TCS CodeVita Mock Test Coding Question 3
Coding Question 3
In this article, we will discuss about the TCS CodeVita Coding Question which is asked in the TCS placement test along with a detailed solution. This will help you excel your coding problems.
TCS CodeVita Coding Question
Problem Description
John is working in high security organization, where he needs to deal with lot of confidential documents. He must encrypt the important number in documents. Please help John to write a program where the system should identify the integer number and encrypt number based on John’s actions.
John’s actions are as follows
| Actions | Denoted by | Remarks |
|---|---|---|
| Right arrow | R | Move to next character |
| Left arrow | T | Move to previous character |
| Up arrow | L | Increment the integer by one |
| Down arrow | D | Decrement the integer by one |
| Swap | Sn | Swap the current position with number in the nth position |
These actions are governed by the rules mentioned below.
Rules
- Initial position is at the first character of the string comprised of numbers.
- When incrementing the value of a number at a current location, if the number is 9 then it remains 9 else it increases by 1.
- When decrementing the value of a number at a current location, if the number is 0 then it remains O else it decreases by 1.
- Once all the actions from the action string are consumed, if any part of the input string remained unprocessed then leave those characters of the input string as they are.
Constraints
0 <n<= I where I is length of the input string.
Input
First line contains a string comprising of numbers. This is referred to as input string in the rest of the document.
Second line contains the actions denoted by values mentioned in the table above. This is referred to as action string in the rest of the document.
Output
Single string comprising of numbers.
Time Limit (secs)
1
Examples
Input
123456
RLTDRRTRS2S1
Output
244156
Explanation
Initial string: 123456
Action: RLTDRRTRS2S1
R: 1[2]3456
L: [1]23456
T: [2]23456
D: [1]23456
R: 1[2]3456
R: 12[3]456
T:12[4]456
R: 124[4]56
S2: 144[2]56
S1: 244[1]56
output string.
Since all the actions from the action string are consumed and only first four characters of the input string are processed leave the last two as they are and make them the part of
#include< bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
string s;
cin>>s;
int i=0;
string a = to_string(n);
int j =0;
for(int i=0;i< s.size();i++){
if(s[i] == 'R') j= j+1;
else if(s[i] == 'L') j = j-1;
else if(s[i] == 'T') (int)a[j]++;
else if(s[i] == 'D') (int)a[i]--;
else{
swap(a[s[i]], a[s[i+1]]);
i++;
}
}
cout<< a;
}
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
String s = scanner.next();
int i = 0;
String a = Integer.toString(n);
int j = 0;
for (int k = 0; k < s.length(); k++) {
if (s.charAt(k) == 'R') {
j = j + 1;
} else if (s.charAt(k) == 'L') {
j = j - 1;
} else if (s.charAt(k) == 'T') {
a = a.substring(0, j) + (char)(a.charAt(j) + 1) + a.substring(j + 1);
} else if (s.charAt(k) == 'D') {
a = a.substring(0, i) + (char)(a.charAt(i) - 1) + a.substring(i + 1);
} else {
char temp = a.charAt(s.charAt(k));
a = a.substring(0, s.charAt(k)) + a.charAt(s.charAt(k + 1)) + a.substring(s.charAt(k) + 2);
a = a.substring(0, s.charAt(k + 1)) + temp + a.substring(s.charAt(k + 1) + 1);
k++;
}
}
System.out.println(a);
}
}
n = int(input())
s = input()
i = 0
a = str(n)
j = 0
for i in range(len(s)):
if s[i] == 'R':
j = j + 1
elif s[i] == 'L':
j = j - 1
elif s[i] == 'U':
a = a[:j] + str((int(a[j]) + 1) % 10) + a[j+1:]
elif s[i] == 'D':
a = a[:j] + str((int(a[j]) - 1) % 10) + a[j+1:]
else:
a = list(a)
a[s[i]], a[s[i+1]] = a[s[i+1]], a[s[i]]
a = ''.join(a)
print(a)
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