ISet<string> uniqueProducts = new HashSet<string>();
for(int i = 0; i < name.Count; i++)
{
uniqueProducts.Add(name[i] + " " + price[i] + " " + weight[i]);
}
return name.Count = uniqueProducts.Count;
Sunday, July 17, 2022
Frequency of Max Value | C# | HackerRank
public static List<int> FrequencyOfMaxValue(List<int> numbers,
List<int> q)
{
List<int> result = new List<int>();
int n = numbers.Count;
int[,] table = new int[2,n];
Dictionary<int, int> counts = new Dictionary<int, int>();
table[0,n-1] = numbers[n-1];
table[1,n-1] = 1;
counts.Add(numbers[n-1], 1);
for(int i = n-2; i >= 0; i--)
{
if(!counts.ContainsKey(numbers[i]))
counts.Add(numbers[i],1);
else
counts[numbers[i]]++;
if(numbers[i] > table[0, i+1])
{
table[0,i] = numbers[i];
table[1,i] = 1;
}
else
{
table[0,i] = table[0, i+1];
table[1,i] = counts[table[0,i]];
}
}
for(int i = 0; i < n; i++)
{
result.Add(table[1,q[i] - 1]);
}
return result;
}
Equal Levels | HackerRank
public static int updateTimes(List<int> signalOne,
List<int> signalTwo)
{
int noOfUpdate = 0;
int maxEqual = int.MinValue;
int length;
int signalOneCount = signalOne.Count;
int signalTwoCount = signalTwo.Count;
if(signalOneCount < signalTwoCount)
length = signalOneCount;
else
length = signalTwoCount;
for(int i = 0; i < length; i++)
{
if(signalOne[i] == signalTwo[i])
{
if(maxEqual < signalOne[i])
{
maxEqual = signalOne[i];
noOfUpdate++;
}
}
}
return noOfUpdate;
}
Sunday, June 5, 2022
Customer List | HackerRank Certification | React
import React, {useState} from "react";
import "./index.css";
function CustomerList(){
const[customer, setCustomer] = useState("");
const[customers, setCustomers] = useState([]);
const handleSubmit = e => {
e.preventDefault();
if(customers.length === 0)
return;
setCustomers([...customers,
{name:customer, count:customer.length}]);
setCustomer("");
}
let rendered = "";
if(customer.length !== 0){
rendered = <ul className="styled mt-50"
data-testid="customer-list">
{customers && customers.map(cus => {
return(
<li className="slide-up-fade-in"
data-testid={"list-item"+cus.count}
key={"list-item"+cus.count}>
{cus.name}
</li>
)
})}
</ul>
}
return(
<div className="mt-75 layout-column
justify-content-center
align-items-center">
<section className="layout-row
align-items-center
justify-content-center">
<input type="text"
className="large"
placeholder="Name" data-testid="app-input"
value={customer}
onChange={e => setCustomer(e.target.value)} />
<button type="submit"
className="ml-30"
data-testid="submit-button"
onClick={handleSubmit}>
Add Customer
</button>
</section>
{rendered}
</div>
);
}
export default CustomerList
Monday, May 30, 2022
Count String Permutations | HackerRank Certification
Count all possible N-length vowel permutations that can be generated based on the given conditions
Given an integer N, the task is to count the number of N-length strings consisting of lowercase vowels that can be generated based the following conditions:
- Each ‘a’ may only be followed by an ‘e’.
- Each ‘e’ may only be followed by an ‘a’ or an ‘i’.
- Each ‘i’ may not be followed by another ‘i’.
- Each ‘o’ may only be followed by an ‘i’ or a ‘u’.
- Each ‘u’ may only be followed by an ‘a’.a
nput: N = 1
Output: 5
Explanation: All strings that can be formed are: “a”, “e”, “i”, “o” and “u”.
Input: N = 2
Output: 10
Explanation: All strings that can be formed are: “ae”, “ea”, “ei”, “ia”, “ie”, “io”, “iu”, “oi”, “ou” and “ua”.
using System;
using System.Collections.Generic;
class StringPermutation {
static int countVowelPermutation(int n)
{
int MOD = (int)(1e9 + 7);
long[,] dp = new long[n + 1, 5];
for (int i = 0; i < 5; i++) {
dp[1, i] = 1;
}
List<List<int>> relation = new List<List<int>>();
relation.Add(new List<int> { 1 });
relation.Add(new List<int> { 0, 2 });
relation.Add(new List<int> { 0, 1, 3, 4 });
relation.Add(new List<int> { 2, 4 });
relation.Add(new List<int> { 0 });
for (int i = 1; i < n; i++)
{
for (int u = 0; u < 5; u++)
{
dp[i + 1, u] = 0;
foreach(int v in relation[u])
{
dp[i + 1, u] += dp[i, v] % MOD;
}
}
}
long ans = 0;
for (int i = 0; i < 5; i++)
{
ans = (ans + dp[n, i]) % MOD;
}
return (int)ans;
}
static void Main() {
int N = 2;
Console.WriteLine(countVowelPermutation(N));
}
}
Condensed List | HackerRank Certification | Remove repeated node from a Singly Linked List
Given a list of integers, remove any nodes that have values that have previously occurred in the list and return a reference to the head of the list.
For e.g:
Linked List
Input : 3 --> 4 --> 3 --> 6
Output: 3 --> 4 --> 6
public static SinglyLinkedListNode(SinglyLinkedListNode head)
{
var hSet = new HashSet<int>();
SinglyLinkedListNode newList = new SinglyLinkedListNode();
while(head != null)
{
hSet.Add(head.data);
head = head.next;
}
foreach(int i in hSet)
{
newList.InsertNode(i);
}
return (newList.head);
}
Sunday, April 17, 2022
Staircase | HackerRank | C#
This is a staircase of size :
#
##
###
####
Its base and height are both equal to . It is drawn using # symbols and spaces. The last line is not preceded by any spaces.
Write a program that prints a staircase of size .
Function Description
Complete the staircase function in the editor below.
staircase has the following parameter(s):
- int n: an integer
Print a staircase as described above.
Input Format
A single integer, , denoting the size of the staircase.
Constraints
.
Output Format
Print a staircase of size using # symbols and spaces.
Note: The last line must have spaces in it.
Sample Input
6
Sample Output
#
##
###
####
#####
######
Explanation
The staircase is right-aligned, composed of # symbols and spaces, and has a height and width of .
Solution
public static void staircase(int n)
{
for(int i=0; i < n; i++)
{
Console.WriteLine(new String('#', i+1).PadLeft(n));
}
}
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