {
result[i] = list.get(i);
}
return result;
}
}
vector asteroidCollision(vector& asteroids) {
int n=asteroids.size();
stack st;
vector nums;
for(int i=0;i0 && valabs(top))
val=val;
else if(abs(val)
koi check karke batao na kya error hai isme. test cases pass nahi ho rahe
class Solution {
public:
vector asteroidCollision(vector& asteroids) {
vector st;
int n = asteroids.size();
for(int i = 0; i0) st.push_back(asteroids[i]);
else{
while(!st.empty() && st.back()>0 && st.back()
for loop should be from 0 to n
what's the code of 3:05 (when -3 is getting eliminated) anyone please??
Cz there is 7 before which is a positive and greater than absolute of -3 i.e 3
Solved by myself without watching video :
class Solution {
public:
vector asteroidCollision(vector& nums) {
int n = nums.size();
stack st;
vector ans;
for (int i=0 ; i 0){
st.push(nums[i]);
}
if (i != n &&(st.empty() && nums[i] < 0 || (!st.empty() && st.top() < 0)) && f){
st.push(nums[i]);
}
}
if (st.empty()){
return {};
}
while (!st.empty()){
ans.push_back(st.top());
st.pop();
}
reverse(ans.begin() , ans.end());
return ans;
}
};
TBH Got cooked by a ton of edge cases !
can you explain in this case [-19,-18, 20] Why is the answer [-19,-18, 20] and not [20].
Bcoz, 1st 2 elements are moving to the left and the last element is moving to the right. So they won't collide
because -19 is going left and 20 is going right so they can never collide
Understood
understood