TCS NextStep Coding Questions and Answers 2023 (Actual Questions): Previous, Expected Q/A, & More

TCS NextStep Coding Questions and Answers 2023 | TCS NextStep Coding Questions | TCS NextStep Coding Asked Questions | TCS NextStep Coding Questions With Answers | TCS NextStep Coding Questions with Solutions

TCS NextStep Coding Questions and Answers 2023 | TCS NextStep Coding Expected Questions: The coding test is of medium complexity, which implies that although it is not impossible to pass, it is also not particularly easy. You would surely have an edge in the next TCS NextStep coding test if you practiced some of the code Questions from last year’s test. Let’s examine a few C, C++, Java, and Python, Test patterns, TCS NextStep Coding Questions Python, code issues, and their remedies. Prior recruiting campaigns included a heavy emphasis on arrays and other basic programming concepts in their coding tests. With enough preparation, one may easily pass the first round and even subsequent rounds. To answer the coding tasks, you must pick a language that you are familiar with, such as C, C++, Java, or even Python, and use your programming skills. To get more private jobs, click here

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TCS NextStep Coding Questions and Solutions

C++  PROGRAM

#include<bits/stdc++.h>
using namespace std;int main()
{
    string s;
    int c=0;
    getline(cin,s);
    for (auto i:s)
    {
        if(i=='(') c++;
        if(i==')') c--;
    }
    cout<<(c==0);
}

PYTHON PROGRAM

s=input()
c=0
for i in s:
    if i =='(':
        c+=1
    elif (i==')') and c>0 :
        c-=1
print(int(c>0))

Q) Write a program to find out and display prime numbers from the given list of integers. The program will accept input in two lines. First-line contains a number indicating the total number of integers in the list and the second line contains integers separated by spaces.

Example 1

• Input: 5
  4 6 9 3 7
• Output:  3 7

Example 2

• Input:  10
  8 10 3 12 7 15 11 2 17 26
• Output:  3 7 11 2 17

C++  PROGRAM

#include<bits/stdc++.h>
using namespace std;
map<int,int> m;
bool ifPrime(int a)
{
    if(m[a]==2||m[a]==1) return m[a]-1;
    if((a&1)==0) {m[a]=1;return false;}
    for(int i=3;i<=sqrt(a);i+=2)
    if(a%i==0) {m[a]=1;return 0;}
    m[a]=2;
    return 1;
}
int main()
{
    m[2]=2;
    int n,a;cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>a;
        if(ifPrime(a)) cout<<a<<" ";
    }
}

Q) Write a program that receives a word A and some texts as input. You need to output the texts (without modifying them) in the ascending order of the number of occurrences of the word A in the texts. The input is as follows: an integer M(between 1 and 100, inclusive), followed by the word A in the next line, and some text in each of the M next lines.

Note: The texts and the word A contain only lowercase Latin letters (a,b,c…,z) and blank spaces (“ ”). The maximum size of the texts and the word A is 100 Characters. Every text has a different number of occurrences of the word A.

Note 2:you must print one text per line without modifying the texts.

Example 1

• Input: 2
  Java
  I hate java
  Python is a good programming language
• Output: Python is a good programming language
  I hate java

Example 2

• Input:  3
  python
  I like to code in python
  python is named after a show name monty python and not after the snake python
  I think python is good i think python is important than php
• Output: i like to code in python
  i think python is good i think python is important than php
  python is named after a show name monty python and not after the snake python

C++ PROGRAM 

#include<bits/stdc++.h>
using namespace std;
string s;
map<string,int> m2;
bool cmp(pair<string, int>& a,
         pair<string, int>& b)
{
    return a.second < b.second;
}
void sort(map<string, int>& M)
{
    vector<pair<string, int> > A;
    for (auto& it : M) {
        A.push_back(it);
    }
    sort(A.begin(), A.end(), cmp);
    for (auto& it : A) {
        for(int i=0;i<m2[it.first];i++)
        cout << it.first <<endl; } } int count(string s1) { istringstream ss(s1); int c=0; while(ss) { string w; ss>>w;
        if(w==s) c++;
    }
    return c;
}
int main()
{
    int n;getline(cin,s);n=stoi(s);
    getline(cin,s);
    transform(s.begin(),s.end(),s.begin(),::tolower);
    vector<string> v(n);
    vector<int> a(n);
    map<string,int> m;
    for(int i=0;i<n;i++)
    {
        getline(cin,v[i]);
        transform(v[i].begin(),v[i].end(),v[i].begin(),::tolower);
        m2[v[i]]++;
        m[v[i]]=count(v[i]);
    }
    sort(m);
}

JAVA PROGRAM 

import java.util.*;
public class Main
{

static int countOccurences(String str, String word)
{
    String a[] = str.split(" ");
 
    int count = 0;
    for (int i = 0; i < a.length; i++)
    {
      if (word.equals(a[i]))
        count++;
    }
 
    return count;
}
  public static void main(String[] args)
{
  Scanner sc=new Scanner(System.in);
  int n=sc.nextInt();
 sc.nextLine();
 String word=sc.next(); 
 //System.out.println(word);
 sc.nextLine();
 String arr[]=new String[n];
 for(int i=0;i<n;i++)
{
  arr[i]=sc.nextLine();
  //System.out.println(arr[i]); 
}
 TreeMap<Integer,String> map=new TreeMap<Integer,String>();	
  for(int i=0;i<n;i++)
{
   map.put(countOccurences(arr[i],word),arr[i]);
}

Set s=map.entrySet();
Iterator itr=s.iterator();
while(itr.hasNext())
{
  Map.Entry m=(Map.Entry)itr.next();

  System.out.println(m.getValue());
}

}
}

PYTHON PROGRAM 

from collections import defaultdict
d=defaultdict(int)
d1=defaultdict(int)
n=int(input())
s=input()
s=s.lower()
l=[]
for i in range(n):
    L=list(map(str,input().split()))
    s1=' '.join(i for i in L)
    c=0
    for j in L:
        if s==j:
            c+=1
    l.append(s1)
    d[i]=c
    d1[i]+=1
    c=0
d=sorted(d.items(),key=lambda a:a[1])
for i in d:
    for j in range(d1[i[0]]):
        print(l[i[0]])

Q) Write a program that will print the sum of diagonal elements of a 10X10 matrix. The program will take a total of 100 numbers as input (10 numbers will be input per line and each number will be separated by a space).

Example 1

• Input:    1  2 3 4 5 6 7 8 9 0
  0 1 2 3 4 5 6 7 8 0
  3 4 5 6 7 8 9 6 4 0
  2 3 4 5 6 7 8 9 3 2
  3 4 5 6 7 4 3 2 1 3
  3 4 5 6 2 4 4 2 4 6
  2 3 4 6 2 4 6 2 3 5
  2 3 5 6 2 4 6 2 3 5
  2 4 6 2 1 4 3 3 5 2
  3 3 5 2 4 6 2 1 4 6
• Output:  42

Example 2

• Input:   1 22 33 44 55 66 77 88 99 100
  100 1 88 77 66 55 44 33 22 11
  88 88 1 66 55 44 33 22 11 100
  88 77 66 1 44 33 22 11 100 99
  77 66 55 44  1 22  11 88 99 100
  66 55 44 33 22 1 77 88 99 100
  44 33 22 11 100 99 1 77 66 55
  33 22 11 100 99 88 77 1 55 44
  22 11 100 99 88 77 66 55 1 33
  100 11 22 33 44 55 99 88 77 1
• Output: 10

C++ PROGRAM

#include<bits/stdc++.h>
using namespace std;
 
int main()
{
    vector<vector<int>> v(10,vector<int>(10));
    for(int i=0;i<10;i++)
    for(int j=0;j<10;j++) cin>>v[i][j];
    int sum=0;
 
    for(int i=0;i<10;i++)
    sum+=v[i][i];
    cout<<sum;
}

import java.util.*;
public class Main
{
 public static boolean isVowel(char ch)
{
  return ch=='a' || ch=='e' || ch=='o' || ch=='i' || ch=='u' || ch=='A' || ch=='E' || ch=='I' || ch=='O' || ch=='U';
}
  public static void main(String[] args)
{
  Scanner sc=new Scanner(System.in);
  String str=sc.next();
  char arr[]=str.toCharArray(); 
 String res="";
 if(!isVowel(arr[0]))
  res+=arr[0];
 for(int i=1;i<arr.length;i++)
 {
    if(isVowel(arr[i])&&isVowel(arr[i-1]))
      res+=arr[i-1]+""+arr[i];
     if(!isVowel(arr[i]))
     res+=arr[i];
 }
System.out.println(res);
}

l=['a','e','i','o','u','A','E','I','O','U']
s=input()
s+=" "
s1=""
if len(s)==0:
    print("")
elif len(s)==1:
    if s[0] not in l:
        print(s)
else:
    if s[0] in l and s[1] in l:
        s1+=s[0]
    elif s[0] not in l :
        s1+=s[0]
 
    for i in range(1,len(s)-1):
        if s[i-1] not in l and s[i] in l and s[i+1] not in l:
            continue
        s1+=s[i]
 
    print(s1)

TCS NextStep Coding Questions Python

Question: Parallel Columbus

Problem Statement: Nobel Prize-winning Austrian-Irish physicist Erwin Schrödinger developed a machine and brought as many Christopher Columbus from as many parallel universes he could. Actually he was quite amused by the fact that Columbus tried to find India and got America. He planned to dig it further.

Though totally for research purposes, he made a grid of size n X m, and planted some people of America in a position (x,y) [in 1 based indexing of the grid], and then planted you with some of your friends in the (n,m) position of the grid. Now he gathered all the Columbus in 1,1 positions and started a race.

Given the values for n, m, x, y, you have to tell how many different Columbus(s) together will explore you as India for the first time.

Remember, the Columbus who will reach to the people of America, will be thinking that as India and hence wont come further.

Function Description:

Complete the markgame function in the editor below. It has the following parameter(s):

Parameters:

Constraints:

• 1 <= n <= 10^2
• 1 <= m <= 10^2
• 1 <= x <= n
• 1 <= y <= m

Input Format:

• The first line contains an integer, n, denoting the number of rows in the grid.
• The next line contains an integer m, denoting the number of columns in the grid.
• The next line contains an integer, x, denoting the American cell’s row.
• The next line contains an integer, y, denoting the American cell’s column.

Sample Cases

Sample Input 1

2

2

2

1

Sample Output 1

1

Explanation

The only way possible is (1,1) ->(2,1) -> (2,2), so the answer is 1.

C++ PROGRAM

#include <bits/stdc++.h>
using namespace std;
unordered_map<int,long long int> f;
 
long long int Fact(int n)
{
  if(f[n]) return f[n];
  return f[n]=n*Fact(n-1);
}
 
int main()
{
  int n,m,x,y;
  cin>>n>>m>>x>>y;
  n-=1;m-=1;x-=1;y-=1;
  f[0]=f[1]=1;
  int p=(Fact(m+n)/(Fact(m)*Fact(n)));
  int imp=((Fact(x+y)/(Fact(x)*Fact(y)))*(Fact(m-x+n-y)/(Fact(m-x)*Fact(n-y))));
  cout<<p-imp;
}

PYTHON PROGRAM

import math
n=int(input())-1
m=int(input())-1
x=int(input())-1
y=int(input())-1

ans=math.factorial(n+m)
ans=ans//(math.factorial(n))
ans=ans//(math.factorial(m))

ans1=math.factorial(x+y)
ans1=ans1//(math.factorial(x))
ans1=ans1//(math.factorial(y))

x1=n-x
y1=m-y

ans2=math.factorial(x1+y1)
ans2=ans2//(math.factorial(x1))
ans2=ans2//(math.factorial(y1))
print(ans-(ans1*ans2))

Q): While playing an RPG game, you were assigned to complete one of the hardest quests in this game. There are n monsters you’ll need to defeat in this quest. Each monster i is described with two integer numbers – poweri and bonusi. To defeat this monster, you’ll need at least poweri experience points. If you try fighting this monster without having enough experience points, you lose immediately. You will also gain bonusi experience points if you defeat this monster. You can defeat monsters in any order.

The quest turned out to be very hard – you try to defeat the monsters but keep losing repeatedly. Your friend told you that this quest is impossible to complete. Knowing that, you’re interested, what is the maximum possible number of monsters you can defeat?

Input:

• The first line contains an integer, n, denoting the number of monsters. The next line contains an integer, e, denoting your initial experience.
• Each line i of the n subsequent lines (where 0 ≤ i < n) contains an integer, poweri, which represents power of the corresponding monster.
• Each line i of the n subsequent lines (where 0 ≤ i < n) contains an integer, bonusi, which represents bonus for defeating the corresponding monster.

Output

• 2

Example Programs:

JAVA PROGRAM

import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;

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

        int monst[] = new int[n];
        int bonus[] = new int[n];

        for (int i = 0; i < n; i++) {
            monst[i] = s.nextInt();
        }
        for (int i = 0; i < n; i++) {
            bonus[i] = s.nextInt();
        }

        class Monster {
            private final int power, bonus;
            public Monster(int power, int bonus){
                this.power = power;
                this.bonus = bonus;
            }
        }

        Monster[] monsters = new Monster[n];

        for(int i = 0; i < n; i++)
            monsters[i] = new Monster(monst[i], bonus[i]);

        Arrays.sort(monsters, Comparator.comparingInt(m -> m.power));

        int count = 0;

        for(Monster m: monsters){
            if(exp < m.power) break;
            exp += m.bonus;
            ++count;
        }
        System.out.println(count);
    }
}

n = int(input())
lev = int(input())
p = []
b = []
a = []
ans = 0
for i in range(n):
   p.append(int(input()))
for j in range(n):
   b.append(int(input()))
for k in range(n):
   a.append([p[k], b[k]])
a.sort()
for z in a:
   if z[0] > lev:
       break
   lev += z[1]
   ans += 1
print(ans)

Q) Henry is extremely keen on history and every one of the ages of his family. He does a ton of exploration and understands that he has plummeted from the incomparable Amaya line. After a ton of looking through old records and the most recent records of the general public, he can discover all the parent-kid connections in his family right from the extraordinary ruler Ming of the tradition to himself.

These connections are given in the structure of a direct exhibit where the emperor is at the main position and his kids are at pos (2i + 1) and (2i + 2)

This is the pattern followed throughout.

Henry needs to sort out every one of the kin of individual X from the information.

Write a program for Henry to figure out all the siblings of person X from the data.
Return the sorted list of all of Henry’s siblings.

If no sibling return {-1}

• input 1: N, the length of the array
• input2: An array representing the ancestral tree
• input 3: X, the person whose siblings are sought.
• output – return the array of all siblings in increasingly sorted order.

Example 1 :

input 1: 5
input 2: {1,2,3,4,5}
input 3: 1
output: {-1}
Explanation: x is the root of the tree and has no siblings

Example 2 :

input 1: 6
input 2: {1,2,3,4,5,6}
input 3: 5
output: {4,6}
Explanation: {2,3 } are the children of {1}. {4,5,6 } are the children of {2,3}, thus the siblings of x= 5 are {4,6}

JAVA PROGRAM

import java.util.*;
 
public class Main
{
    Node root;
    static class Node
    {
        int data;
        Node left, right;
    
        Node (int data) 
        {
            this.data = data;
            this.left = null;
            this.right = null;
        } 
    } 

    public static void main (String[]args)
    {
        Main t2 = new Main ();
        Scanner sc = new Scanner (System.in);
        int length = sc.nextInt ();
        int arr[] = new int[length];
    
        for (int i = 0; i < length; i++)
        {
            arr[i] = sc.nextInt ();
        }
        int target = sc.nextInt ();
        t2.root = t2.insertLevelOrder (arr, t2.root, 0);
    
        Set < Integer > sets = t2.levelOrderBottom (t2.root, target);
        sets.remove (target);
        System.out.println (sets);
    }   
    public static Node insertLevelOrder (int[]arr, Node root, int i) 
    {
        if (i < arr.length)
        {
            Node temp = new Node (arr[i]);
            root = temp;
            root.left = insertLevelOrder (arr, root.left, 2 * i + 1);
            root.right = insertLevelOrder (arr, root.right, 2 * i + 2);
        }
        return root;
    }

    public Set < Integer > levelOrderBottom (Node root, int target)
    {
        if (root == null)
        {
        	return null;
        }
        Queue < Node > q = new LinkedList <> ();
        q.offer (root);
    
        while (!q.isEmpty ())
        {
            int qsize = q.size ();
            Set < Integer > temp = new HashSet <> ();
            
            for (int i = 0; i < qsize; i++)
	        {
                Node child = q.poll ();
                temp.add (child.data);
                
                if (child.left != null)
	            {
                    q.offer (child.left);
                }
                if (child.right != null)
	            {
                    q.offer (child.right);
                }
            }
            if (temp.contains (target))
                return temp;
        }
        return null;
    }
}

PYTHON PROGRAM

def identify_siblings(tree_array, x):
    tree_len = len(tree_array)
    index = tree_array.index(x)
    level = 0
    start_index = level
    number_of_nodes = 0
    while start_index < tree_len:
    	end_index = pow(2, level) + start_index
    	if x in tree_array[start_index:end_index]:
    		break
    	level += 1
    	start_index = (2 * start_index) + 1
    final_array = tree_array[start_index:end_index]
    final_array.remove(x)
    return final_array if final_array else [-1]

n = int(input())
arr = []
for i in range(n):
    arr.append(int(input()))
x = int(input())
print(identify_siblings(arr, x))

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int N, ans, price;
    cin>>N;
   
    int arr[N];
    
    for (int i = 0; i < N; i++)
        cin>>arr[i];
    
    price = 0;
    ans = 0;
    for (int i = 0; i < N; i++)
    {
        price = price + arr[i];
        if (ans > price)
            {
                ans = price;
            }
    }
    cout << ans;
}

C PROGRAM:

#include<stdio.h>
int main()
{
    int N, ans, price;
    scanf("%d", &N);
   
    int arr[N];
    
    for (int i = 0; i < N; i++)
        scanf("%d", &arr[i]);
    
    price = 0;
    ans = 0;
    for (int i = 0; i < N; i++)
    {
        price = price + arr[i];
        if (ans > price)
            {
                ans = price;
            }
    }
    printf("%d", ans);
}

PYTHON PROGRAM

n = int(input())
arr = []
for i in range(n):
   arr.append(int(input()))
min = 0
for i in range(1, n):
   if sum(arr[:i]) < min:
       min = sum(arr[:i])
print(min)

Question: Amusement Park

Problem Statement: Aashay loves to go to WONDERLA , an amusement park. They are offering students who can code well with some discount. Our task is to reduce the cost of the ticket as low as possible.

They will give some k turns to remove the cost of one ticket where the cost of tickets are combined and given as string. Help Aashay in coding as he is not good in programming and get a 50%  discount for your ticket.

Constraints:

• 1 <= number of tickets <= 10^5
• 1 <= K <= number of tickets

Input Format for Custom Testing:

• The first line contains a string,Tickets, denoting the given cost of each ticket.
• The next line contains an integer, K, denoting the number of tickets that is to be removed.

Sample Cases:

• Sample Input 1
  203
  3
• Sample Output 1
  0

C++ PROGRAM

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int smallestNumber (string num, int k)
{
    if(num.length()<=k)
        return 0;
    unordered_map<char,int> pos;
    for(int i=0;i<num.length();i++)
   { pos[num[i]]=i;}
 
    string temp=num;
    sort(num.begin(),num.end());
    string ans=num.substr(0,num.length()-k);
    vector<int> v;
    for(int i=0;i<ans.length();i++)
    v.push_back(pos[ans[i]]);
 
    sort(v.begin(),v.end());
    string ret;
    for(int i=0;i<v.size();i++)
    {ret+=temp[v[i]];}
    int final=stoi(ret);
 
    return final;
}
int main()
{
    string s;
    cin >> s;
    int k;
    cin >> k;
    int ans;
    cout<<smallestNumber(s,k)%(int)(pow(10,9)+7);
    return 0;
}

PYTHON PROGRAM

import sys
n=input()
k=int(input())
n1=len(n)
if len(n)<=k:
 print(0)
 sys.exit()

a=''
i=0
while i<(n1-1) and k>0:
 if int(n[i])>int(n[i+1]):
  i+=1
  k-=1
  continue

 else:
  a+=n[i]
  i+=1

a+=n[i]
i+=1
if k>0:
 a=a[:-k]

if i<=(n1-1):
 while i<n1:
  a+=n[i]
  i+=1
print(int(a)%((10**9)+7))

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