Lecture 30: Trapping Rain Water || 3 SUM || 4 SUM

In Array, Today we learn about Trapping Rain Water || Triplet Sum in Array || 4 SUM
1: Trapping Rain Water: leetcode.com/p...
2: 3 SUM: practice.geeks...
3: 4 SUM :practice.geeks...
4: Array 3 Pointers: www.interviewb...
HomeWork Sheet: drive.google.c...
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Hard Concept ab easy lag rhe hai?

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Brute force to optimised sab yaad rakhna hai ke sirf best wala

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Question karne mei jo maja hai wo kahi nahi 😊 thanks bhaiya❤

[@akarshgoel4990]

48:32 Solution of trapping rain water in only one traversing: class Solution { public: int trap(vector& height) { int unit=0, start=0, end=height.size()-1, maxL=height[start], maxR=height[end]; while(start

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THanks

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[@IndiLores709]

58:00 *Solution using binary search* : #include #include #include using namespace std; vector Possibility(vector& arr, int target) { int n = arr.size(); sort(arr.begin(), arr.end()); for (int i = 0; i < n - 2; i++) { for (int j = i + 1; j < n - 1; j++) { int start = j + 1, end = n - 1, mid; // Corrected the 'end' value while (start target; vector result = Possibility(v, target); if (result.empty()) { cout

[@sandeepsahu2681]

Timestamps 00:02 - Calculating the units of water stored between buildings. 02:08 - Storing water in between buildings 05:52 - Determining the minimum building height to store water 07:50 - Calculating support for trapping rainwater 12:02 - Calculating maximum water trapping 14:11 - Calculating trapped rainwater using a special formula 18:33 - Algorithm for calculating trapped rain water levels. 20:33 - Calculating trapped rainwater using the 3 SUM and 4 SUM method 24:31 - Solving trapping rainwater problem without using extra space 26:30 - Comparing and selecting the maximum number 30:03 - Finding maximum support for trapping rainwater 31:47 - Algorithm for trapping rainwater explained 35:13 - Algorithm explanation for trapping rainwater with 3 SUM and 4 SUM 37:07 - Finding the biggest building and calculating water storage 41:35 - Explanation of trapping rain water problem and left max estimation 43:34 - Algorithm for finding maximum building height and water trapping 47:33 - Optimizing solutions using two pointer concept 49:11 - The three sum problem is about finding three numbers in an array that add up to 13. 53:27 - Using binary search to find elements in a sorted array. 55:18 - Discussing process of finding numbers through explanations and examples. 59:15 - Converting a problem to a two sum problem 1:00:56 - Finding number combinations that add up to 20 1:05:07 - Solving the problem of triplet sum in an array. 1:07:17 - The speaker discusses finding the sum of numbers and applying conditions. 1:11:36 - Converting 4 SUM problems into 3 SUM 1:13:34 - Discussion about proof of work

[@musaifbangi8705]

I have try solve hard problem water trapping before video. I think I won't able to solve this. But guess what I solved it. And I took 0 ms time. Time complexity O(n) and space O(1). This led my code better than the sir's code. So I created the logic in 20 - 30 mins. And the code about 15 mins. Total time 45 mins. And I used very understandable logic.

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The best explanation i have seen for the trapping rain water .

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000q1 method1(m1) 4:15 m2, 36:49 m3 42:57 q1 code by m3, 47:05 q2 m1 54:00 m2 59:25 m3 its code 1:06:33 by m3 1:11:01 homework question

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Day: 40 HARD grinding. Walking slowly to climb up the Mt. Everest Sir. But walking consistently. You all have to believe than one day bhaiya will achieve massive success and he will not leave his students behind, he will make us also to achieve that greatness. Hats off Sir. Standing Ovation... 👏👏👏👏👏👏

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Me too

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[@IndiLores709]

Bhaiya! for 4 sum problem, I wrote this code at my own , int Solution::minimize(const vector &A, const vector &B, const vector &C) { int n = A.size(), m = B.size(), o = C.size(); int minm = INT_MAX; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { for (int k = 0; k < o; k++) { int first = abs(A[i] - B[j]); int second = abs(B[j] - C[k]); int third = abs(C[k] - A[i]); int Q = max(max(first, second), third); // Corrected the max function if (Q < minm) { minm = Q; } } } } return minm; } But this is not optimal solution at all. We are not taught pointers yet.

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35:30 //O(1) space complexity class Solution { public: int trap(vector& height) { int Maxi=0,n = height.size(); for(int i = 0;iheight[Maxi]) Maxi = i; } int leftMax = 0,rightMax = 0,temp,ans=0; for(int i = 0;i0) ans+=temp; leftMax = max(leftMax,height[i]); } for(int i = n-1;i>Maxi;i--) { temp = rightMax-height[i]; if(temp>0) ans+=temp; rightMax = max(rightMax,height[i]); } return ans; } };

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Tagda

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[@joydipsen3057]

When I see the problem first, I do it in n time complexity and 1 space complexity. Here's the program // Trapping rain water(n time complexity & 1 space complexity) #include using namespace std; int main() { int arr[100],n; cin>>n; for(int i=0;i>arr[i]; } int highest=0; for(int k=1;karr[highest]) highest=k; } //break into two part;left side and right side of highest piler int large=0,small=1,store=0; for(int j=0;jarr[small]) { store=store+(arr[large]-arr[small]); small++; } else if(arr[large]highest+1;j--) { if(arr[large]>arr[small]) { store=store+(arr[large]-arr[small]); small--; } else if(arr[large]

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1:06:24 chamak gya bhaiya .

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[@Gyanendra012]

Q. trapping rain water class Solution { public: int trap(vector& height) { int n=height.size(),maxWater=0,minHeight=0; vectorpremax(n); premax[0]=height[0]; vectorsufmax(n); sufmax[n-1]=height[n-1]; for(int i=1;i=0;i--){ sufmax[i]=max(sufmax[i+1],height[i]); } for(int i=1;iheight[i]){ maxWater+=minHeight-height[i]; } } return maxWater; } };

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Day 40/180 . Moving forward with confidence just because of you ♥♥.Thank you

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can we use two pointer ?? by using switch case?

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#40 out of 180 Good morning bhaiya

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Concept++

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class Solution { public int trap(int []height){ if(height. length ==0) return 0; int ans=0,l=0,r=height.length-1; int maxL=height [l]; int maxR=height [r]; while(l

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Day 40/180 done thanku bhaiya🙃

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