Next Greater Element 1 (LeetCode 496) | Full solution with examples and animation | Study Algorithms

[@NerchukoMava]

Underrated channel, sir please make more videos ,you have the potential to be on top.

[@nikoo28]

I am adding one new video every week 😄

[@NerchukoMava]

@@nikoo28 I am so glad for your reply sir😇

[@NerchukoMava]

Sir please do videos on Recursion as soon as possible, which will be so helpful for many students

[@TufanPandu]

After watching my expensive course i come to watch this video and it's OSM!

[@nikoo28]

Amazing!

[@surajsidar3280]

Which course you took and what was the cost? Please tell so that I won't take that course

[@bonafontciel]

I was able to type my own code just with your explanation. Thanks!

[@nikoo28]

Great job!

[@varsshhaa]

omg!! I am happy that I have found this channel !! explanation is simply great !! Looking forward to more such videos !!

[@Hayat26474]

i never thought this could be that simple , thanku sir

[@dharmeshkase6003]

the way you explain everything is just

[@ranjansingh5414]

second if statement will not give correct output for your case this will give correct output, since you need to peek and add to greater element stack then push the element if(element < stack.peek()){ result[i] = stack.peek(); stack.push(element); continue; }

[@nikoo28]

if you check the code available in the video description, that works perfectly and passes on LeetCode.

[@kireetmishra1671]

yeah bro you're correct, the sequence in the video is wrong, if the element is pushed before it will give the top element i.e. the same element from the array

[@lakcai4063]

Great explanation! Clear and intuitive!

[@a3rdtierguy864]

UPDATED SOLUTION(LEETCODE SOLUTION) JUST USE ONLY MAP vector nextGreaterElement(vector& nums1, vector& nums2) { unordered_mapm; stacks; int n=nums2.size(); m[nums2[n-1]]=-1; s.push(nums2[n-1]); for(int i=nums2.size()-2;i>=0;i--) { if(s.size()>0&&s.top()>nums2[i]) { m[nums2[i]]=s.top(); s.push(nums2[i]); continue; } while(s.size()>0&&s.top()

[@pietech1694]

while forming a new array with stack store original array element index into map then traverse num2 and get the index of elements from hash and look the value at that index in fresh Array

[@tanishktripathi8773]

This is such a nice explanation, Here is the Java simplified solution class Solution { public int[] nextGreaterElement(int[] nums1, int[] nums2) { HashMap map = new HashMap(); Stack stack = new Stack(); for(int i=nums2.length-1; i>=0; i--) { while(!stack.isEmpty() && stack.peek()0) { map.put(nums2[i], stack.peek()); } else { map.put(nums2[i], -1); } stack.push(nums2[i]); } int[]res = new int[nums1.length]; for(int i =0; i

[@prasadm8441]

Great explanation. Thank you so much Nikhil

[@srikrishna_ss8263]

Your Way Of Explanation Simply Awesome 🤗

[@sahithd3125]

loved the way you explain the problem!

[@andretheruler714]

thank you, i was a bit confused until you explained it

[@pradheeshkumar-v4p]

That was a crystal clear explanation !!

[@vilakshan.s]

I think time complexity will still be O(n^2) where in worst case we will end up emptying the whole stack in order to find next greater element for ith position

[@sammedpatil3907]

what will be space and time space complexity if we are using stack and Map??

[@MuhammadFahreza]

in 13:40. Why the time complexity is O(n) ? Should we also count, the effort to traverse down the stack for each value ?

[@nikoo28]

you do not pop the entire stack each time. the max pops you will do is (n). So total time complexity is still O(2n) which equates to O(n)

[@mohammedshahed5051]

Okay, in the "optimized solution" as well we have an inner loop to remove all the elements which are lesser in that so even here we have a loop with in a loop, how does the time complexity o(n2)(brute force) != o(n2) (optimized) i guess it is inevitable, so best case o(n) worst case o(n2)

[@mdshafiuddin1234]

amazing explanation!

[@kemarjordan628]

Brilliant man!

[@Rocky-me7qh]

Array is actually filled with distinct numbers. Next greater number of 4 is 12 and 6 what's the right answer? In the leetcode description itself is written, given arrays are distinct

[@mohinischannel9665]

very nice explaination ..clear.

[@hilalkocak_]

Best explanation ever!

[@nikoo28]

Glad it was helpful!

[@deepakkumar-sk4kj]

Please do Solve Leetcode 503 Next Greater Element II

[@Elkhaligy]

Good one

[@stanleyjekwu6810]

Thank you so much, very clear

[@sumitsanu9524]

Same example as striver

[@vinodpaluvuri54]

we need to compare nums1 with nums2..where are we doing that

[@nikoo28]

nums1 is a subset of nums2. Can you please elaborate your concern?

[@amogu_07]

@@nikoo28 i didnt understand how ppl got it , you didnt explain it properly at all u shd see the video urself

[@sanand6084]

Awesome bro ! Subbed.

[@nikoo28]

Thanks for the sub!

[@AdityaRaj-pe3zy]

most of the you tuber have taken the same example and are doing the same thing but in question we have to compare NUMS1 and NUMS2 the find it

[@NidhiChaudhary-u4b]

I never thought this could be that simple thank you

[@kavinsanthosh1306]

you are finding the greater element for every element in nums2 arr. But the question was for only nums1 arr. How to get only the answer for nums1 arr?

[@VinaySharma-pz9tr]

exactly...is jxtu ne aadha question btaya hai aur ek jagha galat logic bhi btaya hai. iski mkc

[@FelixTheForgotten]

Have Fun! If you can't finish the code from there, you need to learn programming better. Here is my solution in c++, copy it if you want. class Solution { public: vector nextGreaterElement(vector& nums1, vector& nums2) { vector ans(nums2.size(),-1); stack num; for(int i = nums2.size()-1;i>=0;i--){ while(!num.empty() && num.top()

[@amogu_07]

@@VinaySharma-pz9tr bro he did NGE 1 for the first half and jumped to NGE 2 next half wtf man🤣🤣

[@nikoo28]

once you have found out the next greater element for each integer in nums2, you literally have everything needed to solve. The remaining part should be fairly simple. You can maintain a map/array for each element of nums1.

[@pratishkumar32]

Sir please can you explain Next greater element II question no. 503

[@nikoo28]

i will create a video on it soon

[@vishaltorgal3372]

Awesome

[@ranjansingh5414]

The should be actual code running for all inputs ''''''''' long[] result = new long[n]; // Your code here long[] result = new long[n]; Stack stack = new Stack(); for(int i = n-1; i>=0; i--){ long element = arr[i]; if(stack.isEmpty()){ stack.push(element); result[i] = -1; continue; } if(element < stack.peek()){ result[i] = stack.peek(); stack.push(element); continue; } while(stack.peek()

[@chinmaypanchal2521]

Thankyou bro

[@Ishowspeed_147]

if i need to print the index of that numbers how we can done with this code please explain me

[@nikoo28]

use an additional stack to store the indices

[@Ishowspeed_147]

@@nikoo28 can u please send me the code

[@nikoo2805]

try it out on your own first.

[@Ishowspeed_147]

@@nikoo2805 huhu

[@ashishkr.229]

you didnt declare helperStack

[@nikoo28]

check out the solution on Github in the video description. That is often more complete.

[@harikrishnas1299]

static int[] nextGreaterElement(int[] arr) { int[] nge = new int[arr.length]; Stack st = new Stack(); for(int i= arr.length-1; i>=0; i--) { int element = arr[i]; if(st.isEmpty()) { st.push(element); nge[i] = -1; continue; } if(st.peek() > element ) { st.push(element); nge[i] = st.peek(); continue; } while(st.peek()

[@Harshii2134]

just woww o my god!!

[@shalini3286]

keep it up broooo :)

[@jackfrost8969]

this wasn't even Leetcode 496 problem

[@nikoo28]

What do you mean?

[@jackfrost8969]

@@nikoo28 your solution was not related to Leetcode 496 question

[@gokulakannan3664]

In your explanation part, In the array, there is two 3.... if we want next higher value of 3 , then which one will consider.... in your explanation, u use map.... it can't allow duplicate key .... explain u explain?

[@pranshusati5115]

Hey nikhil i iknow this video is over 1 year old but could you help me get why dose my code not work i passed 12/16 test cases //code here HashMap mp = new HashMap(); mp.put(nums2[nums2.length -1],-1); int max = nums2[nums2.length-1]; for(int i = nums2.length-2; i>=0; i--){ int current= nums2[i]; if(current < max){ if(i!=nums2.length && nums2[i+1] > current){ max = nums2[i+1]; mp.put(current,max );} } else mp.put(current,-1); max = Math.max(current,max); } int[] res = new int [nums1.length]; for(int i =0; i< nums1.length; i++){ res[i] = mp.get(nums1[i]); } return res;

[@nikoo28]

It will be hard to debug for me. Try some edge cases would be my suggestion