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This is the best quick select explanation indeed. Other videos never explain what each element of the code actually means.

I was so scared of quickselect before this video, thank you so much for breaking it down in the most understandable way. You're a great tutor!

Fantastic explanation. Very clear and concise and thorough. Thank you!

Shouldn't the time complexity be denoted big theta of n and big O of n^2?

brilliant. One of the best algorithm explanations here. fine work thank you!

Perfect explanation! Thank you

This is an unbelievably concise yet helpful explanation - thanks a ton

best video ever for quickselect

Fill the empty cells in the rows with their basic possibilities given the numbers in the sub grid row and column no need to do any more working out. Now go from left to right or right to left ignore the data in your cell and see what numbers ain't covered by the possibilities in the other cells or the cells already solved. If there are no possibilities deduced move to the next empty cell and wipe the info in the cell you just tested. If you do find a possible number or 2 from this test then the solution is that number or is one of those numbers. It's a very quick O n time test that trips the game of sudoku up and ruins it for everyone.

Good work!

Best Quick Sort explaination❤❤❤❤

Man you make such great videos, why'd you stop?

Initially, I thought of using Next Greater to Left & Next Greater to Right concepts, & did dry run as per below written code for the test cases, [1,2,2], [3,2,1,2,1,7] & [4,4,4,4,4]. According to the dry run, I got the correct answers, but when I run the test cases on leetcode, I get correct output for [1,2,2], but for [3,2,1,2,1,7] & [4,4,4,4,4], it shows output of 4, which is wrong. My dry run (as per the same code) yielded correct answer, but on running on platform, it fails. Can anyone help me address this issue ? The code is : class Solution { private: vector<int> NGR(vector<int>& arr, int n){ stack<int> st; vector<int> result(n, -1); for (int i = n - 1; i >= 0; i--) { while (!st.empty() && st.top() <= arr[i]) { st.pop(); } if (!st.empty()) { result[i] = st.top(); } st.push(arr[i]); } return result; } vector<int> NGL(vector<int>& arr, int n){ stack<int> st; vector<int> result(n, -1); for (int i = 0; i < n; i++) { while (!st.empty() && st.top() <= arr[i]) { st.pop(); } if (!st.empty()) { result[i] = st.top(); } st.push(arr[i]); } return result; } public: int minIncrementForUnique(vector<int>& nums) { int n = nums.size(); vector<int> ngr = NGR(nums, n); vector<int> ngl = NGL(nums, n); set<int> st; int minMoves = 0; for(int i = 0; i < n; i++){ int num = nums[i]; if(st.find(num) == st.end()) st.insert(num); else{ int gRight = ngr[i]; int gLeft = ngl[i]; // 4 cases // case - 1 if(gRight == -1 && gLeft == -1){ minMoves++; num++; } // case - 2 else if(min(gRight, gLeft) == -1){ minMoves += max(gRight, gLeft) + 1 - num; num = max(gRight, gLeft) + 1; } // case - 3 & 4 else if(min(gRight, gLeft) != -1){ int diff = abs(gRight - gLeft); if(diff > 1){ minMoves += min(gRight, gLeft) + 1 - num; num = min(gRight, gLeft) + 1; } else if(diff == 1){ minMoves += max(gRight, gLeft) + 1 - num; num = max(gRight, gLeft) + 1; } } } } return minMoves; } };

Wrong solution

One of the better explanations out there. The thing with partition is that there are various ways to do it and so many nuances with the boundary conditions if you are not careful. There is another method using while loops which I have always found tad bit more difficult. This one is more straightforward.

You easily had the best video on it, thank you!

brother you need more views you are saving my algorithms exam fr

lowkey the best quickselect explanation ever what

The time complexity is wrong, its o(n^2) worst case if you select the largest or smallest element in the array

great video , it deserves more views

I can say this video is the best explanation of the quick select algorithm on TH-cam. I read an article on GeeksforGeeks and only found myself more confused because of poor variable naming such as x and y, and weird way of partitioning like right - left, and so on. The method in this video and the explanation make more sense! Thanks CS Robot!

Isn't the third largest element 10 here?

Awesome explanation. Thanks!!!

great explanation

Super good explanation video. This deserves more views

great video!

your videos are really intuitive and easy to understand, you should keep posting!!!

really well explained

this is the first time i can understand algorithms from the first time.

Believe it or not, this is the best quick select explanation on youtube.For the first timein my life, i can write this algorithm without memorization

good explanation, thanks

straight to the point and no nonse explaination!, thanks for explaining the partition function now I can easily remember the solution

Very good explanation, thank you!

really really good video

How do you make these graphics ? Is it a presentation ? Slides , or something like manim ?

Keep posting these king of videos Great usage of example and step by step impl explanation. Keeping it simple

This is best as far as I have seen on YT I was really stuck at I and J pointer. nailed it 🔥

Great explanation, thank you

Selecting the last as pivot is risky, selecting a random is much better.

thank you so much this helped me understand very quickly <3 please post more content

What a great, down to earth explanation.

In line 12 of your code, how's it guaranteed that newValue will be at max only one less than nums[i-1]?

thanks

really appreciate your way of explaining😇😇

Is this works if the last element in an array is the largest?, bcoz the arr[i] and arr[r] swaps at the end of the each iteration right

thank you so much. that was a great explanation

Love from india

You have good explanation, Thanks it's help me a lot

Really good explanation!

Great Explanation, thank you