Therap Software Engineer

Interview Stages

The selection process has 3 stages,

1. Initial screening: This round is taken in written format
2. 1st technical round The first round is taken by the BD team
3. HR Round: This is the final stage before onboarding and typically deals with salary negotiation.

Software Engineering Questions

Given an array of numbers indicating stock price of n consecutive days. If you buy stock at one day and sell at any later day what is the maximum profit that you can get

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💻 Submit Code

int maxProfit(vector<int>& prices) {
    int buy = prices[0];
    int profit = 0;
    for(int i=1;i<prices.size();i++){
        if( prices[i]-buy > profit ) profit = prices[i] - buy;
        if( prices[i] < buy ) buy = prices[i];
    }
    return profit;
}

Given an array of n integers. You need to take all zeroes in array to the end without changing the relative order of remaining element. eg: [2,0,0,3,1,0,5] => [2,3,1,5,0,0,0]

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💻 Submit Code

void moveZeroes(vector<int>& nums) {
    int i = 0;
    for(int j=0;j<nums.size();j++){
        swap(nums[i], nums[j]);
        if( nums[i] != 0 ) i++;
    }
}

Given an array of n integers. Reorder the elements such that all odd numbers occur after even numbers.

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[Answer]

Given an array of strings. Print the sets of strings which are anagram. eg: ["cat","tab","act","bat","taco"] => [{"cat","act"},{"tab","bat"},{"taco"}]

---

💻 Submit Code

vector<vector<string>> groupAnagrams(vector<string>& strs) {
    map<string,int> index;
    vector<vector<string>> vs;
    for(auto str:strs){
        string str2 = str;
        if( str2.size()>1 ) sort(str2.begin(),str2.end());
        if( index.find(str2) == index.end() ){
            vs.push_back(vector<string>());
            index[str2] = vs.size()-1;
        }
        vs[ index[str2] ].push_back(str);
    }
    return vs;
}

Given an array of n integers. Find the kth largest element in the array.

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💻 Submit Code

int findKthLargest(vector<int>& nums, int k) {
    partial_sort(nums.begin(), nums.begin() + k, nums.end(), greater<int>());
    return nums[k-1];
}

Given two very large number in string format. Find the sum of the two number

---

string sum(string &A, string &B){
    reverse(A.begin(),A.end());
    reverse(B.begin(),B.end());
    string sum;
    int c = 0;
    int i=0,j=0;
    while(true){
        int a=0,b=0;
        if( i<A.size() ) a = A[i++]-'0';
        if( j<B.size() ) b = B[j++]-'0';

        int s = (a+b+c)%10;
        c = (a+b+c)/10;
        sum.push_back(s+'0');
        if( i>=A.size() and j>=B.size() and c == 0 ) break;
    }
    reverse(sum.begin(),sum.end());
    return sum;
}

Given two binary tree. Check if they are identical [not isomorphism]

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💻 Submit Code

Normal Solution

bool isSameTree(TreeNode* p, TreeNode* q) {
    if( p == nullptr and q != nullptr ) return false;
    if( p != nullptr and q == nullptr ) return false;
    if( p == nullptr and q == nullptr ) return true;

    if( p->val != q->val ) return false;

    return isSameTree(p->left,q->left) &&
            isSameTree(p->right,q->right);
}

Weird Solution

// ref: https://go.dev/tour/concurrency/7
package main

import (
	"fmt"

	"golang.org/x/tour/tree"
)

// Walk walks the tree t sending all values
// from the tree to the channel ch.
func WalkRecursive(t *tree.Tree, ch chan int) {
	if t.Left != nil {
		WalkRecursive(t.Left, ch)
	}
	ch <- t.Value
	if t.Right != nil {
		WalkRecursive(t.Right, ch)
	}
}

func Walk(t *tree.Tree, ch chan int) {
    WalkRecursive(t, ch)
    close(ch)
}

// Same determines whether the trees
// t1 and t2 contain the same values.
func Same(t1, t2 *tree.Tree) bool {
	ch1 := make(chan int)
	ch2 := make(chan int)
	go Walk(t1, ch1)
	go Walk(t2, ch2)
	for {
		x, ok1 := <-ch1
		y, ok2 := <-ch2

		if ok1 != ok2 || x != y {
			return false
		}
		if !ok1 {
			break
		}
	}
	return true
}

func main() {
	fmt.Println(Same(tree.New(1), tree.New(2)))
}

Given two array of integers. Find the common elements between them.

---

💻 Submit Code Unique
💻 Submit Code Repeats

Return uniques

vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
    set<int> st;
    for(auto num:nums1) st.insert(num);
    set<int> res;
    for(auto num:nums2) if( st.count(num) == 1 ) res.insert(num);
    vector<int> ret;
    for(auto num:res) ret.push_back(num);
    return ret;
}

With repeatation

vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
    sort(nums1.begin(),nums1.end());
    sort(nums2.begin(),nums2.end());

    vector<int> merged;
    int i=0,j=0;
    while(i<nums1.size() and j<nums2.size()){
        if( nums1[i] == nums2[j] ){
            merged.push_back(nums1[i]);
            i++;j++;
        }else if( nums1[i]<nums2[j] ) i++;
        else j++;
    }
    return merged;
}

Find pairs with given target sum in a doubly linked list. Input: 1 <> 2 <> 4 <> 5 <> 6 <> 8 <> 9 target = 7 Output: (1,6), (2,5)

---

💻 Submit Code

class Solution
{
public:
    vector<pair<int, int>> findPairsWithGivenSum(Node *head, int target)
    {
        vector<pair<int,int>> ans;
        
        Node* left = head;
        
        /// traverse to the end of the list
        while(head!= nullptr && head->next!=nullptr){
            head = head->next;
        }
        Node* right = head;
        
        while(left!= right && left->prev != right){
            if(left->data + right->data == target){
                ans.push_back(make_pair(left->data, right->data));
                left = left->next;
                right = right->prev;
            }
            else if(left->data + right->data > target){
                right = right->prev;
            }else{
                left = left->next;
            }
        }
        
        return ans;    
        
    }
};

Solve the problem using Object Oriented Programming

int main(){
    int square1width = 50;
    int square2width = 80;
    int rectangle1width = 30, rectangle1height = 40;
    int rectangle2width = 20, rectangle2height = 40;

    int square1area = square1width* square1width;
    int square2area = square2width* square2width;
    int rectangle1area = rectangle1height*rectangle1width;
    int rectangle2area = rectangle2width* rectangle2height;
}

---

#include <iostream>
using namespace std;

// Abstract base class
class Shape {
public:
    virtual int area() const = 0;  // Pure virtual function for area
};

class Square : public Shape {
private:
    int width;
public:
    Square(int w) : width(w) {}  // Constructor to initialize width

    int area() const override {
        return width * width;  // Area of square
    }
};

class Rectangle : public Shape {
private:
    int width;
    int height;
public:
    Rectangle(int w, int h) : width(w), height(h) {}  // Constructor to initialize width and height

    int area() const override {
        return width * height;  // Area of rectangle
    }
};

int main() {
    
    Square square1(50);
    Square square2(80);
    Rectangle rectangle1(30, 40);
    Rectangle rectangle2(20, 40);

    cout << "Square 1 area: " << square1.area() << endl;
    cout << "Square 2 area: " << square2.area() << endl;
    cout << "Rectangle 1 area: " << rectangle1.area() << endl;
    cout << "Rectangle 2 area: " << rectangle2.area() << endl;
    return 0;
}