TCS NQT Advanced Coding Previous year Questions
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- TCS NQT Advanced Coding Previous Year Questions
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TCS NQT Advanced Coding Videos
TCS NQT Advanced Coding Previous Year Questions
For example :
3
2 3 2
It doesn't have equal elements appear in pairs
7
1 1 2 2 2 3 3
It contains three consecutive instances of an element.
5
2 2 3 1 1
It is valid and the odd occurring element present in it is 3.
Enter only valid inputs.
Sample Input :
5
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;
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
6
2 3 5 6 8 10
10
5
1 2 3 4 5
10
Output :
3
3
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
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 :
2
1
2
Output :
0
1
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
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
Output the answer modulo 100000007.
Refer the sample output for formatting
Sample Input :
4
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)
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:
7
Explanation:
The fake palindrome for the string ABAB are A, B, A, B, ABA, BAB, ABAB.
Sample Input 2:
AAA
Sample output 2:
6
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);
Problem Statement:
You are given a binary tree with N nodes where each node contains a positive integer value. The goal is to find the maximum sum path from the root to any leaf node. However, you need to ensure that the sum of the values along the path is not divisible by a given integer K.
Input:
- The first line of input contains an integer
N, the number of nodes in the tree. - The second line contains
Nspace-separated integers representing the values of the nodes. - The third line contains
N-1pairs of integersuandv, representing the edges of the tree (the tree is rooted at node 1). - The fourth line contains a single integer
K, the divisor.
Output:
- Print the maximum sum of the path from the root to any leaf node such that the sum is not divisible by
K. If no such path exists, print-1.Input:
7 3 4 8 2 1 6 10 1 2 1 3 2 4 2 5 3 6 3 7 5Output:
21Explanation:
-
The possible paths from the root (node 1) to the leaves are:
- 3 -> 4 -> 2 (sum = 9)
- 3 -> 4 -> 1 (sum = 8)
- 3 -> 8 -> 6 (sum = 17)
- 3 -> 8 -> 10 (sum = 21)
-
The path with the maximum sum that is not divisible by 5 is
3 -> 8 -> 10, with a sum of 21. -
Notes:
- Consider edge cases where no valid path exists.
- Optimize the solution for large inputs where
Ncan be as large as100,000.
Solutions:
C++ Solution
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> values;
vector<vector<int>> adj;
int K;
int maxSum = -1;
void dfs(int node, int parent, int currentSum) {
currentSum += values[node - 1];
bool isLeaf = true;
for (int neighbor : adj[node]) {
if (neighbor != parent) {
isLeaf = false;
dfs(neighbor, node, currentSum);
}
}
if (isLeaf && currentSum % K != 0) {
maxSum = max(maxSum, currentSum);
}
}
int main() {
int N;
cin >> N;
values.resize(N);
adj.resize(N + 1);
for (int i = 0; i < N; i++) {
cin >> values[i];
}
for (int i = 0; i < N - 1; i++) {
int u, v;
cin >> u >> v;
adj[u].push_back(v);
adj[v].push_back(u);
}
cin >> K;
dfs(1, -1, 0);
cout << maxSum << endl;
return 0;
}
Java Solution
import java.util.*;
public class Main {
static List<Integer>[] adj;
static int[] values;
static int K;
static int maxSum = -1;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
values = new int[N];
adj = new ArrayList[N + 1];
for (int i = 0; i < N; i++) {
values[i] = sc.nextInt();
}
for (int i = 1; i <= N; i++) {
adj[i] = new ArrayList<>();
}
for (int i = 0; i < N - 1; i++) {
int u = sc.nextInt();
int v = sc.nextInt();
adj[u].add(v);
adj[v].add(u);
}
K = sc.nextInt();
dfs(1, -1, 0);
System.out.println(maxSum);
}
static void dfs(int node, int parent, int currentSum) {
currentSum += values[node - 1];
boolean isLeaf = true;
for (int neighbor : adj[node]) {
if (neighbor != parent) {
isLeaf = false;
dfs(neighbor, node, currentSum);
}
}
if (isLeaf && currentSum % K != 0) {
maxSum = Math.max(maxSum, currentSum);
}
}
}
Python Solution
def dfs(node, parent, current_sum):
global max_sum
current_sum += values[node - 1]
is_leaf = True
for neighbor in adj[node]:
if neighbor != parent:
is_leaf = False
dfs(neighbor, node, current_sum)
if is_leaf and current_sum % K != 0:
max_sum = max(max_sum, current_sum)
N = int(input())
values = list(map(int, input().split()))
adj = {i: [] for i in range(1, N + 1)}
for _ in range(N - 1):
u, v = map(int, input().split())
adj[u].append(v)
adj[v].append(u)
K = int(input())
max_sum = -1
dfs(1, -1, 0)
print(max_sum)
Explanation:
- DFS Traversal: The depth-first search (DFS) algorithm is used to traverse the tree from the root node (node 1) to each leaf node.
- Sum Calculation: For each path from the root to a leaf node, the sum of node values is calculated.
- Divisibility Check: If the sum is not divisible by the given integer
K, it is compared to the current maximum sum, and the maximum is updated if necessary. - Edge Cases: The code handles cases where no valid path exists by setting the initial maximum sum to
-1.
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