Video

thanku dii bobber bhaiya ne es question ko thoda complex bna diya hai so now i understand this explanation btw thanku😃

Nice

Day2

17. Letter Combinations of a Phone Number for Java people class Solution { public List<String> letterCombinations(String digits) { HashMap<Integer,char[]> mp = new HashMap<>(); mp.put(2,new char[]{'a','b','c'}); mp.put(3,new char[]{'d','e','f'}); mp.put(4,new char[]{'g','h','i'}); mp.put(5,new char[]{'j','k','l'}); mp.put(6,new char[]{'m','n','o'}); mp.put(7,new char[]{'p','q','r','s'}); mp.put(8,new char[]{'t','u','v'}); mp.put(9,new char[]{'w','x','y','z'}); List<String>res = new ArrayList<>(); if(digits.length() == 0) return res; findLetterCombination(0,"",mp,res,digits); return res; } void findLetterCombination(int ind,String str,HashMap<Integer,char[]>mp,List<String> res, String dig){ if(ind >= dig.length()){ res.add(str); return; } int a = dig.charAt(ind) - '0'; char[] ch = mp.get(a); for(int i = 0;i < ch.length;i++){ findLetterCombination(ind+1,str+ch[i],mp,res,dig); } } }

Di JAVA? 🥲🥲

DAY1

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tooo good !!

try testing the solution for {2,1}. You are a queen of wrong solutions

queen of wrong solutions

OK YOu explained really well ! Thankyou so much!

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sum to target. Each number in candidates may only be used once in the combination. mam could you make a video to solve this?

Can't be this easy. Thnq so much. ❤

WOW!!! A+ Quality teaching Mam . you know our Mannn Ka doubt ... thats why you make us understand everything so easily . stay Succesfull Mam.

Thank you keerti for wonderful explaination

Hi

super easy explanation mam love it

Additional condition included for this problem in practice problem. if(N <= 0) return {N}; if the number is less than or equal to zero return that number only. Otherwise print the pattern.

I understood the problem the relation part But the mod part and add are trial and error right or are there any further logic to break it down

Help me where my logic is wrong: Split Array Largest Sum: 28/32 test cases passed, Approach Linear search Case: [2,3,1,1,1,1,1] , k = 5 // This test cases failed Please help me in function possibleMinimizedLargestSum(), where my logic is wrong.. class Solution { // //Approach 1: Linear, TC: O((sum-max +1)* n) public int splitArray(int[] nums, int k) { int max = Integer.MIN_VALUE; int sum = 0; int len = nums.length; for(int elm : nums){ max = Math.max(elm, max); sum += elm; } if(k == 1 ) return sum; if(k == len) return max; while(max <= sum){ int countOfSplit = findCountOfSplit(nums, max, len, k); if(countOfSplit == k) return max; // minimized largest sum max++; } return -1; } public int findCountOfSplit(int []arr, int possibleMinimizedLargestSum, int len, int k){ // [7,2,2,15,10,18], k = 3 // Edge Case: [2,3,1,1,1,1,1] , k = 5 int count = 1; int splitSum = 0; for(int i = 0; i < len; i++){ if(splitSum + arr[i] <= possibleMinimizedLargestSum){ splitSum += arr[i]; }else{ count++; if(count > k) break; //Just Optimizing the code splitSum = arr[i]; } } return count; } }

Wonderful explanation mam.

int findMaximum(int[] arr, int n) { // code here int bitonic = 0; for(int i=1; i<arr.length; i++){ if(arr[i]<arr[i-1]){ bitonic = arr[i-1]; break; }else{ bitonic = arr[i]; } } return bitonic; }. no need to have binary search.

thanks mam especially for giving us hints and not directly jumping to solution and i thought about the edge cases

Didi thank you so much we are getting knowledge from you only thankyou guruji

using + with string name adds parenthesis?

Awesome explanation please keep uploading more videos

Why is it necessary that either half will be sorted, not both. let given array is [4, 5, 6, 7, 0, 1,2] (actual sorted array is [0, 1,2, 4,5,6,7] and rotated by 4 units] . This array is not only sorted in both halves but also less than mid value. if the target is 2 then will it work ? how can we achieve it logarithm time complexity

Excellent explanation.

Thank you for posting this. I was able to figure out the small mistake that I had made: return count <= 2*k whereas I put: return count // 2 <= k Which wouldn't work because of integer division

Could you please make another video for the dry run. Just for better visibility.

awesome video ma'am, very clear explanation.

best explaination

Amazing explanation!! , finally I got the gist of it after a day of watching 🤯tutorials

Very good explanation 😊 thanku

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bahut acha samjhaya

thank you ma'am, solved this without referring solution once you explained the logic, so easy it was, thanks a lot ma'am!

jo ata tha wo bhi bhol gya

If you dont pass curr by reference then there is no need of poping

I am unable to understand recursive solution for question- minimum jump to reach end of array question Please could you explain it in video i request keerti and thank for this video ❤

Thanks, cool solution.

what an awesome lecture, thank you for making recursion a lot easier ma'am🙏🙏

Nice 🙂

After referencing 1000 of videos. Finally I got from this video. Thank you🙏

Aaram se bolo kuchh samajh mein nahin aaya kya bol diya😂😂😂😂😂😅😅😅😅😅

Chup kr

Didi, kaha gye aap?

I thought to do solve without binary search and succeeded with it Just a 1 line did this return min(nums) Thats it It worked on leetcode for both the minimum medium problem and the minimum hard as well.

what if the array was A[] = {1, 45, 60, 100} then in first iteration the mid value will be 51 and during isAllocation function, since A[2] i.e 60 is >mid(51) and it will return false?