The Silicon Partners Coding Questions

May 13, 2023

The Silicon Partners Coding Questions with Solutions

In this page, you will find out The Silicon Partners Coding Questions and Answers asked in Online Assessments and Technical Interviews involved in the Hiring Process of the Company.

Go through this page to get more details like Job Profile, Job Location, CTC Offered, Steps involved in the recruitment process, etc. of the company.

About The Silicon Partners

TSP (The Silicon Partners) is a US Based IT Consulting Organization headquartered in California. TSP specialize in providing Consulting Services for enterprise technologies like SAP S/4HANA, SuccessFactors Human Experience Management (HXM), SAP Application Management Service, Robotic Process Automation (UiPath), AI / ML and Cloud.

With a global presence in over 10 countries and offshore development centres in Noida and Columbia, TSP has a global footprint with over 400 employees.

About The Silicon Partners Recruitment Process

The The Silicon Partners Recruitment Process consists of the following steps :

Online Assessment [ MCQ + Coding ]
Technical Interview with the LEADs
Discussion with BU Head
HR Round

We have mentioned further details of the The Silicon Partners Recruitment Process in the following Tabular Form

The Silicon Partners	Related Information
Position :	Technical / Functional Trainee
Course :	B.Tech – CSE, IT,ECE , EEE , ME , CE Eligible Batch – 2023
Eligibility Criteria / Academic Qualification Required :	60% (10th , 12th and graduation) No Current Backlogs.
Salary Offered and Terms	3.55 LPA Service Agreement: 2 Years Notice Period: 3 months
Selection Process :	Online Assessment Technical Interview with head LEADs Discussion with BU Head. HR Interview
Joining Location :	Noida

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The Silicon Partners Coding Questions and Answers

Question 1: Stars Between Bars

Given a string s consisting of stars “*” and bars “|” ,an array of starting indices starIndex,and an array of ending indices endIndex,determine the number of stars between any two bars within the substrings between the two indices inclusive . NOTE that in this problem indexing starts at 1.

A Star is represented as an asterisk [*=ascii decimal 42]
A Bar is represented as a Pipe [“|”=ascii decimal 124]

Example

    s=’|**|*|’
    startIndex=[1,1]
    endIndex=[5,6]

For the first pair of indices (1,5) the substrings is “|**|*” . There are 2 stars between a pair of bars
For the second pair of indices (1,6) the substring is “|**|*|” and there are 2+1=3 stars in between the bars.
Both of the answers are returned to the array [2,3].

Constraints

1<=n<=105
1<=StartInde[i]<=endIndex[i]
Each Character of s is either “*” or “|”

Input Format for Custom testing

First line contains a string S the next line contains an integer n , the no.of elements in startIndex. Each line i of the n subsequent lines contains an integer of startIndex.the next line contains an integer n , the no.of elements in endIndex. Each line i of the n subsequent lines contains an integer of endindex

Sample Input

    *|*| → s=”*|*|”
    1 → startindex[] size=1
    1 → startindex= 1
    1 → endindex[] size=1
    3 → endindex=3

Sample output:

Explanation :

The substring from index =1 to index=3 is “*|*” . there is no consecutive pair of bars in this string.

C++

Java

Python

C++

Run

#include <bits/stdc++.h>
using namespace std;
int main(){
    string s;
    cin >> s;
    int n;
    cin >> n;
    int st[n], ed[n];
    for (int i = 0; i < n; i++){
        cin >> st[i];
        st[i]--;
    }
    cin>>n;
    for (int i = 0; i < n; i++){
        cin >> ed[i];
        ed[i]--;
    }
    vector<int> bars;
    for (int i = 0; i < s.length(); i++){
        if (s[i] == '|')
            bars.push_back(i);
    }
    for (int i = 0; i < n; i++){
        int idx = lower_bound(bars.begin(),bars.end(),st[i]) - bars.begin();
        st[i] = bars[idx];
        idx = lower_bound(bars.begin(),bars.end(),ed[i]) - bars.begin();
        if(idx == 0 || bars[idx]==ed[i])
            continue;
        ed[i] = bars[idx-1];
    }

    int sz = s.length();
    int starCt[sz] = {0};
    if(s[0] == '*')
        starCt[0] = 1;

    for(int i =1;i<sz;i++){
        starCt[i] = starCt[i-1] + (s[i]=='*');
    }

    for(int i =0;i<n;i++){
        if(st[i]>=ed[i]){
            cout<<0<<endl;
        }
        else{
            int ans = starCt[ed[i]];
            if(st[i]>0)
                ans-=starCt[st[i]-1];
            cout<<ans<<endl;
        }
    }    
    return 0;
}

Java

Run

import java.util.*;

public  class Main{
  public static void main(String[] args){
   Scanner sc=new Scanner(System.in);
   String str=sc.next(); 
   int n=sc.nextInt(); 

   int startIndex[]=new int[n];
   int endIndex[]=new int[n];

   for(int i=0;i<n;i++)
     startIndex[i]=sc.nextInt();
   
   for(int i=0;i<n;i++)
     endIndex[i]=sc.nextInt();
   
   int count=0;

   for(int i=0;i<n;i++){
    count=0;
    String sub=str.substring(startIndex[i]-1,endIndex[i]);
    int start= sub.indexOf("|");
    int end=sub.lastIndexOf("|");
    for(int j=start;j<end;j++)
      if(sub.charAt(j)=='*')
         count++; 
    System.out.print(count+" ");
  } 
 }
}

Python

Run

s=input()
start=[]
end=[]
n=int(input())
for i in range(n):  
    start.append(int(input()))
for i in range(int(input())):
    end.append(int(input()))
for i in range(n):
    Str=s[start[i]-1:end[i]]
    print(Str.strip('*').count('*'))

Question 2 : Share Holder (R -> Hard)

Problem Statement :
Ratan is a crazy rich person. And he is blessed with luck, so he always made the best profit possible with the shares he bought. That means he bought a share at a low price and sold it at a high price to maximize his profit. Now you are an income tax officer and you need to calculate the profit he made with the given values of stock prices each day. You have to calculate only the maximum profit Ratan earned.
Note that:
Ratan never goes into loss.

Example 1 :
Price=[1,6,2]
Ratan buys it on the first day and sells it on the second.
Example 2 :
Price=[9,8,6]

The Price always went down, Ratan never bought it.

Input Format:
First line with an integer n, denoting the number days with the value of the stack
Next n days, telling the price of the stock on that very day.

Output Format:
Maximum profit done by Ratan in a single line.
Constraints:
Number of days <=10^8

Sample Input for Custom Testing:

STDIN
7
1
9
2
11
1
9
2

Sample Output :

10

Explanation :

The maximum profit possible is when Ratan buys it in 1 rupees and sells it in 11.

C++

JAVA

Python

C++

Run

#include <bits/stdc++.h>
using namespace std;
int solve(vector < int > v)
{
    int n = v.size();
    if (n == 0)
        return 0;
    int mx = v[0];
    for (int i = 1; i < n; i++)
        mx = max(mx, v[i]);
    if (mx <= 0)
        return 0;
    int mxSum = 0;
    int cSum = 0;
    for (int i = 0; i < n; i++)
    {
        cSum += v[i];
        if (cSum < 0)
            cSum = 0;
        mxSum = max(mxSum, cSum);
    }
    return mxSum;
}
int main()
{
    int n;
    cin >> n;
    int price[n];
    for (int i = 0; i < n; i++) cin >> price[i];
    vector < int > diff;
    for (int i = n-2; i >= 0 ; i--) diff.push_back(price[i+1] - price[i]);
    int ans = solve(diff);
    if(ans < 0) cout << 0 << endl;
    else cout<< ans << endl;
}

JAVA

Run

import java.util.*;
public class PrepInsta {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int price[] = new int[n];
		for(int i=0;i < n;i++) {
			price[i] = sc.nextInt();
		}
		Vector < Integer > diff = new Vector<>();
		for(int i=n-2;i >=0;i--) {
			diff.add(price[i+1]-price[i]);
		}
		int ans = solve(diff);
		if(ans < 0) {
			System.out.println(0);
		}else {
			System.out.println(ans);
		}

	}
	private static int solve(Vector < Integer > v) {
		int n = v.size();
		if(n==0) {
			return 0;
		}
		int mx = v.get(0);
		for(int i=1;i < n;i++) {
			mx = Math.max(mx, v.get(i));
		}
		if(mx <= 0) {
			return 0;
		}
		int mxSum=0,csum=0;
		for(int i=0;i < n;i++) {
			csum+=v.get(i);
			if(csum < 0)
				csum=0;
			mxSum = Math.max(csum, mxSum);
		}
		return mxSum;
	}
}

Python

Run

def func(diff):
    n=len(diff)
    if n==0:
        return 0
    mx=max(diff)
    if mx <= 0:
        return 0
    mxS=0
    cS=0
    for i in diff:
        cS+=i
        if cS <= 0:
            cS=0
        mxS=max(cS,mxS)
    return mxS

n=int(input())
arr=[]
diff=[]
ans=[0]
for i in range(n):
    arr.append(int(input()))c++
for i in range(n-1):
    diff.append(arr[i+1]-arr[i])
ans=func(diff)
if ans < 0:
    print("0")
else:
    print(ans)

Question 3 : Password Creation

Problem Statement: A password manager wants to create new passwords using two strings given by the user, then combined to create a harder-to- guess combination. Given two strings,interleave the characters of the strings to create a new string. Beginning with an empty string, alternately append a character from string a and from string b. If one of the strings is exhausted before the other, append the remaining letters from the other
string all at once. The result is the new password.

Example :

If a = ‘hackerrank’ and b = ‘mountain’,
The result is hmaocuknetrariannk.

Function Description :

Complete the function newPassword in the editor below.

Parameter(s):

Str : string a
Str : string b
Returns:
Str: new password using two strings

Sample input:

abc → a=”abc”
def → b=”def”

Sample output 0:

Adbecf

JAVA

Python

JAVA

Run

import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String str1 = sc.next();
        String str2 = sc.next();
        String password = "";
        int i, j;
        for (i = 0, j = 0; i < str1.length() && j < str2.length(); i++, j++) {
            password = password + str1.charAt(i) + str2.charAt(j);
        }
        if (i < str1.length())
            password = password + str1.substring(i, str1.length());
        if (j < str2.length())
            password = password + str2.substring(j, str2.length());
        System.out.println(password);
    }
}

Python

Run

def newPassword(a, b):
    ans = ""
    if len(a) > len(b):
        m = len(b)
        x = a[m:]
    if len(a) <= len(b):
        m = len(a)
        x = b[m:]
    for i in range(m):
        ans += a[i]
        ans += b[i]
    return ans + x

a = input()
b = input()
print(newPassword(a, b))

Question 4

You just received another bill which you cannot pay because you lack the money.

Unfortunately, this is not the first time to happen, and now you decide to investigate the cause of your constant monetary shortness. The reason is quite obvious: the lion’s share of your money routinely disappears at the entrance of party localities.

You make up your mind to solve the problem where it arises, namely at the parties themselves. You introduce a limit for your party budget and try to have the most possible fun with regard to this limit.

You inquire beforehand about the entrance fee to each party and estimate how much fun you might have there. The list is readily compiled, but how do you actually pick the parties that give you the most fun and do not exceed your budget?

Write a program which finds this optimal set of parties that offer the most fun. Keep in mind that your budget need not necessarily be reached exactly. Achieve the highest possible fun level, and do not spend more money than is absolutely necessary.

Input

The first line of the input specifies your party budget and the number n of parties.
The following n lines contain two numbers each. The first number indicates the entrance fee of each party. Parties cost between 5 and 25 francs. The second number indicates the amount of fun of each party, given as an integer number ranging from 0 to 10.
The budget will not exceed 500 and there will be at most 100 parties. All numbers are separated by a single space.
There are many test cases. Input ends with 0 0.

Output

For each test case, your program must output the sum of the entrance fees and the sum of all fun values of an optimal solution. Both numbers must be separated by a single space.

Example

Sample input:
50 10
12 3
5 8
16 9
16 6
10 2
21 9
18 4
12 4
17 8
18 9
50 10
13 8
19 10
16 8
12 9
10 2
12 8
13 5
15 5
11 7
16 2
0 0

Sample output:
50 29
48 32

Python

Java

C++

Python

Run

def knapSack(W, wt, val, n): 
    K = [[0 for x in range(W + 1)] for x in range(n + 1)] 
    
     
    for i in range(n + 1): 
        for w in range(W + 1): 
        	if i == 0 or w == 0: 
    		    K[i][w] = 0
        	elif wt[i-1] <= w: 
        		K[i][w] = max(val[i-1] 
        				+ K[i-1][w-wt[i-1]], 
        					K[i-1][w]) 
        	else: 
        		K[i][w] = K[i-1][w] 

    res = K[n][W] 
    res1=res 
    x=0  
    w = W 
    for i in range(W+1):
        if(K[n][i]==res):
            x=i 
            break
        
    print(x , res1)

b,n=map(int,input().split())
fun=[]
cost=[]
for i in range(n):
    x,y=map(int,input().split())
    fun.append(y)
    cost.append(x) 
    
(knapSack(b,cost,fun,n))

Java

Run

import java.util.*;
class Solution 
{
    public static void main (String[]args) 
    {
        Scanner sc = new Scanner (System.in);
        while (true)
        {
            int budget = sc.nextInt ();
            int n = sc.nextInt ();
        
            if (budget == 0 && n == 0)
                break;
        
            int cost[] = new int[n + 1];
            int fun[] = new int[n + 1];
            int arr[][] = new int[n + 1][budget + 1];

            for (int i = 0; i < n; i++)
        	{
                cost[i] = sc.nextInt ();
                fun[i] = sc.nextInt ();
            } 
            for (int i = 0; i <= n; i++)
                for (int j = 0; j <= budget; j++)
                    arr[i][j] = 0;
        
            for (int i = 1; i <= n; i++)
    	    {
                for (int j = 1; j <= budget; j++)
                    if (cost[i - 1] <= j)
                        arr[i][j] =Math.max (fun[i - 1] + arr[i - 1][j - cost[i - 1]],arr[i - 1][j]);
        	        else
                        arr[i][j] = arr[i - 1][j];
            }
            int c = 0;
            for (int i = 0; i <= budget; i++)
        	{       
                if (arr[n][i] == arr[n][budget])
    	        {
                    c = i;
                    break;
                }       
            }
            System.out.println (c + " " + arr[n][budget]);
        }
    }
}

C++

Run

#include <bits/stdc++.h>
using namespace std;

int main() {
  int w, n;

  x: cin >> w >> n;

  if (w == 0 and n == 0)
    goto r;

  else {

    int ct[n], val[n];

    for (int i = 0; i < n; i++) { cin >> ct[i] >> val[i];
    }

    int t[n + 1][w + 1];

    for (int i = 0; i <= n; i++) {

      for (int j = 0; j <= w; j++)
        t[i][j] = 0;
    }

    for (int i = 1; i <= n; i++) {

      for (int j = 1; j <= w; j++) {

        if (ct[i - 1] <= j)
          t[i][j] = max(val[i - 1] + t[i - 1][j - ct[i - 1]], t[i - 1][j]);

        else
          t[i][j] = t[i - 1][j];
      }
    }

    int cost = 0;

    for (int i = 0; i <= w; i++) {

      if (t[n][i] == t[n][w]) {
        cost = i;
        break;
      }
    }

    cout << cost << " " << t[n][w] << endl;

    goto x;
  }

  r: return 0;

}

Question 5 :Class Monitor

Problem Statement :

After JEE Mains, some students got admission into an engineering college. Now there is a class consisting of such n students, and the HOD came to say it is time to select the class monitor. But He never gets all of them at one time. So he brought a register, every time he gets someone with less rank than the previous time he cut the name and wrote the name of the student and the rank.

For a given number of ranks he gets each time, you have to predict how many names are cut in the list.

Constraints:

Number of Visiting<=10^9
ranks <=10000

Input Format:

Number of Visiting N in their first line
N space separated ranks the HOD gets each time

Output Format:

Number of ranks cut in the list

Sample Input:

6
4 3 7 2 6 1

Sample Output:

C++

JAVA

Python

C++

Run

#include <bits/stdc++.h>

using namespace std;

int main() {
    int n, p, m = INT_MAX, ans = 0;
    cin >> n;
    vector < int > a(n);
    for (int i = 0; i < n; i++) {
        cin >> p;
        if (p < m) {
            m = p;
            ans++;
        }
    }
    cout << ans - 1;
}

JAVA

Run

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        int n, p, ans = 0, m = Integer.MAX_VALUE;
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();
        for (int i = 0; i < n; i++) {
            p = sc.nextInt();
            if (p < m) {
                m = p;
                ans++;
            }
        }
        System.out.print(ans - 1);
    }
}

Python

Run

n = int(input())
a = list(map(int, input().split()))
m = a[0]
ans = 0
for i in range(1, n):
    if a[i] < m:
        m = a[i]
        ans += 1
print(ans - 1)

FAQs related to The Silicon Partners Coding Questions

Question 1: What should I expect during the interview process?

During the interview process at TSP, you can expect to be asked a series of questions related to your experience, skills, and qualifications . You will be showcasing your skills like:

Programming languages, including C++, Java, and C
Experience in building libraries and APIs.

Question 2: Is Coding questions asked in The Silicon Partner Recruitment Process?

Yes, Coding Questions are included in Online Assessment and Technical Interview of TSP.

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The Silicon Partners Coding Questions

The Silicon Partners Coding Questions with Solutions

About The Silicon Partners

About The Silicon Partners Recruitment Process

Details about Online Assessment

Skill Set Required

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The Silicon Partners Coding Questions and Answers

Question 1: Stars Between Bars

Example

Constraints

Input Format for Custom testing

Sample Input

Sample output:

Explanation :

Question 2 : Share Holder (R -> Hard)

Question 3 : Password Creation

Example :

Function Description :

Parameter(s):

Sample input:

Sample output 0:

Question 4

Question 5 :Class Monitor

FAQs related to The Silicon Partners Coding Questions

Question 1: What should I expect during the interview process?

Question 2: Is Coding questions asked in The Silicon Partner Recruitment Process?

Get over 200+ course One Subscription

Checkout list of all the video courses in PrepInsta Prime Subscription

Checkout list of all the video courses in PrepInsta Prime Subscription

30+ Companies are Hiring

Unlock this article for Free, by logging in

The Silicon Partners Coding Questions

The Silicon Partners Coding Questions with Solutions

About The Silicon Partners

About The Silicon Partners Recruitment Process

Details about Online Assessment

Skill Set Required

Prime Course Trailer

Related Banners

The Silicon Partners Coding Questions and Answers

Question 1: Stars Between Bars

Example

Constraints

Input Format for Custom testing

Sample Input

Sample output:

Explanation :

Question 2 : Share Holder (R -> Hard)

Question 3 : Password Creation

Example :

Function Description :

Parameter(s):

Sample input:

Sample output 0:

Question 4

Question 5 :Class Monitor

FAQs related to The Silicon Partners Coding Questions

Question 1: What should I expect during the interview process?

Question 2: Is Coding questions asked in The Silicon Partner Recruitment Process?

Get over 200+ course One Subscription

Checkout list of all the video courses in PrepInsta Prime Subscription

Checkout list of all the video courses in PrepInsta Prime Subscription

30+ Companies are Hiring

Unlock this article for Free,
by logging in