TCS Coding Questions & Solution 2023

#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] << ' ';
}
}

This code is for rotating an array of numbers to the right by a given number of positions (K positions). Here's a simple explanation:

1. We have an array called Array with some numbers and want to rotate it.
2. N is the number of elements in the array, and K is the number of positions we want to rotate it to the right.
3. We create a function called rotate that takes three arguments: the array nums, its size n, and the number of positions to rotate k.

Inside the rotate function:

4. We check if k is greater than n, and if it is, we set k to be the remainder of k divided by n. This is to handle cases where k is larger than the array size.

5. We create a vector called ans with the same size as the array to store the rotated elements.
6. We copy the last k elements from the original array to the beginning of the ans vector.
7. Then, we copy the remaining elements from the original array to the end of the ans vector.

Finally, in the main function:

8. We call the rotate function with our array Array, its size N, and the number of positions to rotate K.

9. We print the elements of the ans vector, which now contains the rotated array.

So, if you run this code with the given array [1, 2, 3, 4, 5] and rotate it 2 positions to the right, it will print: 4 5 1 2 3, which is the array after rotation.

import java.util.*;
public class Main
{
    static int[] rotate(int nums[], int n, int k) {
if (k > n)
k = k % n;

int[] ans = new int[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;
}
public static void main(String[] args) {
int Array[] = { 1, 2, 3, 4, 5 };
int N = 5;
int K = 2;

int[] ans = rotate(Array, N, K);
for (int i = 0; i < N; ++i) {
System.out.println(ans[i]);
}
}
}

This Java code is for rotating an array of numbers to the right by a given number of positions (K positions). Here's a simple explanation:

1. We have an array called Array with some numbers [1, 2, 3, 4, 5], and we want to rotate it.
2. N is set to 5, which is the number of elements in the array, and K is set to 2, which is the number of positions we want to rotate the array to the right.

Inside the rotate method: 3. We check if k (the number of positions to rotate) is greater than n (the size of the array). If it is, we set k to be the remainder of k divided by n. This is to handle cases where k is larger than the array size.

4. We create an integer array called ans with the same size as the original array to store the rotated elements.
5. We copy the last k elements from the original array to the beginning of the ans array.
6. Then, we copy the remaining elements from the original array to the end of the ans array.

In the main method: 7. We call the rotate method with our array Array, its size N (which is 5), and the number of positions to rotate K (which is 2).

8. We store the result, the rotated array, in the ans array.
9. We then loop through the ans array and print each element using System.out.println().

So, when you run this Java code with the given array [1, 2, 3, 4, 5] and rotate it 2 positions to the right, it will print:

These are the elements of the array after rotation.