INFOSYS EXAM 4 TH AUGUST

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📢 𝐇𝐞𝐲 𝐞𝐯𝐞𝐫𝐲𝐨𝐧𝐞! 𝐁𝐢𝐠 𝐧𝐞𝐰𝐬!

𝐎𝐧 𝐀𝐮𝐠𝐮𝐬𝐭 4𝐭𝐡, 𝐈'𝐥𝐥 𝐛𝐞 𝐨𝐟𝐟𝐞𝐫𝐢𝐧𝐠 𝐅𝐑𝐄𝐄 𝐈𝐧𝐟𝐨𝐬𝐲𝐬 𝐜𝐨𝐝𝐢𝐧𝐠 𝐡𝐞𝐥𝐩 🆓🆓💻.

𝐈'𝐥𝐥 𝐬𝐡𝐚𝐫𝐞 𝐚𝐥𝐥 𝐭𝐡𝐞 𝐜𝐨𝐝𝐞𝐬 𝐫𝐢𝐠𝐡𝐭 𝐡𝐞𝐫𝐞, 𝐬𝐨 𝐦𝐚𝐤𝐞 𝐬𝐮𝐫𝐞 𝐭𝐨 𝐭𝐞𝐥𝐥 𝐚𝐥𝐥 𝐲𝐨𝐮𝐫 𝐟𝐫𝐢𝐞𝐧𝐝𝐬! 🚀

🔄 𝐏𝐥𝐞𝐚𝐬𝐞 𝐬𝐡𝐚𝐫𝐞 𝐨𝐮𝐫 𝐜𝐡𝐚𝐧𝐧𝐞𝐥, 𝐚𝐧𝐝 𝐝𝐨𝐧'𝐭 𝐟𝐨𝐫𝐠𝐞𝐭 𝐭𝐨 𝐭𝐰𝐞𝐚𝐤 𝐭𝐡𝐞 𝐯𝐚𝐫𝐢𝐚𝐛𝐥𝐞 𝐧𝐚𝐦𝐞𝐬 𝐢𝐧 𝐭𝐡𝐞 𝐜𝐨𝐝𝐞𝐬 𝐭𝐨 𝐦𝐚𝐤𝐞 𝐭𝐡𝐞𝐦 𝐲𝐨𝐮𝐫 𝐨𝐰𝐧! 👨‍💻👩‍💻

𝐓𝐡𝐚𝐧𝐤𝐬 𝐟𝐨𝐫 𝐭𝐡𝐞 𝐬𝐮𝐩𝐩𝐨𝐫𝐭 @Coding_000✅

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To remove Plagiarism 👇👇
1) Use your Own template✅
2) Change Variables ✅
3) Try to read input in unique style✅
4) Understand the code and write in your style ✅
@coding_000
5) Write whole code in main function / write in diffrent function✅
6) Change Function name solve()=> unique✅
7) If possible change flow of the problem (if it doesn't impact the logic of code)✅
@coding_000
there are so many other things to avoid Plagiarism🔥🔲

@Coding_000❤️

Everyone pls Follow this To qualify INFOSYS EXAM 💯✅

See u all at 11 am ✅
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Start at 11:20 or 11:25 ✅

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Start at 11:22 ✅

Lost in orange code✅

SHare @coding_000✅❤️

const int MODULO = 1e9 + 7;

int main() {
    int numElements, divisor;
    cin >> numElements >> divisor;
    vector<int> elements(numElements);
    for (int i = 0; i < numElements; i++) {
        cin >> elements[i];
    }
   
    vector<vector<vector<long long>>> dp(divisor, vector<vector<long long>>(divisor, vector<long long>(divisor, 0)));
    dp[0][0][0] = 1; 
    
    for (int value : elements) {
        vector<vector<vector<long long>>> newDp(divisor, vector<vector<long long>>(divisor, vector<long long>(divisor, 0)));
        
        for (int rSum = 0; rSum < divisor; rSum++) {
            for (int gSum = 0; gSum < divisor; gSum++) {
                for (int bSum = 0; bSum < divisor; bSum++) {
                    if (dp[rSum][gSum][bSum] > 0) {
                        newDp[(rSum + value) % divisor][gSum][bSum] = (newDp[(rSum + value) % divisor][gSum][bSum] + dp[rSum][gSum][bSum]) % MODULO;
                        newDp[rSum][(gSum + value) % divisor][bSum] = (newDp[rSum][(gSum + value) % divisor][bSum] + dp[rSum][gSum][bSum]) % MODULO;
                        newDp[rSum][gSum][(bSum + value) % divisor] = (newDp[rSum][gSum][(bSum + value) % divisor] + dp[rSum][gSum][bSum]) % MODULO;
                    }
                }
            }
        }
        
        dp = newDp;
    }
    
    long long result = 0;
    for (int rSum = 0; rSum < divisor; rSum++) {
        for (int gSum = 0; gSum < divisor; gSum++) {
            int bSum = (divisor - rSum - gSum) % divisor;
            if (bSum < 0) bSum += divisor;
            result = (result + dp[rSum][gSum][bSum]) % MODULO;
        }
    }
    
    cout << result << endl;
    return 0;
}

RGB COUNTING CODE C++✅

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from collections import defaultdict, deque

def dfs_count(node, parent, adj, A, mod_target):
    count = 0
    stack = [(node, parent)]
    while stack:
        current, parent = stack.pop()
        if A[current] % 3 == mod_target:
            count += 1
        for neighbor in adj[current]:
            if neighbor != parent:
                stack.append((neighbor, current))
    return count

def solve(N, M, A, E, Q, Queries):
    adj = defaultdict(list)
    for u, v in E:
        adj[u-1].append(v-1)
        adj[v-1].append(u-1)
   
    total_result = 0
    for query in Queries:
        if query[0] == 1:
            _, U, X = query
            U -= 1
            A[U] = X
        elif query[0] == 2:
   
            _, U, X = query
            U -= 1 
            mod_target = X % 3
            count = dfs_count(U, -1, adj, A, mod_target)
            total_result += count
   
    return total_result % (10**9 + 7)


Path Queries On MOD

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python✅

swapping code - python3✅

#include <bits/stdc++.h>
using namespace std;
#define int long long

int32_t main()
{
    int N, M;
    cin >> N >> M;

    vector<vector<int>> grid(N, vector<int>(M));
    for (int i = 0; i < N; ++i)
    {
        for (int j = 0; j < M; ++j)
        {
            cin >> grid[i][j];
        }
    }

    vector<vector<long long>> dp(N, vector<long long>(M, 0));
    vector<vector<long long>> dogDistance(N, vector<long long>(M, LLONG_MAX));

    if (grid[0][0] == 0)
    {
        dogDistance[0][0] = 0;
    }

    dp[0][0] = grid[0][0];

    for (int i = 0; i < N; ++i)
    {
        for (int j = 0; j < M; ++j)
        {
            if (i == 0 && j == 0)
                continue;

            if (i > 0)
            {
                dp[i][j] = max(dp[i][j], dp[i - 1][j] + grid[i][j] - dogDistance[i - 1][j]);
                dogDistance[i][j] = min(dogDistance[i][j], dogDistance[i - 1][j] + 2);
            }

            if (j > 0)
            {
                dp[i][j] = max(dp[i][j], dp[i][j - 1] + grid[i][j] - dogDistance[i][j - 1]);
                dogDistance[i][j] = min(dogDistance[i][j], dogDistance[i][j - 1] + 2);
            }

            if (grid[i][j] == 0)
            {
                dogDistance[i][j] = 0;
            }
        }
    }

    cout << dp[N - 1][M - 1] << endl;

    return 0;
}


Lost in orange ✅- c++

def count_not_divisible(K, L, R):
    def sieve_of_eratosthenes(n):
        is_prime = [True] * (n + 1)
        primes = []
        for p in range(2, n + 1):
            if is_prime[p]:
                primes.append(p)
                for multiple in range(p * p, n + 1, p):
                    is_prime[multiple] = False
        return primes

    def count_divisible_up_to(x, primes):
        from itertools import combinations
        count = 0
        for i in range(1, len(primes) + 1):
            for comb in combinations(primes, i):
                lcm = 1
                for num in comb:
                    lcm *= num
                    if lcm > x:
                        break
                if lcm > x:
                    continue
                if i % 2 == 1:
                    count += x // lcm
                else:
                    count -= x // lcm
        return count

    def count_not_divisible_up_to(x, primes):
        if x == 0:
            return 0
        return x - count_divisible_up_to(x, primes)

    L = int(L)
    R = int(R)

    primes = sieve_of_eratosthenes(K)
   
    count_R = count_not_divisible_up_to(R, primes)
    count_L_minus_1 = count_not_divisible_up_to(L - 1, primes)

    return count_R - count_L_minus_1

K = int(input().strip())
L = input().strip()
R = input().strip()

result = count_not_divisible(K, L, R)
print(result)

Divisible string ✅
Infosys