ll solve(vector<int>& heights) {
    int n = heights.size();
    vector<ll> left(n), right(n, n);
    stack<int> s;
    for (int i = 0; i < n; ++i) {
        while (!s.empty() && heights[s.top()] <= heights[i]) {
            right[s.top()] = i;
            s.pop();
        }
        left[i] = s.empty() ? -1 : s.top();
        s.push(i);
    }
    ll tot = 0;
    for (int i = 0; i < n; ++i) {
        tot += (i - left[i]) + (right[i] - i) - 1;
    }
    return tot;
}.

// CALCULATE REGION
MSCI ✅

#include <iostream>
#include <vector>
using namespace std;
const int MOD = 10000000;
int solve(int N, const string& a) {
    vector<vector<int>> dp(N, vector<int>(3, 0));
    if (a[0] != 'R') dp[0][0] = 1;
    if (a[0] != 'P') dp[0][1] = 1;
    if (a[0] != 'S') dp[0][2] = 1;
    for (int i = 1; i < N; ++i) {
        if (a[i] != 'R') {
            dp[i][0] = (dp[i-1][1] + dp[i-1][2]) % MOD;
        }
        if (a[i] != 'P') {
            dp[i][1] = (dp[i-1][0] + dp[i-1][2]) % MOD;
        }
        if (a[i] != 'S') {
            dp[i][2] = (dp[i-1][0] + dp[i-1][1]) % MOD;
        }
    }
    int ans = (dp[N-1][0] + dp[N-1][1] + dp[N-1][2]) % MOD;
    return ans;
}.

RocK paper scissors ✅

#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>

using namespace std;

const int MOD = 1e9 + 7;
const int MAXN = 101;

vector<int> primes;
vector<vector<int>> adj;
vector<vector<long long>> dp;
bool is_prime[MAXN];

void sieve() {
    fill(is_prime, is_prime + MAXN, true);
    is_prime[0] = is_prime[1] = false;
    for (int i = 2; i < MAXN; ++i) {
        if (is_prime[i]) {
            primes.push_back(i);
            for (int j = 2 * i; j < MAXN; j += i) {
                is_prime[j] = false;
            }
        }
    }
}

void preprocess(int N) {
    sieve();
    adj.clear();
    adj.resize(N);
    dp.clear();
    dp.assign(N, vector<long long>(primes.size(), 1));   
}

void dfs(int node, int par = -1) {
    for (int ne : adj[node]) {
        if (ne == par) continue;
        dfs(ne, node);       
       
        for (int i = 0; i < (int)primes.size(); i++) {
            long long t = 0;
            for (int j = 0; j < (int)primes.size(); j++) {
                if (is_prime[primes[i] + primes[j]]) continue;
                t += dp[ne][j];
                t %= MOD;
            }
            dp[node][i] *= t;
            dp[node][i] %= MOD;
        }
    }
}

int solve(int N, vector<vector<int>> edges) {
    preprocess(N);
   
    if (N == 1) return primes.size();
   
    for (vector<int> edge : edges) {
        adj[edge[0]].push_back(edge[1]);
        adj[edge[1]].push_back(edge[0]);    
    }
   
    dfs(0);
    long long ans = 0;
   
    for (int i = 0; i < (int)primes.size(); i++) {
        ans += dp[0][i];
        ans %= MOD;
    }
   
    return ans;
}

int main() {
    int N;
    cin >> N;
    vector<vector<int>> edges(N-1, vector<int>(2));
    for (int i = 0; i < N-1; i++) {
        cin >> edges[i][0] >> edges[i][1];
        edges[i][0]--;
        edges[i][1]--;
    }

    cout << solve(N, edges) << endl;
    return 0;
}
prime tree code ✅

def getMinimumCost(arr):
    n = len(arr)
    ans = 0
    max_diff = 0

    for i in range(n - 1):
        a, b = arr[i], arr[i + 1]
        max_diff = max(max_diff,abs(a-b))
        ans += ((a-b)** 2)

    if max_diff % 2 == 0:
        ans = ans -max_diff**2/2
    else:
        ans = ans-max_diff 2 + ((max_diff-1) / 2) 2 + ((max_diff+1) / 2) ** 2

    return int(ans) 

MINIMIZE ARRAY COST ✅

class Solution {
public:
    string oddString(vector<string>& words) {
        map<string,int> mp;int k = 3;
        for(int j = 0;j<words.size();j++){
            string temp;
            for(int i = 0;i<words[j].size()-1;i++) temp.push_back(words[j][i+1]-words[j][i]);
            mp[temp]+= k++;
            if(mp.size()==2 and j>2)return words[min(begin(mp)->second,next(begin(mp))->second)-3];

        }
        return words[min(begin(mp)->second,next(begin(mp))->second)-3];
    }
};

// ODD ONE OUT ✅

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

const int MOD = 1e9 + 7;

long long solve(vector<int>& A) {
    int N = A.size();
   
    vector<vector<int>> dp(N + 1, vector<int>(N + 1, 0));
   
    for (int L = 1; L <= N; ++L) {
        stack<int> stack;
        int trees = 0;
        for (int R = L; R <= N; ++R) {
            while (!stack.empty() && A[stack.top() - 1] < A[R - 1]) {
                stack.pop();
            }
            if (stack.empty()) {
                trees++;
            }
            stack.push(R);
            int max_trees = trees;
            if (R > L) {
                if (stack.size() > 1) {
                    max_trees = max(max_trees, trees - 1);
                } else {
                    max_trees = max(max_trees, trees);
                }
               
                if (R - L > trees) {
                    max_trees = max(max_trees, trees + 1);
                }
            }
           
            dp[L][R] = max_trees;
        }
    }
   
    long long result = 0;
    for (int i = 1; i <= N; ++i) {
        for (int j = i; j <= N; ++j) {
            result = (result + dp[i][j]) % MOD;
        }
    }
   
    return result;
}

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

//So many trees

import java.util.Scanner;

public class Miain {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
       
        String S = scanner.nextLine();
       
        StringBuilder S1 = new StringBuilder();
        StringBuilder S2 = new StringBuilder();
       
        for (int i = 0; i < S.length(); i++) {
            if ((i + 1) % 2 == 0) {
                S1.append(S.charAt(i));
            } else {
                S2.append(S.charAt(i));
            }
        }
        System.out.println(S2.toString());
        System.out.println(S1.toString());
       
       
        scanner.close();
    }
}

vector<string> areAlmostEquivalent(vector<string>& s, vector<string>& t) {
    vector<string> result;
    auto countCharacters = [](const string& str) {
        vector<int> count(26, 0);
        for (char c : str) {
            count[c - 'a']++;
        }
        return count;
    };
    for (size_t i = 0; i < s.size(); ++i) {
        const string& str_s = s[i];
        const string& str_t = t[i];
  
        vector<int> count_s = countCharacters(str_s);
        vector<int> count_t = countCharacters(str_t);

        bool isAlmostEquivalent = true;
        for (int j = 0; j < 26; ++j) {
            if (abs(count_s[j] - count_t[j]) > 3) {
                isAlmostEquivalent = false;
                break;
            }
        }

        if (isAlmostEquivalent) {
            result.push_back("YES");
        } else {
            result.push_back("NO");
        }
    }
   
    return result;
}

// Almost equivalent String ✅

SELECT 
    sector, 
    SUM(CAST(REPLACE(REPLACE(capitalization, 'B', '000000000'), 'M', '000000') AS DECIMAL(20, 2))) / 1000000 AS total_capitalization
FROM 
    companies
WHERE 
    sector != 'n/a' AND capitalization != 'n/a'
GROUP BY 
    sector
ORDER BY 
    sector;

SQL stock confirm

def arrayChallenge(arr):
    ans = [0]
    for i in range(1, len(arr)):
        diff = arr[i] - arr[i-1]
        han = (i-1) * diff + diff + ans[i-1]
        ans.append(han)
    return ans.         

wayfair Array challenge ✅

SQL: Calendar Application Events Report 2

SELECT 
    e.dt,
    e.title,
    GROUP_CONCAT(CONCAT(g.full_name, ' <', g.email_address, '>') ORDER BY g.full_name SEPARATOR ', ') AS guests
FROM 
    events e
JOIN 
    events_guests eg ON e.id = eg.event_id
JOIN 
    guests g ON eg.guest_id = g.id
WHERE 
    g.on_vacation = 0
GROUP BY 
    e.id, e.dt, e.title
ORDER BY 
    e.dt ASC
LIMIT 5;

def getMinCores(start, end):
    events = [(t, 1) for t in start] + [(t, -1) for t in end]
    events.sort(key=lambda x: (x[0], -x[1]))
    max_cores = 0
    current_cores = 0
    for time, event_type in events:
        if event_type == 1:
            current_cores += 1
        else:
            current_cores -= 1
        max_cores = max(max_cores, current_cores)
    return max_cores

Process Scheduler ✅

#include <iostream>
#include <vector>
#include <unordered_map>
#include <algorithm>
const int MOD = 1000000007;
using namespace std;
void dfs(int node, const unordered_map<int, vector<pair<int, int>>>& tree, vector<long long>& magnitudes) {
    auto it = tree.find(node);
    if (it != tree.end()) {
        for (const auto& neighbor : it->second) {
            int next_node = neighbor.first;
            int weight = neighbor.second;
            magnitudes[next_node] = (magnitudes[node] * weight) % MOD;
            dfs(next_node, tree, magnitudes);
        }
    }
}

vector<int> findEquivalentMagnitude(int unit_nodes, vector<int> unit_from, vector<int> unit_to, vector<int> unit_weight, int x) {
    unordered_map<int, vector<pair<int, int>>> tree;
    for (int i = 0; i < unit_from.size(); ++i) {
        tree[unit_from[i]].emplace_back(unit_to[i], unit_weight[i]);
    }
    vector<long long> magnitudes(unit_nodes + 1, -1);
    magnitudes[1] = x;

    dfs(1, tree, magnitudes);

    vector<int> result(unit_nodes);
    for (int i = 1; i <= unit_nodes; ++i) {
        result[i - 1] = magnitudes[i];
    }

    return result;
}

int findMaxUses(vector<int>& costs, vector<int>& uses, int budget) {
    int n = costs.size();
   
    if (n < 2) {
        return -1;
    }
   
    vector<pair<int, int>> ft;
    for (int i = 0; i < n; ++i) {
        ft.push_back({costs[i], uses[i]});
    }
   
    sort(ft.begin(), ft.end());
    int mu = -1;

    for (int i = 0; i < n; ++i) {
        for (int j = i + 1; j < n; ++j) {
            int tc = ft[i].first + ft[j].first;
           
            if (tc <= budget) {
                int bg = ft[i].second + ft[j].second;
                mu = max(mu, bg);
            } else {
                break;
            }
        }
    }
   
    return mu;
}

DE Shaw ✅