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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๐ŸšฉMain Group - @SuperExams
๐Ÿ“Job Updates - @FresherEarth

๐Ÿ”ฐAuthentic Coding Solutions(with Outputs)
โš ๏ธDaily Job Updates
โš ๏ธHackathon Updates & Solutions

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๐Ÿš€ Exciting Apprenticeship Opportunity at Microsoft India!

๐ŸŽ‰Microsoft India is now accepting applications for its 12-Month Software Engineering Apprenticeship Program. This is a fantastic opportunity for recent graduates to launch their careers at Microsoft!

๐Ÿ“ข This apprenticeship offers a comprehensive training framework that includes both hands-on project work and structured classroom learning, providing the opportunity to build real-world software, collaborate with global teams, and grow under the guidance of experienced mentors. As a Software Engineering Apprentice, you will gain exposure to cutting-edge technologies, develop in-demand skills, and be part of a vibrant, inclusive communityโ€”all while making meaningful contributions and accelerating your career in tech.

๐ŸŒŸ Eligibility: Graduates of 2025 with no active academic backlogs and no prior full-time employment
๐Ÿ“ Locations: Hyderabad, Bangalore
๐Ÿ“… Duration: 12 months (Expected start: November 2025)
๐Ÿ’ป Apply Now: https://lnkd.in/gRTzJp4g
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 getMaxConsecutiveHidden(commits):
    n = len(commits)
    if n <= 2:
        return 0

    max_hidden = 0
    start = 0


    if commits[-1] != commits[0] + n - 1:
        can_hide_all = True
        for i in range(1, n - 1):
            if commits[i] != commits[0] + i:
                can_hide_all = False
                break
        if can_hide_all:
            return n - 2

    for i in range(1, n):
        if commits[i] == commits[i - 1] + 1:
            continue
        else:
            seq_length = i - start
            if seq_length > 1:
                valid = True
                if start > 0 and i < n:
                    if (commits[start] == commits[start - 1] + 1 and
                        commits[i] == commits[i - 1] + 1):
                        valid = False
                if valid:
                    max_hidden = max(max_hidden, seq_length - 1)
            start = i


    seq_length = n - start
    if seq_length > 1:
        valid = True
        if start > 0:
            if commits[start] == commits[start - 1] + 1:
                valid = False
        if valid:
            max_hidden = max(max_hidden, seq_length - 1)

    return max_hidden
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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#include<bits/stdc++.h>
using namespace std;

int memoize(int l, int r, int residue, const vector<char>& colors) {
    int n = colors.size();
    int dp[n][n][n];
    memset(dp, -1, sizeof(dp));

    if (r < l)
        return 0;
    else if (dp[l][r][residue] != -1)
        return dp[l][r][residue];

    dp[l][r][residue] = (residue + 1) * (residue + 1) + memoize(l, r - 1, 0, colors);

    for (int i = l; i < r; i++) {
        if (colors[i] == colors[r]) {
            dp[l][r][residue] = max(dp[l][r][residue], memoize(l, i - 1, 0, colors) + memoize(i + 1, r, residue + 1, colors));
        }
    }

    return dp[l][r][residue];
}

int solve(int n, vector<char> colors) {
    return memoize(0, n - 1, 0, colors);
}

int main() {
    int n;
    cin >> n;
    vector<char> colors(n);
    for (int i = 0; i < n; i++) {
        cin >> colors[i];
    }
    cout << solve(n, colors);
    return 0;
}

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If you know Python or JavaScript and have some basic software development skills, this could be for you.
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