TCS Digital Coding Questions and Answers 2024 (Freshers/Exp) | TCS Digital Advanced Coding Questions and Answers

TCS Digital Coding Questions and Answers 2024 | TCS Digital Advanced Coding Questions and Answers | Previously Asked TCS Digital Coding Questions | TCS Digital Coding Solved Old Questions | TCS Digital Coding Questions For Freshers

TCS Digital Coding Questions and Answers 2024: TCS Digital Coding is one of the most important components of TCS Digital’s hiring procedure. This section will assess your problem-solving and programming efficiency. This round will consist of four code questions covering a variety of topics. You can supply the solutions in any of the supported programming languages. TCS Digital did not specify a language for hiring, whereas only Python and Java were allowed for TCS Digital Last year. However, we suggest that you focus more on C++, Java, Python, etc.

Users are given a series of TCS Digital coding problems to test their coding abilities. The level of the TCS Digital Coding Questions ranges from easy to moderate. More details like TCS Digital Coding Questions and Answers, TCS Digital Coding Questions and Answers, TCS Digital Interview Questions and Answers, TCS Digital Cut Off, Salary, Hike, Increment, Annual Appraisal, etc. are available on this page

TCS Digital Coding Questions Marking Scheme 2024

TCS Digital Coding Test Pattern 2024 (Updated)

You may use any of the following languages during the test:

• C
• C#
• C++
• Java 7
• Kotlin
• Lua
• Perl
• Python
• Python3
• Ruby
• Scala
• Erlang
• go
• Haskel
• Java

#include<bits/stdc++.h> 
using namespace std;
typedef long long int lld;
#define mod 1000000007
long e_sum(long m,long n,long K,long N)
{
if(K==1)
{
return n;
}
else
{
return (N-(m-n)*e_sum(m,n,K-1,N)%(1000000007));
}
}
int main()
{
long low,high,K,m,n,diff,Out,N,i;
scanf("%ld",&low);
scanf("%ld",&high);
scanf("%ld",&K);
diff=high-low+1;
if(diff%2==0)
{
m=diff/2;
n=m;
}
else
{
if(low%2==0)
{
m=(diff-1)/2;
n=m+1;
}
else
{
m=(diff+1)/2;
n=m-1;
}
}
N=m;
for(i=0;i<K-1;i++)
{
N=(N*(m+n))%1000000007;
}
Out=e_sum(m,n,K,N)%1000000007;
printf("%ld",Out);
return 0;
}

mod=1000000007
def esum(m,n,K,N):
    if(K==1):
       return n
    else:
        return (N-(m-n)*esum(m,n,K-1,N)%mod)
low,high=map(int,input().split())
K=int(input())
diff=high-low+1
if(diff%2==0):
m=diff//2
n=m
else:
if(low%2==0):
m=(diff-1)//2
n=m+1
else:
m=(diff+1)//2
n=m-1
N=m
for i in range(K-1):
N=(N*(m+n))%mod
out=esum(m,n,K,N)%mod
print(out)

import sys
def if_palindrome(s):
    if len(s)==1:
        return True
    s1=s[::-1]
    return s1==s
s=input()
l=len(s) 
for i in range(1,l-1):
    s1=s[:i]
    if if_palindrome(s1):
        for j in range(1,l-i):
            s2=s[i:i+j]
            s3=s[i+j:]
            if if_palindrome(s2) and if_palindrome(s3):
                print(s1)
                print(s2)
                print(s3)
                sys.exit()
print("Impossible")

#include<bits/stdc++.h>
using namespace std;
vector rotate(int nums[], int n, int k) {

if (k > n)
k = k % n;

vector ans(n);

for (int i = 0; i < k; i++) {
ans[i] = nums[n - k + i];
}

int index = 0;
for (int i = k; i < n; i++) {
ans[i] = nums[index++];
}
return ans;
}

int main()
{
int Array[] = { 1, 2, 3, 4, 5 };
int N = sizeof(Array) / sizeof(Array[0]);
int K = 2;

vector ans = rotate(Array, N, K);
for (int i = 0; i < N; ++i) {
cout << ans[i] << ' ';
}
}

import java.util.*;
public class Main
{
    static int find(int n1, int n2) {
        int count = 0;
        for (int i = n1 ; i <= n2 ; i++) { int num = i; boolean[] visited = new boolean[10]; while (num > 0) {
if (visited[num % 10] == true)
break;
visited[num % 10] = true;
num /= 10;
}
if (num == 0)
count++;
}
return count;
}
public static void main(String[] args) {
int n1 = 101, n2 = 200;
System.out.println(find(n1, n2));
}
}

TCS Digital Advanced Coding Questions and Answers 2024

Q) Roco is an island near Africa which is very prone to forest fire. Forest fire is such that it destroys the complete forest. Not a single tree is left.This island has been cursed by God , and the curse is that whenever a tree catches fire, it passes the fire to all its adjacent tree in all 8 directions,North, South, East, West, North-East, North-West, South-East, and South-West.And it is given that the fire is spreading every minute in the given manner, i.e every tree is passing fire to its adjacent tree.Suppose that the forest layout is as follows where T denotes tree and W denotes water.

Your task is that given the location of the first tree that catches fire, determine how long would it take for the entire forest to be on fire. You may assume that the lay out of the forest is such that the whole forest will catch fire for sure and that there will be at least one tree in the forest

Input Format:

• First line contains two integers, M, N, space separated, giving the size of the forest in terms of the number of rows and columns respectively.
• The next line contains two integers X,Y, space separated, giving the coordinates of the first tree that catches the fire.
• The next M lines, where ith line containing N characters each of which is either T or W, giving the position of the Tree and Water in the  ith row of the forest.

Output Format:

Single integer indicating the number of minutes taken for the entire forest to catch fire

Constrains:

• 3 ≤ M ≤ 20
• 3 ≤ N ≤ 20

Sample Input 1:

3 3
W T T
T W W
W T T
Sample Output 1:

5

Explanation:
In the second minute,tree at (1,2) catches fire,in the third minute,the tree at (2,1) catches fire,fourth minute tree at (3,2) catches fire and in the fifth minute the last tree at (3,3) catches fire.
Sample Input 2:
6 6
1 6
W T T T T T
T W W W W W
W T T T T T
W W W W W T
T T T T T T
T W W W W W

Sample Output 2:

16

EXAMPLE

C++ PROGRAM

#include <bits/stdc++.h>
using namespace std;
char f[21][21];
int n,m;
struct node{int a,b;};
bool valid(int x,int y) {return (x>=0&&y>=0&&x<n&&y<m);}
bool step(node temp){return (temp.a==-1&&temp.b==-1);}
int main()
{
cin>>n>>m;
int x,y,i,j,count=0;int ans=1;
cin>>x>>y;x--;y--;
for(i=0;i<n;i++)
for(j=0;j<m;j++)
cin>>f[i][j];
f[x][y]='X';

queueq;
node temp;
temp.a=x;temp.b=y;
q.push(temp);
temp.a=-1;temp.b=-1;
q.push(temp);

while(!q.empty())
{
bool flag=false;
while(!step(q.front()))
{
node count=q.front();
if(valid(count.a+1,count.b)&&f[count.a+1][count.b]=='T')//a+1,b
{
if(flag==false){flag=true;ans++;}
f[count.a+1][count.b]='X';
count.a++;
q.push(count);
count.a--;
}
if(valid(count.a+1,count.b+1)&&f[count.a+1][count.b+1]=='T')//a+1,b+1
{
if(flag==false){flag=true;ans++;}
f[count.a+1][count.b+1]='X';
count.a++;count.b++;
q.push(count);
count.a--;count.b--;
}
if(valid(count.a+1,count.b-1)&&f[count.a+1][count.b-1]=='T')//a+1,b-1
{
if(flag==false){flag=true;ans++;}
f[count.a+1][count.b-1]='X';
count.a++;count.b--;
q.push(count);
count.a--;count.b++;
}
if(valid(count.a,count.b+1)&&f[count.a][count.b+1]=='T')//a,b+1
{
if(flag==false){flag=true;ans++;}
f[count.a][count.b+1]='X';
count.b++;
q.push(count);
count.b--;
}
if(valid(count.a,count.b-1)&&f[count.a][count.b-1]=='T')//a,b-1
{
if(flag==false){flag=true;ans++;}
f[count.a][count.b-1]='X';
count.b--;
q.push(count);
count.b++;
}
if(valid(count.a-1,count.b-1)&&f[count.a-1][count.b-1]=='T')//a-1,b-1
{
if(flag==false){flag=true;ans++;}
f[count.a-1][count.b-1]='X';
count.a--;count.b--;
q.push(count);
count.a++;count.b++;
}
if(valid(count.a-1,count.b+1)&&f[count.a-1][count.b+1]=='T')//a-1,b+1
{
if(flag==false){flag=true;ans++;}
f[count.a-1][count.b+1]='X';
count.a--;count.b++;
q.push(count);
count.a++;count.b--;
}
if(valid(count.a-1,count.b)&&f[count.a-1][count.b]=='T')//a-1,b
{
if(flag==false){flag=true;ans++;}
f[count.a-1][count.b]='X';
count.a--;
q.push(count);
count.a++;
}
q.pop();
}
q.pop();
if(!q.empty())
{
temp.a=-1;
temp.b=-1;
q.push(temp);
}

/*
cout<<endl;
for(i=0;i<n;i++)
{for(j=0;j<m;j++)
{cout<<f[i][j]<<" ";}cout<<endl;}
cout<<endl<<endl;
*/
}
cout<<ans;
}

JAVA PROGRAM

import java.util.HashMap;
import java.util.Scanner;

class Temp {

    static void print(int [][] arr){
        for(int i=0;i<arr.length;i++){
            for (int j=0;j<arr[0].length;j++){
                System.out.print(arr[i][j]+"\t");
            }
            System.out.println();
        }
    }
    public static void main (String[] args) {
        Scanner scanner = new Scanner(System.in);
        int N = scanner.nextInt();
        int [][] array = new int[N][N];
        HashMap<Integer, Integer> map = new HashMap<>();

        int powerpoints = 1 + (N*N)/11;

        int counter = 1;
        int row = 0, col = 0; // for the current row/col that we are end
        int endRow = 0, endColumn = 0; // for the row/col where we need to stop

        map.put(0,0);

        for(int i=0;i<N/2;i++){
            row = col = i;
            endColumn = N-i-1;

            while (col < endColumn) {
                array[row][col] = counter;

                if (counter % 11==0)
                    map.put(row,col);

                counter++;
                col++;
            }
            endRow = N-i-1;
            while (row<endRow){
                array[row][col] = counter;
                if (counter % 11==0)
                    map.put(row,col);
                counter++;
                row++;
            }

            endColumn = i;
            while (col>endColumn){
                array[row][col] = counter;
                if (counter % 11==0)
                    map.put(row,col);
                counter++;
                col--;
            }

            endRow = i;
            while (row>endRow){
                array[row][col] = counter;
                if (counter % 11==0)
                    map.put(row,col);
                counter++;
                row--;
            }


        }
        if (N%2==1)
            array[N/2][N/2] = N*N;

        print(array);
        System.out.println("Total Power Points: " + powerpoints);
        for (Integer key: map.keySet()) {
            System.out.println("("+key+ ","+map.get(key)+")");

        }

    }

}

Problem Statement:- In a Conference ,attendees are invited for a dinner after the conference.The Co-ordinator,Sagar arranged around round tables for dinner and want to have an impactful seating experience for the attendees.Before finalizing the seating arrangement,he wants to analyze all the possible arrangements.These are R round tables and N attendees.In case where N is an exact multiple of R,the number of attendees must be exactly N//R,,If N is not an exact multiple of R, then the distribution of attendees must be as equal as possible.Please refer to the example section before for better understanding.
For example, R = 2 and N = 3
All possible seating arrangements are
(1,2) & (3)
(1,3) & (2)
(2,3) & (1)
Attendees are numbered from 1 to N.

Input Format:

• The first line contains T denoting the number of test cases.
• Each test case contains two space separated integers R and N, Where R denotes the number of round tables and N denotes the number of attendees.

Output Format:

Single Integer S denoting the number of possible unique arrangements.

Constraints:

• 0 <= R <= 10(Integer)
• 0 < N <= 20 (Integer)

#include<bits/stdc++.h>
using namespace std;
typedef long long int ll; 
map <ll,ll> dp;
int fac(int n)
{
if(dp[n])
return dp[n];
dp[n]=n*fac(n-1);
return dp[n];
}
int func(int n)
{
if(n<=3)
return 1;
return fac(n-1);
}

int main()
{
dp[0]=dp[1]=1;
int tests;cin>>tests;
while(tests--)
{int R,N,c=1;
cin>>R>>N;
if(N<=R)
{
cout<<1;
continue;
}
int a=N/R,n=N%R;
ll ans=fac(N)/(pow(fac(a),R-n) * pow(fac(a+1),n) )/fac(n)/fac(R-n);
for(int i=1;i<=n;i++)
c*=func(a);
for(int i=n+1;i<=R;i++)
c*=func(a-1);

cout<<c*ans;}
}

import java.util.*;
public class Stack
{
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int testcases = scanner.nextInt();
        int tables = 0,people;
        while (testcases-->0)
            tables = scanner.nextInt();
            people = scanner.nextInt();
            System.out.println(find(tables,people));
    }
    public static int find(int r,int n)
    {
        int x = n/r;
        int y1 = n%r;
        int x1 = 0;
        int ans1 = 1;
        while(r!=0)
        {
            if(y1>0)
            {
                x1 = x+1;
                y1 = y1-1;
            }
            else
            {
                x1 = x;
            }
            ans1 = ans1*combination(n,x1);
            n = n-x1;
            r--;
        }
        return ans1;
    }
    public static int factorial(int n)
    {
        if(n==0||n==1)
        {
            return 1;
        }
        return n * factorial(n-1);
    }
    public static int combination(int n,int r)
    {
        return factorial(n)/(factorial(n-r)*factorial(r));
    }
}

PYTHON PROGRAM

testcases = int(input())
for i in range(testcases):
    tables, people = map(int, input().split())
    result = 1
if tables >= people:
print(result)
else:
PA = people // tables
PB = PA + 1
TB = people % tables
TA = tables - TB
# Using DP to store factorials pre-hand
fact = [1]
for j in range(1, people + 1):
fact.append(j*fact[j-1])

for i in range(TB):
result = result * (fact[people]//(fact[PB]*fact[people-PB]))
people = people - PB

for i in range(TA):
result = result * (fact[people]//(fact[PA]*fact[people-PA]))
people = people - PA

print(result)

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