Wells Fargo interview experience Real time questions & tips from candidates to crack your interview

Program Associate

Wells Fargo
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3 rounds | 3 Coding problems

Interview preparation journey

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Journey
It was a wonderful session throughout the interview journey, during which I answered properly and hence received the offer via off-campus recruitment. It was a very nice session for both the interview and the test.
Application story
I got this opportunity via off-campus recruitment. I answered very well in my interview and was honest about my project explanation. Don't copy and paste your project if you haven't done it yourself.
Why selected/rejected for the role?
I got selected for the Program Associate role because I answered very well, and the interviewer was satisfied with all of my responses. If I didn't know the answer, I honestly said, "I don't know."
Preparation
Duration: 6 months
Topics: Data Structures and Algorithms, Operating Systems, DBMS, OOP, Computer Networks, Web Development, Splunk Tableau
Tip
Tip

Tip 1: Be very clear when explaining your projects.

Tip 2: Ensure you cover all core Computer Science subjects thoroughly in your explanations.

Application process
Where: Campus
Eligibility: 6.5 CGPA and 65% 12th marks minimum requirement
Resume Tip
Resume tip

Tip 1: Ensure your explanations are clear and demonstrate appropriate knowledge.

Tip 2: Provide truthful information and avoid discussing topics you are unsure about.

Interview rounds

01
Round
Medium
Video Call
Duration60 minutes
Interview date28 Apr 2022
Coding problem1

The interview took place in the morning via Google Meet starting from 10:00 am. The online process was smooth, and the interviewer was friendly.

1. Count Ways To Reach The N-th Stairs

Moderate
30m average time
80% success
0/80
Asked in companies
PayPalOYOMicrosoft

You have been given a number of stairs. Initially, you are at the 0th stair, and you need to reach the Nth stair.


Each time, you can climb either one step or two steps.


You are supposed to return the number of distinct ways you can climb from the 0th step to the Nth step.

Note:

Note: Since the number of ways can be very large, return the answer modulo 1000000007.
Example :
N=3

Example

We can climb one step at a time i.e. {(0, 1) ,(1, 2),(2,3)} or we can climb the first two-step and then one step i.e. {(0,2),(1, 3)} or we can climb first one step and then two step i.e. {(0,1), (1,3)}.
Problem approach

int countWays(int n)
{
if(n == 1) return n;
int mod = 1000000000 + 7;
int prev = 2, prev2 = 1, curr = 0;

for(int i = 2 ; i < n ; i++){
curr = (prev + prev2) % mod;
prev2 = prev;
prev = curr;
}

return prev ;
}

Try solving now
02
Round
Hard
Video Call
Duration55 minutes
Interview date28 Apr 2022
Coding problem1

It was again the morning session and the interviewer was very supportive and gave me the hint too to think about the approach finally I came up with the solution that he was expecting.

1. Kth Smallest Element

Moderate
0/80
Asked in companies
FacebookAmazonWells Fargo

You are given a square matrix ‘NxN’ rows and columns. The matrix is sorted, meaning all rows and columns of the matrix are sorted in ascending order. You are also given an integer ‘K’, and your task is to find the ‘K’th smallest element in the sorted order.

For example:
You are given ‘mat’ = [[1, 2, 2,], [3, 3, 4], [5, 6 ,7]]] and ‘K’ = 5, the elements of the matrix are [1, 2, 2, 3, 3, 4, 5, 6, 7], the 5th smallest element in the matrix is 3. Hence the answer is 3.
Problem approach

#include 
using namespace std;

// Tree Node
struct Node {
int data;
Node *left;
Node *right;

Node(int val) {
data = val;
left = right = NULL;
}
};

// Function to Build Tree
Node* buildTree(string str)

// Corner Case
if(str.length() == 0 || str[0] == 'N')
return NULL;

// Creating vector of strings from input 
// string after spliting by space
vector ip;

istringstream iss(str);
for(string str; iss >> str; )
ip.push_back(str);

// Create the root of the tree
Node* root = new Node(stoi(ip[0]));

// Push the root to the queue
queue queue;
queue.push(root);

// Starting from the second element
int i = 1;
while(!queue.empty() && i < ip.size()) {

// Get and remove the front of the queue
Node* currNode = queue.front();
queue.pop();

// Get the current node's value from the string
string currVal = ip[i];

// If the left child is not null
if(currVal != "N") {

// Create the left child for the current node
currNode->left = new Node(stoi(currVal));

// Push it to the queue
queue.push(currNode->left);
}

// For the right child
i++;
if(i >= ip.size())
break;
currVal = ip[i];

// If the right child is not null
if(currVal != "N") {

// Create the right child for the current node
currNode->right = new Node(stoi(currVal));

// Push it to the queue
queue.push(currNode->right);
}
i++;
}

return root;
}


// } Driver Code Ends
/*The Node structure is defined as
struct Node {
int data;
Node *left;
Node *right;

Node(int val) {
data = val;
left = right = NULL;
}
};
*/

// return the Kth largest element in the given BST rooted at 'root'
class Solution
{public:
void solve(Node*root,int k,int&c,int &ans){
if(root==NULL||c>=k){
return ;
}
solve(root->right,k,c,ans);
c++;
if(c==k){
ans=root->data;
return;
}
solve(root->left,k,c,ans);
}
public:
int kthLargest(Node *root, int k)
{
int c=0;int ans;
solve(root,k,c,ans);
return ans;
}
};

// { Driver Code Starts.

int main()
{
int t;
cin>>t;
getchar();

while(t--)
{
string s;
getline(cin,s);
Node* head = buildTree(s);

int k;
cin>>k;
getchar();

Solution ob;
cout << ob.kthLargest( head, k ) << endl;
}
return 1;
} }

Try solving now
03
Round
Easy
Video Call
Duration30 minutes
Interview date11 May 2022
Coding problem1

It was late evening and happened in a very smooth way.

1. OS Questions

What is a thread in OS? (Learn)

What is deadlock? (Learn)

Problem approach

Tip 1: Try to answer with real-life examples.
Tip 2: Give answers in your language and avoid the bookish language

Here's your problem of the day

Solving this problem will increase your chance to get selected in this company

Skill covered: Programming

What is recursion?

Choose another skill to practice
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