Відео

a great, easy to follow and comprehensive explanation 🔥

Great Explanation!

very nice and clear explanation , thanks :)

const int N = 1e7; int freq[N]; class Solution { public: int countPairs(vector<int>& nums) { int n = nums.size(); memset(freq, 0, sizeof(freq)); int cnt = 0; for(int i = 0; i < n; i++) { // unique numbers that can be created after swaping nums[i] atmost twice cnt += freq[nums[i]]; unordered_set<int> diff_nums; string digits = to_string(nums[i]); if(digits.size() < 7) { int to_add = 7 - digits.size(); string temp = ""; while(to_add--) { temp += '0'; } digits = temp + digits; } // 1 or 2 swaps allowed for(int l1 = 0; l1 < 7; l1++) { for(int r1 = l1+1; r1 < 7; r1++) { // 1 swap swap(digits[l1],digits[r1]); diff_nums.insert(stoi(digits)); swap(digits[l1],digits[r1]); for(int l2 = 0; l2 <7;l2++) { for(int r2 = l2+1; r2 < 7; r2++) { // 2 swap swap(digits[l1],digits[r1]); swap(digits[l2],digits[r2]); diff_nums.insert(stoi(digits)); swap(digits[l2],digits[r2]); swap(digits[l1],digits[r1]); } } } } for(auto it : diff_nums) { freq[it]++; } } return cnt; } }; I written the code using string and stoi method, but getting TLE at 629/630 test cases...

leetcode weekly contest 411 ques D, can you discuss please

How can we use merge tree to solve number of different elements in a segment

Thank you, great explanation!

sir not able to understand whats wrong in my code class Solution { public: const int MOD = 1000000000+7; long long power_mod(long long base, long long exp) { long long result = 1; while (exp > 0) { if (exp % 2 == 1) { result = (result * base) % MOD; } base = (base * base) % MOD; exp = exp / 2; } return result; } vector<int> getFinalState(vector<int>& nums, int k, int multiplier) { if(multiplier ==1) return nums; priority_queue<pair<long long , int>, vector<pair<long long , int>>, greater<>> minHeap; int initialMaximum =0 ; int initialMaximumIndex = -1; for(int i=0 ;i<nums.size();i++ ) { minHeap.push({nums[i],i}); initialMaximum = max(initialMaximum,nums[i]); if(nums[i] > initialMaximum) { initialMaximum = nums[i]; initialMaximumIndex = i; } } while(k> 0 && minHeap.top().first*multiplier <= initialMaximum) { long long mini = minHeap.top().first; int miniIndex = minHeap.top().second; mini = (multiplier * mini) %MOD; minHeap.pop(); minHeap.push({mini,miniIndex}); k--; } int cyclePosition = k/nums.size();// zero based index position int remainingCount = k%nums.size(); long long multiplierKiPowerKaResult = power_mod(multiplier,cyclePosition); while(minHeap.size() > 0) { nums[minHeap.top().second]=(minHeap.top().first*multiplierKiPowerKaResult)%MOD; minHeap.pop(); } for(int i=0 ;i<nums.size();i++) { minHeap.push({nums[i],i}); } for(int i=1 ;i<=remainingCount;i++) { int mini = (minHeap.top().first * multiplier) %MOD; int minIndex = minHeap.top().second; minHeap.pop(); minHeap.push({mini,minIndex}); } while(minHeap.size() >0) { nums[minHeap.top().second] = (minHeap.top().first); minHeap.pop(); } return nums; } };

Perfect explanation by the way

Well made video

So good video. Crystal clear. Please continue making such type of videos very helpful

Can we solve it using disjoint set? Really tried hard but failed at 604/630

THIS QUALITY CONTENT >>>>>>>

First viewer🥰, but sir quality is upto 360p only

Overwhelmed by the explanation :)

Great Solution Bhaiya..

bro mohan is your real name?

god's work

thnxx keep going.

thank you

thank mohan, for this much-needed video i really appreciate it

Nice video. Liked the way you teach things slowly

Can you tell me about your pratice strategy from beginner to leetcode master ?

Great Solution Sir......

can we apply dfs here?

Instead of set for grouping based on index what can be used in Java?

thank you for making such amazing videos....

Accepted

Nice as usual Please when are you dropping next videos?

This channel is treasures. Thank you very much

Best explanation

Great lecture

Can you also explain the last problem from the biweekly 136 contests last problem that is using tree rerouting[DP on Trees] would love to see your explanation on that topic :)

TOP notch content

excellent sir

For each row of your time complexity analysis you are keeping V1i constant. But you are not considering the "cnt" part of your pseudo code wherein you will need to iterate over the bucket of each V1i and V2i pair.

Thanks that condition make this problem easy. Solved it after watching half video.

very nice explanation, thanks

please xplain how you are handling the overflows. Why you are adding the mod while getting the range sum.

bhai last 10 min khud bolke khud smjh liya aapne

Top notch explanation <3

Great explanation...keep up the good work

Thanku bhaiya very good explanation