Random forest Reg

0.0923

K Nearest Neighbours algorithm. By identifying the similarity in physical features of each herb, it can be classified as a given species.

0.0032, computer = "yes"

K-Means Clustering

import pandas as pd

file = pd.read_csv('salary.csv')

q = [0,.25,.50,.75,1]

label = ['0 to 25', '25 to 50', '50 to 75', '75 to 1']

file['Age_q'] = pd.qcut(file['Age'], q = q, labels = label)

Factor of 4

c1 = (data['Grade'] == 'A') & (data['Marks'] > 60)
c2 = (data['Grade'] == 'B') & (data['Marks'] > 70)
c3 = (data['Grade'] == 'C') & (data['Marks'] > 80)

c = pd.Series(np.select([c1,c2,c3], ['Yes','Yes','Yes'], default='No'), index=data.index)
data['Admission'] = data['Admission'].fillna(c)\

From options (0.371,0.971,0.241.0) => correct => 0 ✓

Question=> (Indian,student, male) => answer 1.5 ✓

Linear regression question =>high bais, low variance,underfitting of data

Amazon ML MCQ ✅

0.371✅

C✅

def min_key_presses(input1):
    S=input1
    n = len(S)
    i = 0
    key_presses = 0
   
    while i < n:
        if S[i] == '0':
            zero_count = 0
            while i < n and S[i] == '0':
                zero_count += 1
                i += 1
            key_presses += zero_count // 2
            if zero_count % 2 == 1:
                key_presses += 1
        else:
            while i < n and S[i] != '0':
                key_presses += 1
                i += 1
               
    return key_presses

Amazon ML✅

Matrix value of x    3
Alphabet count   9:45
F'(x) 1/1+x^4     A
James Simon   1/20
John peter tom 1/4
Equal no of chocolates 7
Exams  86
Students weights 60.2
G(3) 13
Perfect sq 1/3
P -1
Restaurant tables 10

3.45 bits
it learns based on unlabbled data
only D
b and d

Amazon ML✅

t1 t2 relation t1>t2
Trigrams c
Age income student comp 1.0
Ek c1 = (data['Grade'] == 'A') & (data['Marks'] > 60)
c2 = (data['Grade'] == 'B') & (data['Marks'] > 70)
c3 = (data['Grade'] == 'C') & (data['Marks'] > 80)
c = pd.Series(np.select([c1, c2, c3], ['Yes', 'Yes', 'Yes'], default='No'), index=data.index)
data['Admission'] = data['Admission'].fillna(c)
Reinforcement learing scenario all A, B, C, D
Yactual ypredicted iska B
Nouns and parts C
Perceptron model 0

Amazon ML✅

int solve(int idx, int n, int countZero, vector<vector<int>>& dp) {
   int mod=1e4+7;
    if (idx >= n) {
        if (countZero >= 2)
            return 1;
        return 0;
    }
   
    if (dp[idx][countZero] != -1)
        return dp[idx][countZero];
   
    int count = 0;
   
    for (int i = 0; i <= 9; ++i) {
        if (idx == 0 && i == 0)
            continue;
       
        if (i == 0)
            count = (count + solve(idx + 1, n, countZero + 1, dp)) % mod;
        else
            count = (count + solve(idx + 1, n, countZero, dp)) % mod;
    }
   
    return dp[idx][countZero] = count;
}

int bounty(int input1) {
    vector<vector<int>> dp(input1 + 1, vector<int>(input1 + 1, -1));
   
    if (input1 < 2)
        return 0;
    if (input1 == 2)
        return 1;
   
    return solve(0, input1, 0, dp);
}

Bounty
Amazon ML✅

Mod 1e4+7 kro

John sam 0.88

8 red balls 0.35



Differential b

Alphabet mean 9.45

2 digit number 12

r nonzero k=1

Mean median mode range A


max diff 162


G(3) =13


AA' x=2

profit margin 0.1667


Red 2nd ball black 1st ball 8/13

Apti Answers
Amazon ML✅

Player disqualification ✅

Pair problem ✅

class Compute:
    @classmethod
    def compute(cls, input1, input2):
        n = input1
        arr = input2
        result = []
       
        for i in range(n):
            m = 0
            k = n - 2
            left = i - 1
            right = i + 1
           
            if i == 0:
                for i in range(1, k + 1):
                    m += abs(arr[0] - arr[i])
                result.append(m)
            else:
                while left >= 0 and right < n:
                    lside = abs(arr[left] - arr[i])
                    rside = abs(arr[right] - arr[i])
                   
                    if lside < rside:
                        m += lside
                        left -= 1
                        k -= 1
                        if k == 0:
                            break
                    else:
                        m += rside
                        right += 1
                        k -= 1
                        if (k == 0):
                            break
               
                while left >= 0 and k > 0:
                    m += abs(arr[left] - arr[i])
                    left -= 1
                    k -= 1
               
                while right < n and k > 0:
                    m += abs(arr[right] - arr[i])
                    k -= 1
                    right += 1
                result.append(m)
        return result

Nearest house ✅

I can't believe how far we've come together on this journey. A big thank you to each and every one of you for supporting me, leaving kind comments, and pushing me to keep creating❤✌🏻

Rupeek is hiring for Android Engineer (2021, 2022 and 2023 grads)

JD: https://docs.google.com/document/u/0/d/1RYlGWEH31KPt5-YZ4G5PSMgrmaMywv-KQpQbPz1mBA4/mobilebasic

Apply Link: https://docs.google.com/forms/d/e/1FAIpQLSfz7XamMLri24X64d-B58m2H70YZl9aItL_o3Hv4N5r-PnVNg/viewform?usp=send_form

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For 2024 grads

Check out this job at setlmint.com: https://www.linkedin.com/jobs/view/3957631749

#include<bits/stdc++.h>
using namespace std;
#define int long long
signed main()
{
   int t;
   cin>>t;
   while(t--)
   {
    int m;
    cin>>m;
    int l;
    cin>>l;
    int a[l];
    for(int i=0;i<l;i++) cin>>a[i];
    int r;
    cin>>r;
    int b[r];
    for(int i=0;i<r;i++) cin>>b[i];

     int i=0;
     int j=0;int price=0;int cnt=0;
    while(i<l and j<r and price<=m)
    {
      if(a[i]>=a[j]) {price+=a[j];cnt++;j++;}
      else {price+=a[i];cnt++;i++;}
    }
    cout<<cnt<<endl;
   }

}

Sheldon And Penny ✅

public static long[] countSentences(String[] wordSet, String[] sentences) {
        long[] counts = new long[sentences.length];
        Map<String, Long> wordCountMap = new HashMap<>();

        for (String word : wordSet) {
            String sortedWord = sortString(word);
            wordCountMap.put(sortedWord, wordCountMap.getOrDefault(sortedWord, 0L) + 1);
        }

        for (int i = 0; i < sentences.length; i++) {
            String[] words = sentences[i].split(" ");
            long sentenceCount = 1;
            for (String word : words) {
                String sortedWord = sortString(word);
                sentenceCount *= wordCountMap.getOrDefault(sortedWord, 1L);
            }
            counts[i] = sentenceCount;
        }

        return counts;
    }

    private static String sortString(String s) {
        char[] charArray = s.toCharArray();
        Arrays.sort(charArray);
        return new String(charArray);
    }
}

How many Sentences ✅
BlackRock