Video

Thik Hai.....

why we need to define next() and hasNext() as we are not using it here?? Is it required ??

Dammmmmmmmmmm❤❤❤❤❤

Appreciate the way of explaining 👏 👌

Very very well explanation

code in c++ class Solution { public: int bagOfTokensScore(vector<int>& tokens, int power) { sort(tokens.begin(),tokens.end()); int s=0,e=tokens.size()-1; int maxi=0; int curr=0; while(s<=e){ if(power>=tokens[s]){ curr+=1; … } maxi=max(maxi,curr); } return maxi; } };

btw i liked your video and its truely better then others explanation it was just like my big brother explaining me things

sir tell me one thing the second condition was when the score is atleast greater then 1 then why are you taking max >0?

Believe me you are explainig very good keep it up..

Bhai, mai ekdum yahi so raha tha par aapne kitni asaani se bta diya. Well done 👍

great explanation brother

your explanation is really good . keep it up. please upload more videos if you can!

Try this test case : 2,1,5,0,6,7 The compiler accept this test case but if we dry run this test case by our own the rule of i<j<k is not satisfied here

"Promo sm" 😭

Bhai why did you stop making these videos?

You explained this the best way on entire CZcams ❤Now I m really looking forward to get solutions of entire leetcode from you 😆

Let's Connect: linktr.ee/udyanojha

leetcode.com/problems/k-th-symbol-in-grammar/

Can you tell Space Complexity here?

class Solution { public boolean validateBinaryTreeNodes(int n, int[] leftChild, int[] rightChild) { HashMap<Integer, ArrayList<Integer>> parentChild = new HashMap<>(); HashMap<Integer, Integer> childParent = new HashMap<>(); for(int node = 0; node< n; node++) { int left = leftChild[node]; if(left != -1) { parentChild.putIfAbsent(node, new ArrayList<>()); parentChild.get(node).add(left); if(childParent.containsKey(left)){ return false; } else { childParent.put(left, node); } } int right = rightChild[node]; if(right != -1) { parentChild.putIfAbsent(node, new ArrayList<>()); parentChild.get(node).add(right); if(childParent.containsKey(right)){ return false; } else { childParent.put(right, node); } } } // Root? Multiple Hai ya nhi h // int root = -1; Integer root = null; for(int node = 0; node< n; node++) { if(!childParent.containsKey(node)) { if(root == null) { root = node; } else { return false; } } } if(root == null) { return false; } HashSet<Integer> visited = new HashSet<>(); Queue<Integer> que = new LinkedList<>(); que.add(root); while(!que.isEmpty()) { int node = que.remove(); visited.add(node); if(parentChild.containsKey(node)) { for(int child : parentChild.get(node)) { que.add(child); } } } return visited.size() == n; } }

Awesome explanation

class Solution { Integer[][] dp = new Integer[501][501]; public int numWays(int steps, int arrLen) { return helper(0, steps, arrLen); } private int helper(int i, int k, int len) { if(dp[i][k] != null) return dp[i][k]; if(i == 0 && k == 0) return 1; if(k==0) return 0; int left = (i > 0) ? helper(i-1, k-1, len) : 0; int stay = helper(i, k-1, len); int right = (i<len-1) ? helper(i+1, k-1, len) : 0; int total = ((left + stay) % (int)(1e9+7) + right) % (int)(1e9+7); return dp[i][k] = total; } }

Thanks for being here Coding JoySoul!

Bhaiya, aap ab lectures nhin banaoge kya?

Bhaiya aapne video hi nhin daali Aaj, Main apne friends ko aapka channel recommend karta hun

Bhaiya, Congratulations for 700 subscribers

I felt that in this video, explanation was not beginner friendly

Tabulation: class Solution { public int minCostClimbingStairs(int[] cost) { int[] dp = new int[cost.length]; int n = cost.length-1; dp[0] = cost[0]; dp[1] = cost[1]; for(int i = 2; i < cost.length; i++) { dp[i] = cost[i] + Math.min(dp[i-1], dp[i-2]); } return Math.min(dp[n-1], dp[n]); } }

JAVA Memoization: class Solution { public int minCostClimbingStairs(int[] cost) { int[] dp = new int[cost.length]; Arrays.fill(dp, -1); return Math.min(helper(cost, 0, dp), helper(cost, 1, dp)); } private int helper(int[] cost, int idx, int[] dp) { if(idx>= cost.length) return 0; if(dp[idx]!= -1) return dp[idx]; return dp[idx] = cost[idx] + Math.min(helper(cost, idx+1, dp), helper(cost, idx+2, dp)); } }

Just a request. Please make all the Leetcode Solution videos of graph topic. Itna awesome padha rahe ho aap

Glad to know that you are back.... frequent viewer of your channel. One thing is for sure... Your explanation is good.

class Solution { public int findInMountainArray(int target, MountainArray mountainArr) { // Find Peak of Mountain int peakIdx = findPeak(mountainArr); // Now BS in left_ASC and right_DSC range int res = -1; // // EDGE CASE if(target > mountainArr.get(peakIdx)) return -1; res = BS_ASC(target, mountainArr, 0, peakIdx); if(res != -1) return res; return BS_DSC(target, mountainArr, peakIdx, mountainArr.length()-1); } private int BS_ASC(int target, MountainArray arr, int st, int ed) { while(ed>=st) { int mid = st + (ed-st)/2; if(arr.get(mid) == target) { return mid; } else if(arr.get(mid) > target) { ed = mid - 1; } else { st = mid + 1; } } return -1; } private int BS_DSC(int target, MountainArray arr, int st, int ed) { while(ed>=st) { int mid = st + (ed-st)/2; if(arr.get(mid) == target) { return mid; } else if(arr.get(mid) > target) { st = mid + 1; } else { ed = mid - 1; } } return -1; } private int findPeak(MountainArray arr) { int st = 0; int ed = arr.length() - 1; while(ed>=st) { int mid = st + (ed-st)/2; if(mid == 0) return 1; if(arr.get(mid-1) < arr.get(mid) && arr.get(mid) > arr.get(mid+1)) { return mid; } else if(arr.get(mid-1) < arr.get(mid) && arr.get(mid) < arr.get(mid+1)) { st = mid+1; } else { ed = mid-1; } } return -1; } }

class Solution { public int[] fullBloomFlowers(int[][] flowers, int[] people) { // Clone arr // Sort Flowers in St ASC // Sort Flowers in Ed ASC // O(n + nlogn + nlogn) // Do Search in st Just <= peoplei AS stIdx // Do Search in ed Just > peoplei // O(n + nlogn + nlogn + logn + logn) // In Res arr push 1 + stIdx - edIdx // it means number of flowers that bloom before = to people time - number of flowers that deBloom before people time int[] res = new int[people.length]; int[][] stFlowers = flowers.clone(); Arrays.sort(stFlowers, (a, b) -> a[0]>b[0] ? 1 : -1); int[][] edFlowers = flowers.clone(); Arrays.sort(edFlowers, (a, b) -> a[1]>b[1] ? 1 : -1); // for(int[] arr : stFlowers) System.out.println(arr[0] +" "+ arr[1]); // for(int[] arr : edFlowers) System.out.println(arr[0] +" "+ arr[1]); for(int i = 0; i<people.length; i++) { res[i] = 1 + stBS(stFlowers, people[i]) - edBS(edFlowers, people[i]); } // O(n + nlogn + nlogn + logn + logn + n) return res; } private int stBS(int[][] flowers, int x) { int stIdx =-1; int st = 0; int ed = flowers.length - 1; while(ed>=st) { int mid = st + (ed-st)/2; if(flowers[mid][0] <= x) { stIdx = mid; st = mid+1; } else { ed = mid-1; } } return stIdx; } private int edBS(int[][] flowers, int x) { int edIdx = -1; int st = 0; int ed = flowers.length - 1; while(ed>=st) { int mid = st + (ed-st)/2; if(flowers[mid][1] < x) { edIdx = mid; st = mid+1; } else { ed = mid-1; } } return edIdx+1; } }

Good work bhai 👍

Java Soln: class Solution { public int[] searchRange(int[] nums, int target) { int leftIdx = findLeft(nums, target); int rightIdx = findRight(nums, target); return new int[] {leftIdx, rightIdx}; } private int findLeft(int[] nums, int target) { int res = -1; int st = 0; int ed = nums.length-1; while(ed>=st) { int mid = st + (ed-st)/2; if(nums[mid]==target) { res = mid; ed = mid-1; } else if(nums[mid] > target) { ed = mid-1; } else { st = mid+1; } } return res; } private int findRight(int[] nums, int target) { int res = -1; int st = 0; int ed = nums.length-1; while(ed>=st) { int mid = st + (ed-st)/2; if(nums[mid]==target) { res = mid; st = mid+1; } else if(nums[mid] > target) { ed = mid-1; } else { st = mid+1; } } return res; } }

Please continue making the Leetcode solution videos again

Why have you stopped making Leetcode solution videos. Please start making videos again, you are doing a good work. Please, please ,please

Your approach is good but initially got below approach in my mind and implemented but after watching your solution it good class Solution { public static boolean checkToeplitz(int i, int j,int[][] matrix){ while(i < matrix.length && j < matrix[0].length){ if(matrix[i][j] != matrix[i-1][j-1]){ return false; } i++; j++; } return true; } public boolean isToeplitzMatrix(int[][] matrix) { int n = matrix.length; int m = matrix[0].length; //My approach is first check the first row whether it is diagonally perfect then check first column whether it is diagonally perfect. //row checking int currentRow = 0; for(int i=0;i<matrix[0].length;i++){ boolean res = checkToeplitz(currentRow+1,i+1,matrix); if(!res){ return false; } } //column checking //for this iteration i am considering as row int currentCol = 0; for(int j=1;j<matrix.length;j++){ boolean res = checkToeplitz(j+1,currentCol+1,matrix); if(!res){ return false; } } return true; } }

Great Explanation

Very nice explanation ! Thank you so muchhhhh <3

bhadiya video thi, hai, nhi rahegi

Bhai mujhe alog basic se learn krni hai.

Excellent bro

Wow what an explanation bro! 🤩🤩🤩 God bless you 🥰🥰🥰

I just found that It's similar to partion sort (card game)

this code is not working bro 😢😢

Thanks buddy for making such a nice solution video.

class Solution { public int[] answerQueries(int[] nums, int[] queries) { int[] prefixSum = new int[nums.length]; Arrays.sort(nums); TreeMap<Integer, Integer> map = new TreeMap<>(); prefixSum[0] = nums[0]; map.put(prefixSum[0], 1); for(int i = 1; i<nums.length; i++) { prefixSum[i] = prefixSum[i-1] + nums[i]; map.put(prefixSum[i], i+1); } int[] ans = new int[queries.length]; int k = 0; for(int query : queries) { Map.Entry<Integer, Integer> entry = map.floorEntry(query); if(entry != null) { ans[k] = entry.getValue(); } k++; } return ans; } }

Great explanation bro👌🏻