def countPairs(numbers, k):
    numbers_set = set(numbers)
    pairs = set()
    for number in numbers:
        if (number + k) in numbers_set:
            pairs.add((number, number + k))
        if (number - k) in numbers_set:
            pairs.add((number - k, number))
    return len(pairs)

Counting Pairs ✅

int numConns(const vector<vector<int>>& grid) {
    vector<int> cnt;
    int rows = grid.size();
    int cols = grid[0].size();

    for (int i = 0; i < rows; ++i) {
        int v = 0;
        for (int j = 0; j < cols; ++j) {
            if (grid[i][j] == 1) {
                v++;
            }
        }
        if (v != 0) {
            cnt.push_back(v);
        }
    }

    int conns = 0;
   
    if (cnt.size() >= 2) {
        for (size_t i = 0; i < cnt.size() - 1; ++i) {
            conns += cnt[i] * cnt[i + 1];
        }
    }

    return conns;
}


//Grid climing ✅

#include <bits/stdc++.h>
using namespace std;
long solve(vector<vector<int>>& grid) {
    long long  prev = 0;
    long long  connections = 0;
    for (vector<int>& row : grid) {
        int sum = 0;
        for (int n : row) {
            sum += n;
        }

        if (sum > 0) {
            connections += prev * sum;
            prev = sum;
        }
    }

    return connections;
}

Grid climbing ✅

#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll solve(ll n, vector<vector<ll>>& flights, ll src, ll dst, ll k)
{
      vector<pair<ll,ll>> adj[n];
        for(auto it : flights){
            adj[it[0]].push_back({it[1],it[2]});
        }
        queue<pair<ll,pair<ll,ll>>>q;
        q.push({0,{src,0}});
        vector<ll>dist(n,1e9);
        dist[src]=0;
        while(!q.empty()){
            auto it=q.front();
            q.pop();
            ll stops=it.first;
            ll node=it.second.first;
            ll wt=it.second.second;
            if(stops>k)continue;
            for(auto iter:adj[node]){
                ll adjnode=iter.first;
                ll dis=iter.second;
                if(wt+dis<dist[adjnode]&&stops<=k){
                    dist[adjnode]=wt+dis;
                    q.push({stops+1,{adjnode,wt+dis}});
                }
            }
        }
        if(dist[dst]==1e9)return -1;
        return dist[dst];
}
signed main()
{
    ll n,src,des,k,m; cin>>n>>src>>des>>k>>m;
    vector<vector<ll>>a(m,vector<ll>(3));
    for(ll i=0;i<m;i++)
    {
       ll x,y,w; cin>>x>>y>>w;
       a[i]={x,y,w};
    }
    cout <<solve(n,a,src,des,k)<<endl;
    return 0;
}

Trip planner ✅

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

using ll = long long; 
const ll INF = LLONG_MAX;

vector<ll> dijkstra(int n, int src, const vector<vector<pair<int, int>>>& adj) {
    vector<ll> dist(n, INF);
    dist[src] = 0;
    priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<>> pq;
    pq.push({0, src});

    while (!pq.empty()) {
        auto [d, u] = pq.top();
        pq.pop();

        if (d > dist[u]) continue;

        for (auto [v, cost] : adj[u]) {
            if (dist[u] + cost < dist[v]) {
                dist[v] = dist[u] + cost;
                pq.push({dist[v], v});
            }
        }
    }

    return dist;
}

vector<vector<ll>> shortest(int nodes, const vector<int>& deliveries, const vector<vector<pair<int, int>>>& adj) {
    int k = deliveries.size();
    vector<vector<ll>> dist(k, vector<ll>(nodes));

    for (int i = 0; i < k; ++i) {
        dist[i] = dijkstra(nodes, deliveries[i], adj);
    }

    return dist;
}

ll minTime(int nodes, int k, const vector<ll>& startDist, const vector<vector<ll>>& dist, const vector<int>& deliveries) {
    vector<vector<ll>> dp(1 << k, vector<ll>(k, INF));

    for (int i = 0; i < k; ++i) {
        dp[1 << i][i] = startDist[deliveries[i]];
    }

    for (int mask = 0; mask < (1 << k); ++mask) {
        for (int i = 0; i < k; ++i) {
            if (mask & (1 << i)) {
                for (int j = 0; j < k; ++j) {
                    if (!(mask & (1 << j))) {
                        dp[mask | (1 << j)][j] = min(dp[mask | (1 << j)][j], dp[mask][i] + dist[i][deliveries[j]]);
                    }
                }
            }
        }
    }

    ll minTime = INF;
    for (int i = 0; i < k; ++i) {
        minTime = min(minTime, dp[(1 << k) - 1][i] + dist[i][0]);
    }

    return minTime;
}

int main() {
    int n, e, c;
    cin >> n >> e;

    vector<int> from(e), to(e), cost(e);
    for (int i = 0; i < e; ++i) {
        cin >> from[i] >> to[i] >> cost[i];
    }

    cin >> c;
    vector<int> deliveries(c);
    for (int i = 0; i < c; ++i) {
        cin >> deliveries[i];
    }

    vector<vector<pair<int, int>>> adj(n);
    for (int i = 0; i < e; ++i) {
        adj[from[i]].emplace_back(to[i], cost[i]);
        adj[to[i]].emplace_back(from[i], cost[i]);
    }

    vector<ll> start = dijkstra(n, 0, adj);
    vector<vector<ll>> dist = shortest(n, deliveries, adj);

    ll mini = minTime(n ,c, start, dist, deliveries);
    cout << mini << endl;

    return 0;
}

//optimising delivery✅

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def dfs(node, par, depth, graph, signal_speed):
    global cnt
    if depth % signal_speed == 0 and depth != 0:
        cnt += 1
    for neighbor, weight in graph[node]:
        if neighbor == par:
            continue
        dfs(neighbor, node, depth + weight, graph, signal_speed)

def getNumPairs(server_nodes, server_from, server_to, server_weight, signal_speed):
    graph = {}
    for i in range(server_nodes - 1):
        if server_from[i] not in graph:
            graph[server_from[i]] = []
        if server_to[i] not in graph:
            graph[server_to[i]] = []
        graph[server_from[i]].append((server_to[i], server_weight[i]))
        graph[server_to[i]].append((server_from[i], server_weight[i]))
    result = []
    for i in range(1, server_nodes + 1):
        global cnt
        cnt = 0
        dfs(i, i, 0, graph, signal_speed)
        result.append(cnt * (cnt - 1))
    return result

Oracle ✅

long long result = 0;

// Helper function to calculate
// the maximum path sum using DFS
long long findMaximumPathSum(int currentNode,
                        int previousNode,
                        const vector<vector<int>> &adj,
                        const vector<int> &A)
{
    // Nodes to which currentNode is
    // connected to
    const vector<int> &v = adj[currentNode];
    int maximumBranchSum1 = 0;
    int maximumBranchSum2 = 0;
    for (auto vtx : v) {
        // checking whether the branch is
        // visited already
        if (vtx == previousNode) {
            continue;
        }
        long long bs = findMaximumPathSum(vtx, currentNode,
                           adj, A);

        // Storing the maximum of value of
        // branch path sums
        // maximumBranchSum1 will store the
        // maximum value
        // maximumBranchSum2 will store the
        // 2nd most maximum value
        if (bs >= maximumBranchSum1) {
            maximumBranchSum2 = maximumBranchSum1;
            maximumBranchSum1 = bs;
        }
        else {
            maximumBranchSum2
                = max(maximumBranchSum2, bs);
        }
    }
    result = max(result,
                 A[currentNode] + maximumBranchSum1
                     + maximumBranchSum2);

    // updating the value of current value
    // with maximum path sum including
    // currentNode
    return A[currentNode] + maximumBranchSum1;
}


long long bestSumAnyTreePath(vector<int>& parents, vector<int>& values) {
    int n = parents.size();
    vector<vector<int>> mp(n);

    for (int i = 0; i < n; i++) {
        if (parents[i] != -1) {
            mp[parents[i]].push_back(i);
        }
    }

    result = 0;
    findMaximumSumPath(0, -1, mp, values);

    return result;
}

Best sum many treepath ✅
Oracle

vector<long> bitwiseEquations(vector<long> a, vector<long> b)
{

    // long long x = 0, y = 0;
    int n = a.size();
    vector<long> ans(n);
    for (int i = 0; i < n; i++)
    {
        if (a[i] < b[i])
        {
            ans[i] = 0;
            continue;
        }
        long x = 0, y = 0;
        long diff = a[i] - b[i];
        diff /= 2;
        // cout << diff << endl;
        for (int j = 0; j < 64; j++)
        {
            if (b[i] & (1 << j))
            {
                if ((diff & (1 << j)) == 0)
                {
                    // x |= (1 << j);
                    y |= (1 << j);
                }
                else
                {
                    x = 0, y = 0;
                    break;
                }
            }
            else
            {
                if ((diff & (1 << j)))
                {
                    x |= (1 << j);
                    y |= (1 << j);
                }
            }
        }
        ans[i] = 2 * x + 3 * y;
    }
    return ans;
}

The bitwise Equation ✅