Was getting queries on this, yes even as students you can deploy your projects on Google Cloud ⬇️

If you work or want to work on Google Cloud, they are offering 300 USD free for all for a limited period of time, so go ahead and grab the same  asap.

Link here : https://ad.doubleclick.net/ddm/trackclk/N5295.4347315TECHGIG/B28999036.394199262;dc_trk_aid=585472451;dc_trk_cid=183313769;dc_lat=;dc_rdid=;tag_for_child_directed_treatment=;tfua=;ltd=;dc_tdv=1

This is officially by Google Cloud and you can run,test and deploy on GCP free for 300 USD 🫡

Register for Hackathon DevX JMI - https://unstop.com/hackathons/devx-jmi-jamia-millia-islamia-1055794?lb=Uy5zoiXM&utm_medium=Share&utm_source=shortUrl

A good opportunity to build some good projects for your resume!

class UserMainCode(object):
    @classmethod
    def theLastChocolate(cls, input1, input2):
        dp = [[0] * (input1 + 1) for _ in range(input1 + 1)]
        dp[1][1] = 1
       
        for i in range(2, input1 + 1):
            for j in range(1, i + 1):
                dp[i][j] += dp[i - 1][j - 1] if j - 1 >= 1 else 0
                dp[i][j] += dp[i - 1][j] if j <= i - 1 else 0
       
        return dp[input1][input2]. 

//   Amazon ml  THE LAST Chocolate

int gcd(int a, int b) {
    while (b != 0) {
        int t = b;
        b = a % b;
        a = t;
    }
    return a;
}

int calculatesun(int input1, int input2[]) {
    // Read only region end
    int N = input1;
    int max_so_far = input2[0];
   
    for (int i = 0; i < N; ++i) {
        max_so_far = std::max(max_so_far, input2[i]);
        input2[i] = gcd(input2[i], max_so_far);
    }
   
    std::sort(input2, input2 + N);
   
    int sum = 0;
    for (int i = 0; i < N / 2; ++i) {
        sum += gcd(input2[i], input2[N - 1 - i]);
    }
   
    return sum;
}

Sum in Array ✅
Amazon ML

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❤✌🏻