Next Greater Element | Animation | Coding Interview
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- Опубліковано 3 жов 2024
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Want to land a software engineering job in the IT industry? This course - 'Visualizing Data Structures and Algorithms' is here to help. The course walks you through multiple Java algorithms, data structures problems, and their solutions with step by step visualizations, so that you are actually learning instead of blindly memorizing solutions.
The course covers in and outs of Data Structures and Algorithms in Java. Java is used as the programming language in the course. Students familiar with Javascript, Python, C#, C++, C, etc will also get to learn concepts without any difficulty. The implementation of various Algorithms and Data Structures have been demonstrated and implemented through animated slides. It covers many interview room questions on Algorithms and Data Structures. The questions and solutions are demonstrated by -
1. Animated slide. (To make visualization of algorithms faster)
2. Coding algorithm on IDE.
The course covers topics such as -
0. Algorithm Analysis
1. Arrays
2. Matrix
3. Singly Linked List
4. Doubly Linked List
5. Circular Singly Linked List
6. Stacks
7. Queues
8. Binary Tree
9. Binary Search Tree
10. Graphs
11. Priority Queues and Heaps
12. Recursion
13. Searching
14. Sorting
15. Strings
16. Trie Data Structure
17. Dynamic Programming
and many more ...
#dsa #algorithms #coding
![[Java] Leetcode 503. Next Greater Element II [Stack & Queue #10]](http://i.ytimg.com/vi/BGEOnmu3PjI/mqdefault.jpg)
![[UA] NAVI проти G2 Esports | Blast Premier Fall Final](http://i.ytimg.com/vi/QWlIm4FGESQ/mqdefault.jpg)
Best series for DS & Algo. I am also 10 yrs java exp guy. Was looking for DS & Algo free course over UA-cam with java implementation and found this. Hats Off To You Man...Excellent Work. GOD BLESS YOU :)
Thanks !!!
Hi 10 yrs java exp guy. I didn't see your comment on previous video I thought you gave up on this playlist. Glad to see you here again )))
@@ascar66 :)
level is increasing. Very helpful. Thanks
welcome !!!
Excellent work Dinesh describing the solution using a stack. I'm still struggling to fully understand the use of the stack without memorizing the solution. I think this technique can be used in other solutions as well. A "brute-force" two 'for' loop solution is easy, but this is much harder. Need to keep working on this...
//yeah i tried the two for loop approach, that works as well
int m = 0;
int arr[] = { 4, 7, 3, 4, 8, 1 };
int result[] = new int[arr.length];
for (int i = 0; i < arr.length; i++) {
for (int j = i + 1; j < arr.length; j++) {
if (arr[i] < arr[j]) {
result[m] = arr[j];
break;
}
else {
result[m] = -1;
}
}
m++;
}
// for the last element
result[arr.length-1]=-1;
Thanks a lot sir!!! . Keep uploading more videos like this.............
Sure !!!
I think you should do Implementation to this video it is much better when we have both animation and implementation , BTW THANKS FOR ALL THE VIDEOS THIS IS BEST EXPLENATION I COULD FIND FOR ADS
Welcome
Great visualization and explanation.
Thanks !!!
Please upload operating system videos in java
operating system videos in Java ????
@@itsdineshvaryani various algorithm as well as codes like Peterson’s, dekker,etc
ok
What is time complexity for the solution?
Hello Sir. I'm following your DSA playlist consistently. Here when you described about the problem I'm just paused the video and I thought about the solution with pen and paper. after getting an idea I just implemented it to code. I solved this problem in such way and it still taking only one while loop while with two pointers, which is O(n) Time complexity. The output what i'm getting is correct.
result is = [7, 8, 4, 8, -1, -1];
The two pointers concept I've learned from your DSA course on youtube. This algorithm also giving correct result with O(n) TC
what is your opinion about this algo, kindly correct me if I'm wrong.
public int[] nextGreaterElement(int[] arr) {
int[] result = new int[arr.length];
//two pointer
int i = 0;
int j = i + 1;
int k = arr.length - 1;
int notExists = -1;
while (i < arr.length) {
// edge cases
if (j >= arr.length)
{
result[i] = notExists;
break;
}
if (arr[i] > arr[j] && j == k)
{
result[i] = notExists;
i++;
j = i + 1;
continue;
}
//assigning the next greater element to the result array
if (arr[i] < arr[j])
{
result[i] = arr[j];
i++;
j = i + 1;
}
else
{
j++;
}
}
return result;
}
I was actually thinking about the same code but in edge cases i was doing mistake so thanks for the code
bro its not O(n)..... calculate the complexity of this :- [8,7,6,5,4,3,2,1,9]
Very well explained
thanks !!!
Hi, For [2,4,75,8,76,90,77,54] input should be passed in this code output should be wrong.
this metthod doesnt pass all test cases and had to use int i = 2* arr.length-1 and arr[i%n]
Sir by much time will this course be completed
I will keep adding to this course ... DSA is very vast and never ending topic ...
@@itsdineshvaryani ok sir for sure
Why did we not sort the array in ascending order ? or do we need to maintain the order of the nos?
If we sort the array, the order of elements will be change and we will not be able to find any elements next greater to right !!!
Thank You so much sir !
Sir I practice a lot on GFG, Hacker rank but still struggle to solve medium level problems and build logic. Please help. How should I prepare for fresher role interviews? Please Help?
Even I struggle to solve ... the idea still remains in practice ... practice and practice ... Be focused ... Be better than yesterday and soon after few months you will see change ... Dnt loose hope ... Keep solving problems ...
@@itsdineshvaryani thanks sir surely i will
we can solve this by using 2 for loops and just arrays right.....without using stacks..
nested for loops will increase the TC
Using stack to solve this problem is hard, why don't we make things easy and use nested for loop?
TC will increase ...
Hello, sir. I tried this algorithm at myself......
My Solution..
public int[] nextGreater(int[] arr){
int[] result=new int[arr.length];
for(int i=0;i
super explanation
thanks
Sir i have the same question. Till where should we think that now It's time for us to stop dsa and learn development?
look knowledge is never measured by time ... you cannot stop dsa at any point and learn development or vice versa ... i hv more than 11 years of exp but i still struggle to solve leetcode questions ... but am happy with what i knw .... its a journey dnt count it on time .... you will realize this when u will hv exp ... my words will sound correct .... practice both and dnt stop anything to learn other ... do both simultaneously ...
Thanks sir.i got it.you're great
thanks !!! keep supporting !!!
will the timing complexity be O(n*n)?? [cannot type superscript here]
Nope !!!
@@itsdineshvaryani how much then😬??
I solved it using a "brute force" method first, using two 'for' loops, which is clearly O(n*n). Dinesh's way using a stack with a single 'for' loop is O(n) by my evaluation. It's a tradeoff of time vs space complexity. Please let me know if wrong. I've went thru Dinesh's solution a few times using the stack. It's quite tricky for me, and I'm still struggling how to think about it, without memorizing the solution.
🙏
hi
i tried with this input value int arr[] = {2, 8, 9, 6, 7}; and getting the wrong output as {8 9 -1 7 -1} . I think it should be { 6, 9, -1, 7, -1}
can you please check and help me. what I am doing wrong. i pasted the whole code below.
int arr[] = {2, 8, 9, 6, 7};
int result[] = new int[arr.length];
Stack stack = new Stack();
for ( int i = arr.length-1; i >=0; i-- ) {
if(!stack.isEmpty()) {
while(!stack.isEmpty() && stack.peek() System.out.print(a+" "));
Output is correct boss!!
I think you understood the problem wrong.
For 2-8, for 8-9, for 9- -1, for 6-7, and for 7 - -1.
@@yuva0101 Thanks Yuvasai... I am new to the coding...learning steps by steps
Output is correct !!! Its next greater element towards right !!!