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๐Ÿ”ฐAuthentic Coding Solutions(with Outputs)
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๐—–๐—ฆ ๐—”๐—น๐—ด๐—ผ ๐Ÿ’ป ๐ŸŒ ใ€Ž๐—–๐—ผ๐—บ๐—ฝ๐—ฒ๐˜๐—ถ๐˜๐—ถ๐˜ƒ๐—ฒ ๐—ฃ๐—ฟ๐—ผ๐—ด๐—ฟ๐—ฎ๐—บ๐—บ๐—ถ๐—ป๐—ดใ€
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def valueOfExpression(n):
    if n <= 1:
        return 0
    count = 0
    d = 2
    while d * d <= n:
        if n % d == 0:
            count += 1
            while n % d == 0:
                n //= d
        d += 1
    if n > 1:
        count += 1
    return count

def fun(arr):
    maxFactors = 0
    result = 0
    for num in arr:
        factors = valueOfExpression(num)
        if factors > maxFactors or (factors == maxFactors and num > result):
            maxFactors = factors
            result = num
    return result

def solve(x, n, a):
    result = float('inf')
    for i in range(n - x + 1):
        result = min(result, fun(a[i:i+x]))
    return result


Value Of an expressionโœ…
Ericsson Hiring Software Developer

Qualification required B.E/ B.Tech/M.tech- 2023 or 2024 passed out with good communication
โ€ข    6 months- 1 year of experience into Java : Core java, Spring, Springboot, Hibernate, Rest Web Services, Java Script, JQuery, Maven
โ€ข    Knowledge of Application Server, Deployment, troubleshooting

https://jobs.ericsson.com/careers/job/563121760839673?jobPipeline=LinkedIn&domain=ericsson.com
๐Ÿ‘1
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int solution(vector<int>& start, vector<int>& dest, vector<int>& limit) {
    int N = start.size();
    int total_cost = 0;
    int max_station = 0;
    for (int i = 0; i < N; i++) {
        int ride_cost = 1;
        ride_cost += 2 * abs(dest[i] - start[i]);
        total_cost += ride_cost;
        max_station = max(max_station, max(start[i], dest[i]));
    }
        int capped_cost = min(total_cost, limit[max_station]);
    return capped_cost;
}


MS TASK 1โœ…
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int solution(vector<int> &A, vector<int> &B) {
    int N = A.size();
    vector<vector<int>> dp(2, vector<int>(N, 0));
    dp[0][0] = A[0];
    dp[1][0] = max(A[0], B[0]);
   
    for (int j = 1; j < N; ++j) {
        dp[0][j] = max(dp[0][j - 1], A[j]);
        dp[1][j] = min(max(dp[0][j], B[j]), max(dp[1][j - 1], B[j]));
    }
   
    return dp[1][N - 1];
}

MS TASK 2โœ