# TCS NQT Previous Year Coding Questions

10 May 2023

10 May 2023

TCS NQT Advanced Coding Previous year Questions

Q1) Given an array of integers where every element appears even number of times except one element which appears odd number of times, write a program to find that odd occurring element in O(log n) time. The equal elements must appear in pairs in the array but there cannot be more than two consecutive occurrences of an element.

For example :

2 3 2

It doesn't have equal elements appear in pairs

1 1 2 2 2 3 3

It contains three consecutive instances of an element.

2 2 3 1 1

It is valid and the odd occurring element present in it is 3.

Enter only valid inputs.

Sample Input :

2 2 3 1 1

Sample Output :

3

Solutions:

C Language

#include <stdio.h>

int main()

{

int n;  scanf("%d",&n);

int arr[n];

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

scanf("%d",&arr[i]);

}

int left=0,right=n-1,mid,pre,nxt;

if(arr[0] != arr[1]){

printf("%d ",arr[0]);

}

else if(arr[n-1] != arr[n-2]){

printf("%d ",arr[n-1]);

}

else{

while(left <= right){

mid = ((right-left)/2)+left;

pre = mid-1;

nxt = mid+1;

if((arr[pre] != arr[mid]) && (arr[nxt] != arr[mid])){

printf("%d ",arr[mid]);

break;

}

else if(mid%2==0){

if(arr[pre] == arr[mid])

right = mid - 1;

else

left = mid + 1;

}

else{

if(arr[pre] == arr[mid])

left = mid + 1;

else

right = mid - 1;

}

}

}

return 0;

}

C++

#include <iostream>

#include <vector>

using namespace std;

int main()

{

int n;  cin >> n;

vector<int> v(n);

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

cin >> v[i];

}

int left=0,right=n-1,mid,pre,nxt;

if(v[0] != v[1]){

cout << v[0];

}

else if(v[n-1] != v[n-2]){

cout << v[n-1];

}

else{

while(left <= right){

mid = ((right-left)/2)+left;

pre = mid-1;

nxt = mid+1;

if((v[pre] != v[mid]) && (v[nxt] != v[mid])){

cout << v[mid];

break;

}

else if(mid%2==0){

if(v[pre] == v[mid])

right = mid - 1;

else

left = mid + 1;

}

else{

if(v[pre] == v[mid])

left = mid + 1;

else

right = mid - 1;

}

}

}

return 0;

}

JAVA

import java.util.*;

public class Main

{

public static void main(String[] args) {

Scanner sc = new Scanner(System.in);

int n = sc.nextInt();

int arr[] = new int[n];

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

arr[i] = sc.nextInt();

}

int left=0,right=n-1,mid,pre,nxt;

if(arr[0] != arr[1]){

System.out.print(arr[0]);

}

else if(arr[n-1] != arr[n-2]){

System.out.print(arr[n-1]);

}

else{

while(left <= right){

mid = ((right-left)/2)+left;

pre = mid-1;

nxt = mid+1;

if((arr[pre] != arr[mid]) && (arr[nxt] != arr[mid])){

System.out.print(arr[mid]);

break;

}

else if(mid%2==0){

if(arr[pre] == arr[mid])

right = mid - 1;

else

left = mid + 1;

}

else{

if(arr[pre] == arr[mid])

left = mid + 1;

else

right = mid - 1;

}

}

}

}

}

PYTHON

n = int(input())

a = list(map(int,input().split()))

left=0

right=n-1

if a[0] != a[1]:

print(a[0])

elif a[n-1] != a[n-2]:

print(a[n-1])

else:

while left <= right:

mid = ((right-left)//2)+left

pre = mid-1

nxt = mid+1

if (a[pre] != a[mid])  and (a[nxt] != a[mid]):

print(a[mid])

break

elif mid%2==0 :

if a[pre] == a[mid] :

right = mid - 1

else :

left = mid + 1

else :

if a[pre] == a[mid] :

left = mid + 1

else :

right = mid - 1;

Q2) Given an array of integers and a sum, the task is to count all subsets of given array with sum equal to given sum.

Input :

The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case contains an integer n denoting the size of the array. The next line contains n space separated integers forming the array. The last line contains the sum.

Output :

Count all the subsets of given array with sum equal to given sum.

NOTE: Since result can be very large, print the value modulo 109+7.

Constraints :

1<=T<=100

1<=n<=103

1<=a[i]<=103

1<=sum<=103

Example :

Input :

2 3 5 6 8 10

10

1 2 3 4 5

10

Output :

Explanation :

Testcase 1: possible subsets : (2,3,5) , (2,8) and (10)

Testcase 2: possible subsets : (1,2,3,4) , (2,3,5) and (1,4,5)

Solutions:

C++

#include <bits/stdc++.h>

using namespace std;

void printBool(int n, int len)

{

while (n) {

if (n & 1)

cout << 1;

else

cout << 0;

n >>= 1;

len--;

}

while (len) {

cout << 0;

len--;

}

cout << endl;

}

void printSubsetsCount(int set[], int n, int val)

{

int sum;

int count = 0;

for (int i = 0; i < (1 << n); i++) {

sum = 0;

for (int j = 0; j < n; j++)

if ((i & (1 << j)) > 0) {

sum += set[j];

}

if (sum == val) {

count++;

}

}

if (count == 0) {

cout << ("No subset is found") << endl;

}

else {

cout << count << endl;

}

}

int main()

{

int t,n,sum;

cin>>t;

while(t--)

{

cin>>n;

int set[n];

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

cin>>set[i];

cin>>sum;

printSubsetsCount(set, n, sum);

}

}

JAVA

import java.io.*;

import java.util.*;

class Main {

static void printBool(int n, int len)

{

while (n>0) {

if ((n & 1) == 1)

System.out.print(1);

else

System.out.print(0);

n >>= 1;

len--;

}

while (len>0) {

System.out.print(0);

len--;

}

System.out.println();

}

static void printSubsetsCount(int set[], int n, int val)

{

int sum;

int count = 0;

for (int i = 0; i < (1 << n); i++) {

sum = 0;

for (int j = 0; j < n; j++)

if ((i & (1 << j)) > 0) {

sum += set[j];

}

if (sum == val) {

count++;

}

}

if (count == 0) {

System.out.println("No subset is found");

}

else {

System.out.println(count);

}

}

public static void main(String[] args)

{

Scanner sc = new Scanner(System.in);

int t = sc.nextInt();

int n,sum;

while(t>0)

{

n=sc.nextInt();

int set[] = new int[n];

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

set[i] = sc.nextInt();

sum = sc.nextInt();

printSubsetsCount(set, n, sum);

t--;

}

}

}

PYTHON

def printBool(n, len):

while n:

if n & 1:

print("1 ")

else:

print("0 ")

n = n >> 1

len -= 1

while len:

print("0 ")

len -= 1

print()

def printSubsetsCount(set, n, val):

sum = 0

count = 0

for i in range(0, 1 << n):

sum = 0

for j in range(0, n):

if (i & 1 << j) > 0:

sum += set[j]

if (sum == val):

count += 1

if (count == 0):

print("No subset is found")

else:

print(count)

t=int(input())

set = []

while t>0:

n = int(input())

set = list((map(int,input().strip().split())))[:n]

Sum = int(input())

printSubsetsCount(set, n, Sum)

t = t-1

Q3) Before the outbreak of corona virus to the world, a meeting happened in a room in Wuhan. A person who attended that meeting had COVID-19 and no one in the room knew about it! So everyone started shaking hands with everyone else in the room as a gesture of respect and after meeting unfortunately every one got infected! Given the fact that any two persons shake hand exactly once, Can you tell the total count of handshakes happened in that meeting?

Input Format :

The first line contains the number of test cases T, T lines follow.

Each line then contains an integer N, the total number of people attended that meeting.

Output Format :

Print the number of handshakes for each test-case in a new line.

Constraints :

1 <= T <= 1000

0 < N < 106

Sample Input :

Output :

Explanation :

Case 1 : The lonely board member shakes no hands, hence 0.

Case 2 : There are 2 board members, 1 handshake takes place.

Solutions:

C Language

#include <stdio.h>

int main()

{

int t,n;  scanf("%d",&t);

long int sum = 0;

while(t--){

scanf("%d",&n);

n--;

sum = (n*(n+1))/2;

printf("%ld\n",sum);

}

return 0;

}

C++

#include <iostream>

using namespace std;

int main()

{

int t,n,sum;  cin >> t;

long long sum;

while(t--){

cin >> n;

n--;

sum = (n*(n+1))/2;

cout << sum << endl ;

}

return 0;

}

JAVA

import java.util.*;

public class Main

{

public static void main(String[] args) {

Scanner sc = new Scanner(System.in);

int t,n;

long sum;

t = sc.nextInt();

while(t > 0){

n = sc.nextInt();

n--;

sum = (n*(n+1))/2;

System.out.println(sum);

t--;

}

}

}

PYTHON

t = int(input())

while t > 0 :

n = int(input())

n = n-1;

print((n*(n+1))//2)

t = t-1

Q4) For enhancing the book reading, school distributed story books to students as part of the Children’s day celebrations. To increase the reading habit, the class teacher decided to exchange the books every weeks so that everyone will have a different book to read. She wants to know how many possible exchanges are possible.

If they have 4 books and students, the possible exchanges are 9. Bi is the book of i-th student and after the exchange, he should get a different book, other than Bi.

B1 B2 B3 B4 – first state, before exchange of the books

B2 B1 B4 B3

B2 B3 B4 B1

B2 B4 B1 B3

B3 B1 B4 B2

B3 B4 B1 B2

B3 B4 B2 B1

B4 B1 B2 B3

B4 B3 B1 B2

B4 B3 B2 B1

Find the number of possible exchanges, if the books are exchanged so that every student will receive a different book.

Constraints

1<= N <= 1000000

Input Format

Input contains one line with N, indicates the number of books and number of students.

Output Format

Refer the sample output for formatting

Sample Input :

Sample Output :

9

Solutions:

C Language

#include <stdio.h>

int main()

{

int mod = (int)1e7+7;

long int n,a=0,b=1,c=2,d,i;

scanf("%ld",&n);

if(n==1 || n==2)

printf("%ld",n-1);

else{

for(i=3 ; i<=n ; i++){

d = (c * (a + b)) % mod ;

a = b;

b = d;

c++;

}

printf("%ld",d);

}

return 0;

}

C++

#include <iostream>

using namespace std;

int main()

{

int mod = (int)1e7+7;

long int n,a=0,b=1,c=2,d,i;

cin >> n;

if(n==1 || n==2)

cout << (n-1);

else{

for(i=3 ; i<=n ; i++){

d = (c * (a + b)) % mod ;

a = b;

b = d;

c++;

}

cout << d;

}

return 0;

}

JAVA

import java.util.*;

public class Main

{

public static void main(String[] args) {

Scanner sc = new Scanner(System.in);

int mod = (int)1e8+7;

long n,a=0,b=1,c=2,d=0;

n = sc.nextLong();

if(n==1 || n==2)

System.out.println(n-1);

else{

for(int i=3 ; i<=n ; i++){

d = (c * (a + b)) % mod ;

a = b;

b = d;

c++;

}

System.out.println(d);

}

}

}

PYTHON

mod = 100000007;

a=0

b=1

c=2

d=0

n = int(input())

if n==1 or n==2:

print(n-1)

else :

for i in range(1,n) :

d = (c * (a + b)) % mod

a = b

b = d

c = c + 1

print(d)

Q5) You are given a string A and you have to find the number of different sub-strings of the string A which are fake palindromes.

Note:

1. Palindrome: A string is called a palindrome if you reverse the string yet the order of letters remains the same. For example, MADAM.

2. Fake Palindrome: A string is called as a fake palindrome if any of its permutations is a palindrome. For example, AAC is fake palindrome, but ACD is not.

3. Sub-string: A sub-string is a contiguous sequence (non-empty) of characters within a string.

4. Two sub-strings are considered same if their starting indices and ending indices are equal.

Input Format:

First line contains a string S

Output Format:

Print a single integer (number of fake palindrome sub-strings).

Constraints:

1 <= |S| <= 2 * 105

The string will contain only Upper case 'A' to 'Z'

Sample Input 1:

ABAB

Sample Output 1:

Explanation:

The fake palindrome for the string ABAB are A, B, A, B, ABA, BAB, ABAB.

Sample Input 2:

AAA

Sample output 2:

Explanation:

The fake palindrome for the string AAA are A, A, A, AA, AA, AAA

Solutions:

C++

#include <bits/stdc++.h>

using namespace std;

void countSubstring(string s)

{

int res = 0;

for (int i = 0;

i < s.length(); i++) {

int x = 0;

for (int j = i;

j < s.length(); j++) {

int temp = 1 << s[j] - 'a';

x ^= temp;

if ((x & (x - 1)) == 0)

res++;

}

}

cout << res;

}

int main()

{

string str;

getline(cin,str);

countSubstring(str);

return 0;

}

JAVA

import java.util.*;

class Main{

static void countSubString(String s)

{

int res = 0;

for (int i = 0; i < s.length(); i++)

{

int x = 0;

for (int j = i; j < s.length(); j++)

{

int temp = 1 << s.charAt(j) - 'a';

x ^= temp;

if ((x & (x - 1)) == 0)

res++;

}

}

System.out.print(res);

}

public static void main(String[] args)

{

Scanner sc = new Scanner(System.in);

String str = sc.nextLine();

countSubString(str);

}

}

PYTHON

def countSubString(s):

res = 0;

for i in range(len(s)):

x = 0;

for j in range(i, len(s)):

temp = 1 << ord(s[j]) - ord('a');

x ^= temp;

if ((x & (x - 1)) == 0):

res += 1;

print(res);

if __name__ == '__main__':

str = input();

countSubString(str);