//Traffic Flow
//Codevita

XIRR(40/100) CodeVita

Board game✅

def getSubsequenceCount(s1, s2):
    def count_subsequences(s1, s2, index1, index2):
        if index1 == 3:
            return 1
        if index2 == len(s2):
            return 0

        count = count_subsequences(s1, s2, index1, index2 + 1)

        if s1[index1] == s2[index2]:
            count += count_subsequences(s1, s2, index1 + 1, index2 + 1)

        return count

    return count_subsequences(s1, s2, 0, 0)

Meesho ✅

def maximum_learning(iv, articles, p):
    size = len(articles)
    art = [x * 2 for x in articles]
    ivs = iv

    return knap_sack_top_down(ivs, art, p, size)

def knap_sack_top_down(val, wt, W, n):
    mat = [[0] * (W + 1) for _ in range(n + 1)]

    for i in range(1, n + 1):
        for j in range(1, W + 1):
            if i == 0 or j == 0:
                mat[i][j] = 0
            elif wt[i - 1] <= j:
                mat[i][j] = max(val[i - 1] + mat[i - 1][j - wt[i - 1]], mat[i - 1][j])
            else:
                mat[i][j] = mat[i - 1][j]

    return mat[n][W]

Meesho ✅

import heapq
def getSeatsAllocation(arr):
    n = len(arr)
    ppl = [[] for i in range(2 * n + 4)]
   
    for i, x in enumerate(arr):
        ppl[x].append(i)

    h = []
    ans = [-1] * n
   
    for pos in range(1, 2 * n + 1):
        for i in ppl[pos]:
            heapq.heappush(h, (-pos, i))
       
        if len(h) > 0:
            ans[heapq.heappop(h)[1]] = pos
   
    return ans

Meesho ✅

def max_tie_moments(n, moments):
    max_tie_count = 0
    current_difference = 0

    for moment in moments:
        bob_goals, noddy_goals = moment
        difference = bob_goals - noddy_goals
        if difference == current_difference:
            max_tie_count += 1
        elif difference > current_difference:
            current_difference = difference
            max_tie_count = 0 

    return max_tie_count

def alternatingParityPermutations(n):
    result = []

    def backtrack(permutation, used):
        if len(permutation) == n:
            result.append(permutation.copy())
            return

        for i in range(1, n + 1):
            if not used[i]:
                if not permutation or (permutation[-1] % 2 != i % 2):
                    permutation.append(i)
                    used[i] = True
                    backtrack(permutation, used)
                    used[i] = False
                    permutation.pop()

    used_numbers = [False] * (n + 1)
    backtrack([], used_numbers)
    return result

Meesho ✅

def binary_exponentiation(base, exponent, mod=None):
    result = 1
    base = base % mod if mod else base

    while exponent > 0:
        if exponent % 2 == 1:
            result = (result * base) % mod if mod else result * base
        base = (base * base) % mod if mod else base * base
        exponent //= 2

    return result

def calculate_expression(n):
    mod = 10**9 + 7  # You can use a different modulus if needed
    return (binary_exponentiation(26, n, mod) -
            3 * binary_exponentiation(25, n, mod) -
            n * binary_exponentiation(25, n-1, mod) +
            3 * binary_exponentiation(24, n, mod) +
            2 * n * binary_exponentiation(24, n-1, mod) -
            binary_exponentiation(23, n, mod) -
            n * binary_exponentiation(23, n-1, mod)) % mod

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Role : Engineering Internship - 6 Months Internship

Batch : 2024/2023 passouts

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Company Name : Fractal

Role : Summer Internship

Batch : 2025 passouts

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#include <bits/stdc++.h>
#define ll long long

using namespace std;

ll solve(ll k, ll s) {
    auto comb = [](ll n) {
        return n * (n - 1) / 2;
    };

    if (k > 3 * s) return 0;

    ll ans = comb(k + 2);

    if (k > s) ans -= 3 * comb(k - s + 1);
    if (k - 2 >= 2 * s) ans += 3 * comb(k - 2 * s);

    return ans;
}

int main() {
    ll n, s;
    cin >> n >> s;
    cout << solve(s,n) << endl;
    return 0;
}

Distribute Car Toy
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#include <bits/stdc++.h>
using namespace std;

int main() {
    int n;
    cin >> n;
    vector<vector<int>> freq(n, vector<int> (26, 0));
    string s;
    for(int i = 0; i < n; i++) {
        string s;
        cin >> s;
        for(auto &ch : s)  
            freq[i][ch - 'a']++;
    }
    long long ans = 0;
    for(int i = 0; i < 26; i++) {
        int minValue = INT_MAX;
        for(int j = 0; j < n; j++)
            minValue = min(minValue, freq[j][i]);
        ans += minValue;
    }
    cout << ans << endl;
}

Service Now ✅

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

int main() {
    int n;
    cin >> n;
    vector<int> nums(n);
    for(int i = 0; i < n; i++) {
        cin >> nums[i];
    }
    nums.erase(unique(nums.begin(), nums.end()), nums.end());
    int sum = 0;
    for(int i = 0; i < nums.size(); i++) {
        sum += pow(2, nums[i]);
    }
    cout << sum << endl;
    return 0;
}


Special sum ✅
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#include<bits/stdc++.h>
using namespace std;

int main() {
    int N;
    cin >> N;
    multiset<int, greater<int>> s;
    for(int i = 0; i < N; i++) {
        int x;
        cin >> x;
        s.insert(x);
    }
    int count = 0;
    while(!s.empty()) {
        int x = *s.begin();
        s.erase(s.begin());
        if(s.empty() || *s.begin() != x) {
            count++;
            if(x / 2 > 0)
                s.insert(x / 2);
        } else {
            s.erase(s.lower_bound(x), s.upper_bound(x));
        }
    }
    cout << count << endl;
    return 0;
}


Counting problem ✅
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Diamond Boxes ✅
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#include<bits/stdc++.h>
using namespace std;

string solve(string s, int k) {
    string vowels = "aeiou";
    string result = "";
    for (char c : s) {
        if (vowels.find(c) != string::npos) {
            k += 2;
        } else {
            k += 1;
        }
        result += to_string(k * int(c)) + " ";
    }
    return result;
}

int main() {
    string s;
    int k;
    cin >> s >> k;
    cout << solve(s, k) << endl;
    return 0;
}


Enciphering a String ✅
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