Today Anyone give American Express OA?

Company Name: Anuta Network

Role: Software Engineer Intern

Batch eligible: 2022 and 2023 grads.

Apply: https://www.linkedin.com/jobs/view/3677244584

public static int find(int[] arr) {
// Sort the array in ascending order
Arrays.sort(arr);
HashSet<Double> hm=new HashSet<>();

int n = arr.length;
int count = 0;
int left = 0;
int right = n - 1;

while (left < right) {

double avgMaxMin = (arr[left] + arr[right]) / 2.0;

hm.add(avgMaxMin);

left++;
right--;
}

return hm.size();
}

Amazon OA✅

    public static String lexicographicallyMaxString(String inputStr) {
        char[] chars = inputStr.toCharArray();
        int n = chars.length;
        HashSet<Character> c=new HashSet<>();

    
        Map<Character, Integer> charCount = new HashMap<>();
        for (char ch : chars) {
            charCount.put(ch, charCount.getOrDefault(ch, 0) + 1);
        }

      
        PriorityQueue<Character> pq = new PriorityQueue<>((a, b) -> a-b);
        for (char ch : charCount.keySet()) {
            pq.add(ch);
            c.add(ch);
        }

      
        StringBuilder sb = new StringBuilder();
        while (!pq.isEmpty()) {
            char currentChar = pq.poll();
            if(c.contains(currentChar)){
            sb.append(currentChar);
            c.remove(currentChar);
            }
        }

        return sb.toString();
    }

Modify the string ✅

Company Name: Jade Global Software

Role: Associate Analyst

Batch eligible: 2023 grads only

Apply: https://myanatomy.in/events-and-hackathon/jade-global-software/jade-global-software-1970-01-01

P.S: Only apply if you don’t have any offers.

BOX ID Systems hiring software developer

Apply : https://www.showwcase.com/job/8243-software-developer

Company Name: American Express

Role: Software Engineer

Batch eligible: 2024 grads only.

Apply: https://aexp.eightfold.ai/careers/job/17212750

Note: M.Tech, B.Tech + M.Tech, MCA and M.Sc passing out in 2024 grads are eligible only. (Only Master degree)

Company Name: GE Healthcare

Role: Software Engineer Specialist

Batch eligible: 2022 and 2023 grads.

Apply: https://jobs.gecareers.com/healthcare/global/en/job/R3724264/Software-Engineering-Specialist

Cisco Interview Experience :

1)First introduction
2)Then she asked about course like i have done cisco ccna course that
3)Then about web dev internship
4)College Project Explain
5)About python
6)Why python
7)She gave a task like CAT BAT PAT DAT
output should be like

["CA", "BA", "PA", "DA"]
8)Then 2 questions about dependency
9)Global and local variables
10)The she gave me 2 strings and asked me how to print them in one line
11)Without using concat
12)OOPS
13)Different between python and java
14)Reverse of a number
15)Swapping
16)Why reactjs
17)API
18)Backend coding
19)If u are from ECE then they will ask logic of code
Any Questions

Ridecell hiring for various roles like Backend, iOS, Android, DevOps, and QA interns

Batch : 2024 / 2025

Apply : https://lnkd.in/dy3_VaNq

KoinBasket hiring Growth Product Management Intern

Batch ; 2025 / 2026

Apply : https://www.linkedin.com/jobs/view/3671860271

Ground truth exam help done ✅ those who will be give this exam send questions in group

Uber ✅

long long solve(int N,int M,vector<int>Arr){
    vector<int>v(N+1,INT_MAX);
    int n=N;
    for(int i=n-1;i>=0;i--){
        v[i]=min(v[i+1],Arr[i]);
    }
    long long ans=INT_MAX;
    for(int i=0;i<N;i++){
        if((i+M-1)<N){
            ans=min(ans,1ll*Arr[i]*v[i+M-1]);
        }
    }
    return ans;
}

Array transfer
Uber ✅

void dfs(int curr, int p, vector<int>* e, vector<int>& v, vector<int>& par, vector<int>& A){
    par[curr] = p;
    v[curr] = A[curr - 1];
    for(auto it : e[curr]){
        if(it == p) continue;
        dfs(it, curr, e, v, par, A);
        v[curr] += v[it];
    }
}
int solve(int N, vector<vector<int>> edges, vector<int> A){
    vector<int> e[N+1];
    map<vector<int>, int> mp;
    int c = 0;
    for(auto it : edges){
        e[it[0]].push_back(it[1]);
        e[it[1]].push_back(it[0]);
        mp[{it[0], it[1]}] = mp[{it[1], it[0]}] = c;
        c++;
    }
    int sum = 0;
    for(auto it : A) sum += it;
    vector<int> v(N+1, 0), par(N+1, 0);
    dfs(1, 0, e, v, par, A);
    int ans = sum;
    int id = edges.size();

    for(int i = 1; i <= N; i++){
        int a = sum - v[i];
        if(abs(a - v[i]) < ans){
            ans = abs(a - v[i]);
            id = mp[{par[i], i}] + 1;
        }
        else if(abs(a - v[i]) == ans){
            id = min(id, mp[{par[i], i}] + 1);
        }
    }
    return id;
}

Division Nodes ✅
Uber

typedef long long ll;

void solve() {

  ll n = N;
  vector<vector<ll>>arr(n, vector<ll>(3));
  for (int i = 0; i < n; i++) {
    arr[i] = {[0] = people[i] , [1] = starting[i] , [2] = ending[i]};
  }
  sort(arr.begin() , arr.end() , [&](vector<ll>& a , vector<ll>& b) {
    if (a[1] == b[1]) return a[2] < b[2];
    return a[1] < b[1];
  });

  ll sum = 0;
  for (auto& x : arr) sum += x[0];

  vector<ll>dp(n, -1);
  auto recurr = [&](ll i , auto && recurr)->ll{
    if (i >= n) return 0;

    ll& ans = dp[i];
    if (ans != -1) return ans; ans = 0;

    ans = recurr(i + 1 , recurr);
    ll low = i , high = n , val = arr[i][2];
    while (high - low > 1) {
      ll mid = (low + high) >> 1;
      if (val >=  arr[mid][1]) low = mid;
      else high = mid;
    }

    ans = max(ans, arr[i][0] + recurr( low + 1 , recurr) );

    return ans;
  };



  return sum - recurr(0, recurr);

}

Meeting Room ✅
Uber

Tomorrow anyone give exam Oracle?