string solution(string &forth){
int xmin = 0 , xmax = 0;
int x = 0 , y = 0;
for(char d : forth){
if(d == 'E') xmax = max(xmax , ++x);
else if(d == 'W') xmin = min(xmin , --x);
else y++;
}
string L = string(x + 1 - xmin , 'W') + string(y , 'S') + string(1 - xmin , 'E');
string R = string(xmax + 1 - x , 'E') + string(y , 'S') + string(xmax + 1, 'W');
return L.size() < R.size() ? L : R;
}
MS 1
int xmin = 0 , xmax = 0;
int x = 0 , y = 0;
for(char d : forth){
if(d == 'E') xmax = max(xmax , ++x);
else if(d == 'W') xmin = min(xmin , --x);
else y++;
}
string L = string(x + 1 - xmin , 'W') + string(y , 'S') + string(1 - xmin , 'E');
string R = string(xmax + 1 - x , 'E') + string(y , 'S') + string(xmax + 1, 'W');
return L.size() < R.size() ? L : R;
}
MS 1
โค1
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int solution(vector<int>& A, vector<int>& B, int X, int Y) {
int N = A.size();
vector<int> dpA(N), dpB(N);
// Initial condition: Start at position 0 on either line
dpA[0] = A[0];
dpB[0] = B[0];
// Fill the dp arrays
for (int i = 1; i < N; ++i) {
dpA[i] = min(dpA[i - 1] + A[i], dpB[i - 1] + Y + A[i]);
dpB[i] = min(dpB[i - 1] + B[i], dpA[i - 1] + X + B[i]);
}
// The result is the minimum time to complete the car on either line at the last position
return min(dpA[N - 1], dpB[N - 1]);
}
int main() {
int N, X, Y;
cin >> N >> X >> Y;
vector<int> A(N), B(N);
for (int i = 0; i < N; ++i) {
cin >> A[i];
}
for (int i = 0; i < N; ++i) {
cin >> B[i];
}
cout << solution(A, B, X, Y) << endl;
return 0;
}
Ms task2
#include <vector>
#include <algorithm>
using namespace std;
int solution(vector<int>& A, vector<int>& B, int X, int Y) {
int N = A.size();
vector<int> dpA(N), dpB(N);
// Initial condition: Start at position 0 on either line
dpA[0] = A[0];
dpB[0] = B[0];
// Fill the dp arrays
for (int i = 1; i < N; ++i) {
dpA[i] = min(dpA[i - 1] + A[i], dpB[i - 1] + Y + A[i]);
dpB[i] = min(dpB[i - 1] + B[i], dpA[i - 1] + X + B[i]);
}
// The result is the minimum time to complete the car on either line at the last position
return min(dpA[N - 1], dpB[N - 1]);
}
int main() {
int N, X, Y;
cin >> N >> X >> Y;
vector<int> A(N), B(N);
for (int i = 0; i < N; ++i) {
cin >> A[i];
}
for (int i = 0; i < N; ++i) {
cin >> B[i];
}
cout << solution(A, B, X, Y) << endl;
return 0;
}
Ms task2
๐1
int solve(vector<int>& sequence) {
int n = sequence.size();
vector<int> dp(n + 1, INT_MAX);
dp[0] = 0;
for (int i = 0; i < n; ++i) {
if (dp[i] == INT_MAX) continue;
int groupSize = sequence[i];
if (i + groupSize < n) {
dp[i + groupSize + 1] =min(dp[i + groupSize + 1], dp[i]);
}
dp[i + 1] = min(dp[i + 1], dp[i] + 1);
}
return dp[n];
}
Uber โ
int n = sequence.size();
vector<int> dp(n + 1, INT_MAX);
dp[0] = 0;
for (int i = 0; i < n; ++i) {
if (dp[i] == INT_MAX) continue;
int groupSize = sequence[i];
if (i + groupSize < n) {
dp[i + groupSize + 1] =min(dp[i + groupSize + 1], dp[i]);
}
dp[i + 1] = min(dp[i + 1], dp[i] + 1);
}
return dp[n];
}
Uber โ
#include <bits/stdc++.h>
using namespace std;
#define ll long long
ll n;
ll minFreq=0;
unordered_map<ll,pair<ll,ll>> keyVal;
unordered_map<ll,list<ll>> freqList;
unordered_map<ll,list<ll>::iterator> pos;
ll get(ll key)
{
if(keyVal.find(key) == keyVal.end()) return -1;
freqList[keyVal[key].second].erase(pos[key]);
keyVal[key].second++;
freqList[keyVal[key].second].push_back(key);
pos[key] = --freqList[keyVal[key].second].end();
if(freqList[minFreq].empty()) minFreq++;
return keyVal[key].first;
}
void put(ll key, ll value)
{
if(!n) return;
if(keyVal.find(key) != keyVal.end())
{
keyVal[key].first = value;
freqList[keyVal[key].second].erase(pos[key]);
keyVal[key].second++;
freqList[keyVal[key].second].push_back(key);
pos[key] = --freqList[keyVal[key].second].end();
if(freqList[minFreq].empty())
minFreq++;
return;
}
if(keyVal.size()==n)
{
ll delKey = freqList[minFreq].front();
keyVal.erase(delKey);
pos.erase(delKey);
freqList[minFreq].pop_front();
}
keyVal[key] = {value,1};
freqList[1].push_back(key);
pos[key] = --freqList[1].end();
minFreq = 1;
}
signed main()
{
cin>>n;
ll q; cin>>q;
while(q--)
{
ll x,y,z; cin>>x>>y>>z;
if(x==1) cout<<get(y)<<" ";
else put(y,z);
}
return 0;
}
Catch it โ
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
Connected Components -> vector<int>vis(1e5+2,0);
vector<vector<int>>facs(1e5+2);
vector<int>par(1e5+2,-1),rnk(1e5+2,1);
int find(int n){
if(par[n]<0){
return n;
}
return par[n]=find(par[n]);
}
void unite(int a, int b){
a=find(a);
b=find(b);
if(a==b){
return;
}
if(rnk[a]>rnk[b]){
swap(a,b);
}
par[a]=b;
rnk[b]+=rnk[a];
}
void sieve(){
for(int i=2;i<=1e5;i++){
for(int j=2;j*j<=i;j++){
if(i%j==0){
facs[i].pb(j);
if(j*j!=i){
facs[i].pb(i/j);
}
}
}
sort(facs[i].begin(),facs[i].end());
}
}
void solve(){
sieve();
int n;
cin>>n;
vector<int>v(n);
for(int i=0;i<n;i++){
cin>>v[i];
}
for(int i=1e5;i>=2;i--){
for(auto x:facs[i]){
bool flag=0;
int val;
for(int j=0;j<n;j++){
if(v[j]%x==0){
if(!flag){
flag=1;
val=j;
}
else{
unite(j,val);
}
}
}
}
}
int cc=0;
for(int i=0;i<n;i++){
if(vis[find(i)]){
continue;
}
else{
vis[find(i)]=1;
cc++;
}
}
cout<<cc<<endl;
}
Connected Components โ
vector<vector<int>>facs(1e5+2);
vector<int>par(1e5+2,-1),rnk(1e5+2,1);
int find(int n){
if(par[n]<0){
return n;
}
return par[n]=find(par[n]);
}
void unite(int a, int b){
a=find(a);
b=find(b);
if(a==b){
return;
}
if(rnk[a]>rnk[b]){
swap(a,b);
}
par[a]=b;
rnk[b]+=rnk[a];
}
void sieve(){
for(int i=2;i<=1e5;i++){
for(int j=2;j*j<=i;j++){
if(i%j==0){
facs[i].pb(j);
if(j*j!=i){
facs[i].pb(i/j);
}
}
}
sort(facs[i].begin(),facs[i].end());
}
}
void solve(){
sieve();
int n;
cin>>n;
vector<int>v(n);
for(int i=0;i<n;i++){
cin>>v[i];
}
for(int i=1e5;i>=2;i--){
for(auto x:facs[i]){
bool flag=0;
int val;
for(int j=0;j<n;j++){
if(v[j]%x==0){
if(!flag){
flag=1;
val=j;
}
else{
unite(j,val);
}
}
}
}
}
int cc=0;
for(int i=0;i<n;i++){
if(vis[find(i)]){
continue;
}
else{
vis[find(i)]=1;
cc++;
}
}
cout<<cc<<endl;
}
Connected Components โ
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int calcDist(int x) {
if (x == 0) return -1;
int max_dist = -1;
int curr_dist = -1;
int prev_pos = -1;
int pos = 0;
while (x > 0) {
if (x & 1) {
if (prev_pos != -1) {
curr_dist = pos - prev_pos;
max_dist = max(max_dist, curr_dist);
}
prev_pos = pos;
}
x >>= 1;
pos++;
}
return max_dist;
}
vector<int> getTopKDistances(vector<int>& nums, int k) {
vector<pair<int, int>> dists;
for (int num : nums) {
int d = calcDist(num);
dists.push_back({d, num});
}
sort(dists.begin(), dists.end(), [](const pair<int, int>& a, const pair<int, int>& b) {
if (a.first != b.first) {
return a.first > b.first;
} else {
return a.second > b.second;
}
});
vector<int> top_k;
for (int i = 0; i < k; ++i) {
top_k.push_back(dists[i].second);
}
return top_k;
}
Maximum bit differenceโ
#include <vector>
#include <algorithm>
using namespace std;
int calcDist(int x) {
if (x == 0) return -1;
int max_dist = -1;
int curr_dist = -1;
int prev_pos = -1;
int pos = 0;
while (x > 0) {
if (x & 1) {
if (prev_pos != -1) {
curr_dist = pos - prev_pos;
max_dist = max(max_dist, curr_dist);
}
prev_pos = pos;
}
x >>= 1;
pos++;
}
return max_dist;
}
vector<int> getTopKDistances(vector<int>& nums, int k) {
vector<pair<int, int>> dists;
for (int num : nums) {
int d = calcDist(num);
dists.push_back({d, num});
}
sort(dists.begin(), dists.end(), [](const pair<int, int>& a, const pair<int, int>& b) {
if (a.first != b.first) {
return a.first > b.first;
} else {
return a.second > b.second;
}
});
vector<int> top_k;
for (int i = 0; i < k; ++i) {
top_k.push_back(dists[i].second);
}
return top_k;
}
Maximum bit differenceโ
WITH june_events AS (
SELECT *
FROM events
WHERE dt >= '2022-06-01' AND dt < '2022-07-01'
),
grouped_data AS (
SELECT
mime,
GROUP_CONCAT(DISTINCT SUBSTRING_INDEX(filename, '.', -1) ORDER BY SUBSTRING_INDEX(filename, '.', -1) SEPARATOR ', ') AS extension,
COUNT(*) AS files,
SUM(filesize) / (1024 * 1024) AS total_mib
FROM june_events
GROUP BY mime
)
SELECT
mime,
extension,
files,
CASE
WHEN total_mib >= 1024 THEN CONCAT(ROUND(total_mib / 1024, 2), ' GiB')
ELSE CONCAT(ROUND(total_mib, 2), ' MiB')
END AS total
FROM grouped_data
ORDER BY total_mib DESC;
Download file type Report โ
import java.util.*;
public class LongestSequenceOfFollowers {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
long n = sc.nextLong();
long[] dp = new long[(int)(n + 1)];
long m = 1000000007;
dp[1] = 0; // Starting condition
for (int i = 2; i <= n; i++) {
if (i % 2 == 0) {
dp[i] = (dp[i-1] * 3 + 3) % m;
} else {
dp[i] = (dp[i-1] * 3 - 3 + m) % m; // Ensure non-negative result before modulo
}
}
System.out.println(dp[(int)n]);
}
}
Number of paths โ
public class LongestSequenceOfFollowers {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
long n = sc.nextLong();
long[] dp = new long[(int)(n + 1)];
long m = 1000000007;
dp[1] = 0; // Starting condition
for (int i = 2; i <= n; i++) {
if (i % 2 == 0) {
dp[i] = (dp[i-1] * 3 + 3) % m;
} else {
dp[i] = (dp[i-1] * 3 - 3 + m) % m; // Ensure non-negative result before modulo
}
}
System.out.println(dp[(int)n]);
}
}
Number of paths โ
#include <bits/stdc++.
}
Count the Arrays โ
h>
using namespace s
td;
typedef long long
ll;
const int MOD = 1e9 +
9;
struct testca
se{
testcase
(){
int n; cin >>
n;
vector<int> a(
n);
for (int i=0; i<n; i++) cin >> a[
i];
sort(a.begin(), a.end(
));
vector<int> b(
n);
for (int i=0; i<n; i++) cin >> b[
i];
sort(b.begin(), b.end(), greater<>(
));
ll result =
1;
for (int i=0; i<n; i+
+){
int geq_count = a.size() - (upper_bound(a.begin(), a.end(), b[i]) - a.begin(
));
result = result * max(geq_count - i, 0) % M
OD;
}
cout << result << "\
n";
}
}
;
int main
(){
cin.tie(0)->sync_with_stdio(
0);
int t; cin >>
t;
while (t--) testcase
();
}
Count the Arrays โ
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
vector < vector < int > > adj;
string letters;
int dfs(int u, int p, int& best_path){
// visit the current node
int best_path_1 = 0, best_path_2 = 0;
auto update_max = [&](int x){
if(x >= best_path_1)
best_path_2 = best_path_1, best_path_1 = x;
else if(x >= best_path_2)
best_path_2 = x;
};
for(auto& v : adj[u]){
// if the adjacent node not visited and not the same letter let's calc it
if(v == p) continue;
int v_ans = dfs(v, u, best_path);
if(letters[u] != letters[v])
update_max(v_ans);
}
// take two branches
best_path = max({best_path, best_path_1 + best_path_2 + 1});
// the maximum path from the current node
return 1 + best_path_1;
}
void add_edge(int u, int v){
adj[u].push_back(v), adj[v].push_back(u);
}
int longestPath(vector<int>& parent, string& s) {
// global variable initialization
int n = parent.size();
adj = vector < vector < int > > (n);
letters = s;
// add edges between i and parent of i
for(int i = 1; i < n; i++)
add_edge(i, parent[i]);
int best_path = 0;
dfs(0, -1, best_path);
// the length of the longest path with the required conditions.
return best_path;
}
int main()
{
int n; cin >> n;
vector<int> parent(n);
for (auto & x : parent) cin >> x;
string s; cin >> s;
cout << longestPath(parent, s) << endl;
}
Fantasy forest Pathfinding โ
#include <bits/stdc++.h
>
using namespace s
td;
#define forn(i, n) for(int i = 0; i < int(n); i+
+)
const int INF = 1
e9;
int main
(){
int
n;
cin >>
n;
vector<string> s(
2);
forn(i, 2) cin >> s[
i];
vector<array<array<int, 2>, 2>> dp(n +
1);
forn(i, n + 1) forn(j, 2) forn(k, 2) dp[i][j][k] = -I
NF;
dp[0][0][s[1][0] == '1'] = s[1][0] == '
1';
dp[0][0][0] =
0;
forn(i, n - 1) forn(j,
2){
int nxtj = s[j][i + 1] == '
1';
int nxtj1 = s[j ^ 1][i + 1] == '
1';
dp[i + 1][j ^ 1][0] = max(dp[i + 1][j ^ 1][0], dp[i][j][1] + nxtj
1);
dp[i + 1][j][nxtj1] = max(dp[i + 1][j][nxtj1], dp[i][j][0] + nxtj1 + nxt
j);
dp[i + 1][j][0] = max(dp[i + 1][j][0], dp[i][j][0] + nxt
j);
}
cout << max({dp[n - 1][0][0], dp[n - 1][0][1], dp[n - 1][1][0], dp[n - 1][1][1]}) << '\
n';
}
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
#include <bits/stdc++.h> using namespace std; #define forn(i, n) for(int i = 0; i < int(n); i++) const int INF = 1e9; int main(){ int n; cin >> n; vector<string> s(2); forn(i, 2) cin >> s[i]; vector<array<array<int, 2>, 2>> dp(n + 1); forn(i, n + 1) forn(jโฆ
Forwarded from OffCampus Jobs | OnCampus Jobs | Daily Jobs Updates | Lastest Jobs | All Jobs | CSE Jobs | Fresher Jobs โฅ (Dushyant)
Canonical is hiring for Cloud Support Associate Engineer
Expected Salary: 5-10 LPA
Apply here:
https://boards.greenhouse.io/canonicaljobs/jobs/6110366?gh_src=2a09c4971us
Expected Salary: 5-10 LPA
Apply here:
https://boards.greenhouse.io/canonicaljobs/jobs/6110366?gh_src=2a09c4971us
Forwarded from OffCampus Jobs | OnCampus Jobs | Daily Jobs Updates | Lastest Jobs | All Jobs | CSE Jobs | Fresher Jobs โฅ (Dushyant)
Miamin Systems Inc. is hiring for Miamin Systems Inc.
Expected Salary: 4-6 LPA
Apply here:
https://linkedin.com/jobs/view/3978436871/?alternateChannel=searchq
Expected Salary: 4-6 LPA
Apply here:
https://linkedin.com/jobs/view/3978436871/?alternateChannel=searchq
Linkedin
Miamin Systems Inc. hiring Associate Software Engineer in Bengaluru, Karnataka, India | LinkedIn
Posted 11:28:53 AM. Overview: We are looking for enthusiastic and talented Fresher Software Engineers to join ourโฆSee this and similar jobs on LinkedIn.
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
class Solution {
int ring_size;
unordered_map<char,vector<int>> mp;
int clockwise(int curr, int new_pos){
if(new_pos >= curr){
return new_pos-curr;
}
return ring_size - (curr - new_pos);
}
int anti_clockwise(int curr, int new_pos){
if(curr >= new_pos){
return curr - new_pos;
}
return ring_size - (new_pos - curr);
}
int solve(string &key, int idx, int pos, vector<vector<int>>& dp){
if(idx == key.size()){
return 0; //end of key
}
if(dp[idx][pos] != -1){
return dp[idx][pos];
}
int steps = INT_MAX;
int key_value = key[idx];
//going to all indexes
for(int i = 0; i < mp[key_value].size(); i++){
int new_pos = mp[key_value][i];
int taken = solve(key,idx+1,new_pos,dp);
//clockwise
steps = min(steps,1+clockwise(pos,new_pos)+taken);
//anticlockwise
steps = min(steps,1+anti_clockwise(pos,new_pos)+taken);
}
return dp[idx][pos] = steps;
}
public:
int findRotateSteps(string& ring, string& key) {
ring_size = ring.size();
for(int i = 0; i < ring_size; i++){
mp[ring[i]].push_back(i);
}
vector<vector<int>> dp(key.size(),vector<int>(ring.size(),-1));
return solve(key,0,0,dp);
}
};
Uber โ