Find the Power of K-Size Subarrays I | Simple Explanation | Dry Run| Leetcode 3254 |codestorywithMIK

[@joydeep-halder]

Initially i thought i can not solve this problem. But applying brute force, i had solved it on first try. Then I came to your video. Thanks for the optimized approach ❤

[@swastikakundu8418]

same

[@codestorywithMIK]

❤️🙏

[@dhruvchopra26]

Tag krne waalo me se Ek Success story meri thi mik bhaiya Knight bn gya mai😁😁Thank you so much for your awesome explanations!!

[@gui-codes]

congrats man.

[@dhruvchopra26]

@gui-codes Thanks broo

[@universalcosmologist3675]

bhai mai to 1 ya 2 hi solve kar paata hun contest me pls help

[@dhruvchopra26]

@@universalcosmologist3675 Koi ni bhai Lage rho dheere dheere hojayga

[@069_Souvik]

​@@dhruvchopra26Congrats bro...

[@pankaj_kumar_malik]

Able to solve in my own, Intuition apne ap banne laga bhaiya... MIK magic✨❤

[@codestorywithMIK]

Way to go ❤️

[@pankaj_kumar_malik]

@@codestorywithMIK ❤🙌

[@Ankitpal-y6h]

keep doing such a kind of series (if possible contest discussion as well),god bless you good health and life .

[@codestorywithMIK]

Means a lot ❤️🙏

[@bd1a0573]

Solved the problem in n*k TC but still not satisfied. SO watching your video now😁😁😁

[@madmaxgaming5864]

loved your optimised approach ♥

[@VineetDixit]

i solved it using the optimized method,thanks to your previous videos

[@codestorywithMIK]

Great 👍

[@nishantdehariya5769]

to not support brute force the length of array should be >= 10^5, great problem

[@ugcwithaddi]

Was able to solve 👌🏻

[@mohammadaftabansari6882]

Thanks for the videos.

[@Engineering.Wallah]

Unique class Solution { public: vector resultsArray(vector& nums, int k) { int n=nums.size(); vectorans(n-k+1,-1); int i=0,j=0; while(j0 && nums[j]-nums[j-1]!=1){ i=j; } while(ik){ i++; } if(j-i+1==k) ans[j-k+1]=nums[j]; j++; } return ans; } };

[@vedanthbaliga7686]

Solved before the video using same concept!

[@amanpaliwalvlogs6860]

Radhe Radhe ❤

[@shivaye08]

Chalo bhaiya Aaj logic toh ban gaya tha but kuch mistake thi magar you me or saath saath aur acchi practice se bhi mistake bhi aana band ho jayegi ❤

[@codestorywithMIK]

Way to go ❤️

[@piyush0mandloi]

thank you bhiay

[@kaat6663]

class Solution { public: vector resultsArray(vector& nums, int k) { int n = nums.size(); vector ans(n-k+1); for(int i = 0 ; i

[@pranildhutraj6038]

class Solution { public: vector resultsArray(vector& nums, int k) { vector ans; // [1,2,3,4,3,2,5], k = 3 vector temp(nums.size(), 1); // 1 2 3 4 1 1 2 for (int i = 1; i < nums.size(); ++i) { if (nums[i] == nums[i - 1] + 1) { temp[i] = temp[i - 1] + 1; } } int currMax = INT_MIN; for (int i = 0; i < k; ++i) { currMax = max(currMax, nums[i]); } int i = 0, j = k - 1; while (j < nums.size()) { if (temp[i] + (k - 1) != temp[j]) { ans.push_back(-1); } else { ans.push_back(currMax); } if (currMax == nums[i]) { currMax = INT_MIN; } if (j + 1 < nums.size()) { currMax = max(currMax, nums[j + 1]); } i++, j++; } return ans; } }; This is Accepted Solution for problem. Can someone suggest me any improvements in the code quality

[@gui-codes]

can be done in one loop also. class Solution { public int[] resultsArray(int[] nums, int k) { int n=nums.length; int[] res=new int[n-k+1]; Arrays.fill(res,-1); int counter=1; for(int i=0;i0 && nums[i]==nums[i-1]+1){ counter++; } else{ counter=1; } if(counter>=k){ res[i+1-k]=nums[i]; } } return res; } }

[@harleenkaur8540]

pls solve using monotonic queue as well

[@codestorywithMIK]

kzbin.info/www/bejne/banOdHRuZc1ol5Ysi=BTZmLnId5xMES3gg Hope this helps ❤️🙏

[@harleenkaur8540]

@codestorywithMIK thanks a lot ❤

[@mohammadaftabansari6882]

@@codestorywithMIK Much appreciated.

[@AS-gf3ci]

Brute-Force : T.C --> O(n^2) || S.C --> O(1) class Solution { int checkSorted(vector &nums, int idx, int k) { for(int i = idx; i < (idx + k - 1); i++) { if(nums[i+1] != (nums[i] + 1)) return -1; } return nums[idx + k - 1]; } public: vector resultsArray(vector& nums, int k) { int n = nums.size(); vector ans; for(int i = 0; i

[@rickdutta942]

you can easily solve in O(n)

[@AS-gf3ci]

@@rickdutta942 Yeah, but this approach is just for intuition building for beginners. 1st approach to discuss in the interview.

[@rickdutta942]

​@@AS-gf3ci Agreed.

[@gui-codes]

@@AS-gf3ci yes bro agree.

[@hellsterkun8764]

Easy Question Tha Aaj Ka. Was able to solve it on my own. 100% faster. class Solution { public: vector resultsArray(vector& nums, int k) { int n = nums.size(); vector results(n - k + 1, -1); int j = 0; while (j < n - k + 1) { int i = j; bool isValid = true; while (i < j + k - 1) { if (nums[i] < nums[i + 1] && nums[i + 1] == nums[i] + 1) { i++; } else { isValid = false; break; } } if (isValid) { results[j] = nums[j + k - 1]; } j++; } return results; } };

[@codestorywithMIK]

Awesome♥️

[@rickdutta942]

TC: O(n), SC: O(1) class Solution { public: vector resultsArray(vector& nums, int k) { int power = 0, prevNum = INT_MAX, n = nums.size(); vector ans; for (int i = 0; i < n; i++) { if (nums[i] (prevNum + 1)) power = 1; else power++; if (i + 1 >= k && power != k) ans.push_back(-1); if(power == k) { ans.push_back(nums[i]); power--; } prevNum = nums[i]; } return ans; } }; //Brute force

[@tauheedahmed4073]

I have done till sliding but when i saw this count was like aha...

[@ArnabBhadra02]

Brute Force: T.C: O(N*K) class Solution { public: int isposs(vector&temp){ for(int i=0;i0) res.push_back(ans); else res.push_back(-1); } } return res; } };

[@AbhiGarg-u1w]

bhaiya can you please make a video for post contest solution

[@aizad786iqbal]

how is it n-k+1 ??

[@gui-codes]

The size of the result array is ( n - k + 1) because you are calculating the power of a subarray of length k for every possible starting position in the array nums where a subarray of length k can fit. Here’s a detailed explanation: To determine the number of subarrays of length k, consider: - The first subarray starts at index 0 and ends at index k - 1 - The second subarray starts at index 1 and ends at index k - ... - The last subarray starts at index n - k and ends at index n - 1 You can't start next subarray from n-k+1 because it will not make a subarray of size k as it will go out of bound. This gives you a total of n-k+1 subarrays. Why n - k + 1 ??? To fit a subarray of size k - The starting index of the subarray, i, can range from 0 to n - k - Therefore, there are n - k + 1 possible values of i corresponding to the number of subarrays you can form.

[@ArnabBhadra02]

at first glance its too much difficult to think this otimized version

[@Resham298]

pls tell me why we set consecutive count to 1 when a non consecutive element is seen

[@codestorywithMIK]

Because when you are at jth element, it’s one element and we keep its count as 1 and we expect that in future we will see more consecutive elements to this jth element

[@Resham298]

@codestorywithMIK thankyou bhaiya understood

[@AlgoUniverse01]

Please use a black screen instead of a white one, as the high contrast of white is uncomfortable for the eyes.

[@codestorywithMIK]

Sure thing. Actually I use black screen only. Actually I am currently travelling and it slipped my mind to use black screen.

[@vidhut-rau]

Why 1st window is preprocessed ??

[@codestorywithMIK]

You can choose to not preprocess it. I did it for getting a count value before hand. No need to preprocess. Check in the comments, someone has posted a code without preprocessing ❤️

[@dayashankarlakhotia4943]

Brute forces public int[]resultsArray(int[]nums,int k){ int n=nums.length; int[]ans=new int[n-k+1]; for(int i=0;i

[@vidhut-rau]

Not able to do own my own feeling low...

[@tauheedahmed4073]

How long have you been dng bro

[@vidhut-rau]

@@tauheedahmed4073 from may 2024 started Dsa

[@codestorywithMIK]

Don’t feel low. This is the process of learning. Such phase will always come. With consistency, things will improve slowly but definitely ❤️

[@Playvish]

finally after 3 wrong submissions I've successfully solved it in O(n) runtime and beats 100%. Feeling happy Thanks to you #MIK bhaiya here is my solution class Solution { public: vector resultsArray(vector& nums, int k) { int n = nums.size(); if (n == 1) return {nums[0]}; int f = -1; vector ans(n - k + 1, -1); for (int i = 1; i < k; i++) { if (nums[i] != nums[i - 1]+1) { f = (i - 1); } } if (f == (-1)) ans[0] = nums[k - 1]; int i = 1; for (int j = k; j < n; j++) { if (k == 1) { ans[j - k + 1] = nums[j]; } else { if (i > f && nums[j] == nums[j - 1]+1) ans[j - k + 1] = nums[j]; else if(nums[j]!=nums[j-1]+1) f = (j - 1); } i++; } return ans; } };

[@codestorywithMIK]

Glad to see this ❤️

[@KeshavKumar-jl1ub]

monotonic and please ... and why are we processing the first window .... seperately why not in that loop only

[@gui-codes]

you can do it in same loop as well. Both work fine. Here is the one loop code class Solution { public int[] resultsArray(int[] nums, int k) { int n=nums.length; int[] res=new int[n-k+1]; Arrays.fill(res,-1); int counter=1; for(int i=0;i0 && nums[i]==nums[i-1]+1){ counter++; } else{ counter=1; } if(counter>=k){ res[i+1-k]=nums[i]; } } return res; } }

[@codestorywithMIK]

Using Monotonic deque - kzbin.info/www/bejne/banOdHRuZc1ol5Ysi=BTZmLnId5xMES3gg Hope this helps ❤️