ll solve(ll n, ll m, ll k) {
    vector<int>a(n, 1);
    k--;
    ll sum = n, maxjob = 1;

    while(sum <= m) {
        a[k]++;
        maxjob = a[k];

        for(ll i = k - 1; i >= 0; i--) {
            if(a[i + 1] - a[i] > 1) a[i]++;
        }

        for(ll i = k + 1; i < n; i++) {
            if(a[i - 1] - a[i] > 1) a[i]++;
        }
        sum = 0;
        for(ll i = 0; i < n; i++) sum += a[i];
    }
}

Job one
Atlassian ✅

Company Name: MasterCard

Role: Software Engineer 1

Batch eligible: 2022 and 2023 grads

Apply: https://careers.mastercard.com/us/en/job/MASRUSR201744EXTERNALENUS/Software-Engineer-I

ZS is hiring for DAA

Qualifications:

• 0-2 years of experience.

• Bachelor's degree in any engineering discipline, MIS, operations management, or other relevant degree.

• Proficiency in MS Excel, Python, SQL and regression techniques.

if you interested send ur resume &Referral

varunvj181@gmail.com

#include <iostream>
#include <vector>
#include <cstring>
using namespace std;

const int MAXN = 105;
const int LOGN = 20;

int n;
vector<int> adj[MAXN];
char label[MAXN];
int parent[MAXN][LOGN];
int depth[MAXN];
int freq[MAXN][26];

void dfs(int node, int par, int d) {
    parent[node][0] = par;
    depth[node] = d;

    for (int i = 1; i < LOGN; ++i) {
        if (parent[node][i - 1] != -1) {
            parent[node][i] = parent[parent[node][i - 1]][i - 1];
        }
    }

    for (int child : adj[node]) {
        if (child != par) {
            dfs(child, node, d + 1);
        }
    }
}

void preprocess() {
    memset(parent, -1, sizeof(parent));
    dfs(1, -1, 0);

    for (int node = 1; node <= n; ++node) {
        freq[node][label[node] - 'a']++;
        for (int i = 0; i < 26; ++i) {
            freq[node][i] += freq[parent[node][0]][i];
        }
    }
}

int lca(int u, int v) {
    if (depth[u] < depth[v]) {
        swap(u, v);
    }

    int diff = depth[u] - depth[v];
    for (int i = 0; i < LOGN; ++i) {
        if ((diff >> i) & 1) {
            u = parent[u][i];
        }
    }

    if (u == v) {
        return u;
    }

    for (int i = LOGN - 1; i >= 0; --i) {
        if (parent[u][i] != parent[v][i]) {
            u = parent[u][i];
            v = parent[v][i];
        }
    }

    return parent[u][0];
}

int main() {
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        cin >> label[i];
    }

    for (int i = 0; i < n - 1; ++i) {
        int u, v;
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }

    preprocess();

    int q;
    cin >> q;
    while (q--) {
        int u, v;
        char c;
        cin >> u >> v >> c;

        int l = lca(u, v);
        int ans = freq[u][c - 'a'] + freq[v][c - 'a'] - 2 * freq[l][c - 'a'] + (label[l] == c);
        cout << ans << endl;
    }

    return 0;
}

Colored Vertex ✅

int lengthOfLIS(vector<int>& nums) {
        vector<int>dp(nums.size(),1);
        for(int i = 0; i < nums.size(); i++)
            for(int j = i -1 ; j >= 0; j--)
               if(nums[i] > nums[j]) dp[i] = max(dp[i], 1 + dp[j]);
        return *max_element(dp.begin(),dp.end());
    }
};

Good subsequence✅
(Oracle)

string getLongestRegex(string a, string b, string c)
{
  const size_t n = a.size();
  int idx = -1;
  for (int i = 0; i < n; i++) {
    if (c[i] != a[i] && c[i] != b[i]) { idx = i; }
  }
  if (idx == -1) return "-1";
  string res;
  for (int i = 0; i < n; i++) {
   
    if (i == idx) {
      string cur = "[";
      for (int j = 'A'; j <= 'Z'; j++)  if (j != c[i]) cur += j;
      cur += "]";
      res += cur;
    } else {
      res += "[ABCDEFGHIJKLMNOPQRSTUVWXYZ]";
    }
  }
  return res;
}

Amazon ✅

Geopits DBA - Freshers Bangalore

Batch: 2023/2022

https://www.geopits.com/careers/dba-fresher-walk-in-interview.html?fbclid=IwAR2iYpwE-QyfahLoEGXJ5e0YaEl5d4RPNa_1fm5dzK3XDU9urTZ3gf-txnc

Company Name: EdgeVerve

Role: Software Engineer

Batch eligible: 2022 and 2023 grads

Apply: https://sjobs.brassring.com/TGnewUI/Search/home/HomeWithPreLoad?PageType=JobDetails&partnerid=26209&siteid=5179&Areq=1972BR#jobDetails=1192953_5179

#include<bits/stdc++.h>
using namespace std;
    vector<int> smallestRange(vector<vector<int>>& nums) {
        vector<int>res={-100000,100000};
        int k=nums.size();
        int maxi=INT_MIN;
        priority_queue<vector<int>,vector<vector<int>>,greater<vector<int>>>pq;
        for(int i=0;i<k;i++){
            int num=nums[i][0];
            vector<int>vec{num,0,i};
            //0th minimum element of list
            //1st index of element from the vector
            //2nd index of the vector
            //1st <4,0,0> 2nd <0,0,1> 3rd <5,0,2>
            maxi=max(maxi,num);//took the max of members //5
            pq.push(vec);
            //<0,0,1><4,0,0><5,0,2>
        }
        while(true){//until condition breaks
            vector<int>minval=pq.top();//<0,0,1>
            pq.pop();//<4,0,0><5,0,2>
            if(res[1]-res[0]>maxi-minval[0]){//5-0=5
                res[0]=minval[0];//res[0]=0
                res[1]=maxi;//res[1]=5
            }
            minval[1]++;//<0,1,1>,now from which we poped element in queue that min element containing vector , we should choose another element
            vector<int>vec2=nums[minval[2]];//minval[2]=1,so <0,9,12,20>
            if(minval[1]==vec2.size()){//1!=4
                break;
            }
            else{
                minval[0]=vec2[minval[1]];//minival[0]=vec2[1]=9
                maxi=max(maxi,vec2[minval[1]]);//maxi=max(5,9)=9
                pq.push(minval);//push <9,1,1>
            }
        }
return res;
    }
    int main(){
        int m,n;
        cin>>m>>n;
        vector<vector<int>>v(m,vector<int>(n));
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                cin>>v[i][j];
            }
            sort(v[i].begin(),v[i].end());
        }
        vector<int> ans = smallestRange(v);
        cout<<ans[0]<<" "<<ans[1];
    }

E comm
Amazon ✅

#include <bits/stdc++.h>
  using namespace std;
 
const int N = 1e5+1;
queue<int>q[N];

void solve(){
 
  int n,k;
  cin>>n;
 
  // 1 2 3
 
  vector<int>v(n);

 
  int ans = 1e9;
 
  for(int i=0;i<n;i++){
    cin>>v[i];
  }
 
  cin>>k;
 
    for(int i=0;i<n;i++){
     
      if(q[v[i]].size()==k)q[v[i]].pop();
     
      if(q[v[i]].size()==k-1){
        ans = min(i-q[v[i]].front(),ans);
       
      }
      q[v[i]].push(i);
  }
 
 
  if(ans == 1e9){cout<<"-1"<<endl;return;}
  
 
  cout<<ans-k+1<<endl;
  return;
 
 

 
 
 
}

  signed main() {

    solve();

    return 0;

  }

this was video streaming ✅
Amazon

int64_t getDiscountPairs(int x, vector<int> price)
{
  const size_t n = price.size();
  map<int, int64_t> cnt;
  int64_t res = 0;
  for (int i = 0; i < n; i++) {
    int uu = price[i];
    int v = uu % x;
    if (v == 0) res += cnt[0];
    else res += cnt[x - v]; 
    cnt[v]++;
  }
  return res;
}

use long instead of int_64

Amazon ✅