I wrote a method that does return sum of decreasing sequence.
For example:
if sequence is 13 9 7 12 13 15, then sum is 29 (13 + 9 + 7).
Now I want to optimize it. Cause the same code is repeating muliple times. But I don't now how. Maybe someone can help me?
int sumMax = 0;
for (int i = 0; i < array.Length; i++)
{
if (i == 0)
{
if (array[i] % 2 == 1)
{
sum += array[i];
if (sumMax < sum)
{
sumMax = sum;
}
}
}
else
{
if (array[i] % 2 == 0)
{
sum = 0;
}
else
{
if (sum != 0)
{
if (array[i - 1] > array[i])
{
sum += array[i];
if (sumMax < sum)
{
sumMax = sum;
}
}
}
else
{
sum += array[i];
if (sumMax < sum)
{
sumMax = sum;
}
}
}
}
}
return sumMax;
Seems like a fairly simple problem. Could simply loop over the array and keep track of the current count and current max value no?
var values = new int[] {13, 9 ,7 ,12 ,13, 15};
var max = Int32.MaxValue;
var total = 0;
foreach (int value in values){
if(value >= max)continue;
max = value;
total += value;
}
This loops over every element in the array so is O(n), might be some way to make this look nicer using LINQ?
I misread what he was asking, something like this would work instead would it not?
var values = new int[] {13, 9 ,7 ,12 ,13, 15};
int previous = int.MaxValue;
var total = 0;
var max = 0;
foreach(int value in values){
if(value >= previous){
if(total > max) max = total;
total = 0;
}
total += value;
previous = value;
}
if(total > max) max = total;
Console.WriteLine(max);
You can still probabbly make this look nicer with some LINQ, but performance wise its still O(n)
Related
Given any natural number N> 1 (previously assigned). Print out the successful development of prime numbers from small to large.
Example:
9 --> 3 * 3
12 --> 2 * 2 * 3
My idea is find all GCD and add to list int, and write a function isPrimeNumber(int n), browse List< int > and check if isPrimeNumber().
But I can't solve problem print out the successful development of prime numbers from small to large
Here is what I tried
static void Main(string[] args)
{
Console.WriteLine("Enter n: ");
int n = Convert.ToInt32(Console.ReadLine());
List<int> arr = new List<int>();
for (int i = 1; i <= n; i++)
{
if (n % i == 0)
{
arr.Add(i);
}
}
/* I need Print out the successful development of prime numbers from small to large here */
}
static bool isPrimeNumber(int n)
{
if (n < 2)
{
return false;
}
for (int i = 2; i <= Math.Sqrt(n); i++)
{
if (n % i == 0)
{
return false;
}
}
return true;
}
As you posted your working solution for that, let me share a different implementation for that that is still simple to understand, but more efficient, because it only tests primes until it reaches the square root of n. After that, there will not be any other divisor, except the number n itself, if n is prime.
static IList<int> Factors(int num) {
var result = new List<int>();
// avoid scenarios like num=0, that would cause infinite loop
if (num <= 1) {
return result;
}
// testing with 2 outside the main loop, otherwise we would skip factor 3
// (2 * 2 > 3)
while (num % 2 == 0) {
result.Add(2);
num /= 2;
}
// only test primes until sqrt(num)
int i = 3;
while (i * i <= num) {
if (num % i == 0) {
result.Add(i);
num /= i;
} else {
i++;
}
}
// if not 1 here, num is prime
if (num > 1) {
result.Add(num);
}
return result;
}
I solved it
Here is code
static void lesson6()
{
Console.WriteLine("Enter n: ");
int n = Convert.ToInt32(Console.ReadLine());
int a = n;
List<int> arr = new List<int>();
for (int i = 2; i <= n; i++)
{
while (n % i == 0)
{
arr.Add(i);
n /= i;
}
}
Console.Write($"{a} = ");
int lastIndex = arr.Count - 1;
for (int i = 0; i < arr.Count; i++)
{
if (i == lastIndex)
{
Console.Write(arr[i]);
}
else
{
Console.Write(arr[i] + "*");
}
}
}
As pointed by derpirscher in the comment, there are several sources online with different approaches for integer factorization.
I recommend you to look for Trial Division algorithm, as it is the easier to understand, and is similar to your approach.
Based on the code you shared, there are some thing you should consider:
for (int i = 1; i <= n; i++)
{
if (n % i == 0)
{
arr.Add(i);
}
}
After finding that a prime is a divisor and appending to the list, you are going to the next number. However, a prime can figure many times in the factorization of a number. E.g: 12 -> { 2, 2, 3 }.
You need divide n by the prime and continue testing the until it is not a divisor anymore, then you can go test the next prime.
This way, your n is shrinking down each time you find a prime divisor, until it eventually become 1. Then you know you found all prime divisors.
Write a console app in C# to find an index i in an array that is the maximum number in the array.
If the maximum element in the array occurs several times, you need to display the minimum index.
If the array is empty, output -1.
Please let me know what is wrong in my code?
If I input the array a = { 1, 2, 46, 14, 64, 64 };, for instance, it returns 0, while it should be returning 4.
public static void Main()
{
double[] a = { 1, 9, 9, 8, 9, 2, 2 };
Console.WriteLine(MaxIndex(a));
}
public static double MaxIndex(double[] array)
{
var max = double.MinValue;
int maxInd = 0, maxCount = 0;
int minIndex = 0;
var min = double.MaxValue;
for (var i = 0; i < array.Length; i++)
{
if (min > array[i])
{
min = array[i];
minIndex = i;
}
if (max == array[i])
maxCount++;
if (max < array[i])
{
maxCount = 1;
max = array[i];
maxInd = i;
}
}
if (array.Length == 0)
return -1;
if (maxCount > 1)
return minIndex;
return maxInd;
}
Your calculation is correct, but you return the wrong variable minIndex instead of maxIndex. Don't do more than necessary in a method. This calculates also the min-index and the count how often it appears, then it does nothing with the results. Here is a compact version:
public static int MaxIndex(double[] array)
{
var max = double.MinValue;
int maxInd = -1;
for (int i = 0; i < array.Length; i++)
{
if (max < array[i])
{
max = array[i];
maxInd = i;
}
}
return maxInd;
}
It also sets maxInd = -1 which was part of your requirement. Since MatthewWatson had an objection regarding repeating double.MinValue in the array, here is an optimized version:
public static int MaxIndex(double[] array)
{
if(array.Length == 0)
return -1;
else if (array.Length == 1)
return 0;
double max = array[0];
int maxInd = 0;
for (int i = 1; i < array.Length; i++)
{
if (max < array[i])
{
max = array[i];
maxInd = i;
}
}
return maxInd;
}
If code-readability/maintainability is more important and you don't care of few milliseconds more or less, you could use LINQ (a version that enumerates only once):
int minIndexOfMaxVal = a.Select((num, index) => new {num, index})
.OrderByDescending(x => x.num)
.Select(x => x.index)
.DefaultIfEmpty(-1)
.First();
This works with every kind of sequence and types not only with arrays and doubles.
Try this simple line of code:
int[] numbers = { 1, 2, 3, 4, 5, 4, 3, 2, 1 };
var index = numbers.ToList().IndexOf(numbers.Max());
I think code is simple enough to be self-explanatory.
However, if you aim for performance, conversion to List could be omitted and you could wirte own code to find index of maximum number.
Or even simplier, proposed by #fubo:
Array.IndexOf(numbers, numbers.Max());
A lot of simplification is possible in this code:
int? maxVal = null; // null because of array of negative value;
int index = -1;
for (int i = 0; i < array.Length; i++)
{
int current = array[i];
if (!maxVal.HasValue || current > maxVal.Value)
{
maxVal = current ;
index = i;
}
}
Will get you the first index of the max value. If you just need a short code and Don't mind iterating twice a simple linq can do the trick
var index = array.ToList().IndexOf(array.Max());
It returns zero, because minIndex is indeed zero:
Change minIndex to maxIndex:
if (maxCount > 1) return maxIndex;
In order to compare doubles use the following code instead of ==:
if (Math.Abs(a-b)<double.Epsilon)
Your code in general is way too complicated with way too many variables for what it has to do.
You want the index of the first occurence of the highest value, so there isn't any real reason to store anything but a MaxValueIndex and a MaxValue.
All other variables should be local so garbage collection can dispose of them.
As for what is actually wrong with your code, I can't help out, since the variable names are confusing to me.
But here is some code that achieves the goal you want to achieve.
double[] a = { 1, 9, 9, 8, 9, 2, 2 };
var highestValueInArray = a.Max();
var arrayLength = a.Length;
for (int i = 0; i < arrayLength; i++)
{
if (a[i] == highestValueInArray)
{
Console.WriteLine(i);
break;
}
}
instead of manually calculating the max value, you can just use YourArray.Max() to get the highest value.
then just iterate through it like normal, but at the first occurence, you break; out of the loop and return the index it is at.
I'm pretty sure there's also a way to use FirstOrDefault in combination with IndexOf, but couldn't get that to work myself.
I have a
list<int> input = //from db of one million records
of 1 million records .
Although I know that using input.OrderByDescending().Skip(1).Take(1) will solve the problem but if I have 1 million records , it is not a best practice to use OrderBY.
In this scenario which solution is the best one when we have more records?
Scan the list while tracking the max, as well as max2 and you'll get O(N) versus O(N * log(N)) time complexity:
// Maximum value
int max = Math.Max(input[input.Count - 1], input[input.Count - 2]);
// Second greatest
int max2 = Math.Min(input[input.Count - 1], input[input.Count - 2]);
// i >= 0: Comparing with 0 is slightly faster then with Count
for (int i = input.Count - 3; i >= 0; --i) {
int v = input[i];
if (v >= max) {
max2 = max;
max = v;
}
else if (v > max2)
max2 = v;
}
Edit: In case duplicates should be ignored (see comments below) and thus the answer for [1, 2, 3, 4, 4, 4, 4] should be 3, not 4:
// Maximum value
int max = int.MinValue;
// Second greatest
int max2 = int.MinValue;
// i >= 0: Comparing with 0 is slightly faster then with Count
for (int i = input.Count - 1; i >= 0; --i) {
int v = input[i];
if (v > max) {
max2 = max;
max = v;
}
else if (v > max2 && v != max)
max2 = v;
}
If your list contains only distinct values, you can also do:
var max = input.Max();
input.Remove(max);
var result = input.Max();
You can do it by keeping track of the next max and iterating once through the list.
int max = Int32.MinValue;
int nextMax = Int32.MinValue;
for(int i=0; i<list.Count(); i++)
{
if(list[i] > max)
{
nextMax = max;
max = list[i];
}
else if(list[i] > nextMax)
{
nextMax = list[i];
}
}
This is a solution iterating just once and using LINQ and C# 7 tuples:
var input = Enumerable.Range(0, 101);
var topTwo = input.Aggregate((Biggest: Int32.MinValue, SecondBiggest: Int32.MinValue),
(acc, next) =>
{
if (next > acc.Biggest)
{
acc.SecondBiggest = acc.Biggest;
acc.Biggest = next;
}
if (next < acc.Biggest && next > acc.SecondBiggest)
acc.SecondBiggest = next;
return acc;
});
WriteLine(topTwo.SecondBiggest); // 99
I took a look at the following question from project euler:
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
What is the 10 001st prime number?
I tried to take the square root of the number and than find all the prime numbers below the square root of the number and then divide the number by all the square roots and see if there is 0 left each time. If the number is not divisible by all the primes under its square root its a prime number. I did this to lower the itterations the programm has to make. Here is what I have now, I am not sure why it isn't working. Anybody knows what i did wrong?
List<int> primeNumbers = new List<int>();
bool prime = true;
bool MainPrime = true;
int check = 1;
for (long i = 3; i < long.MaxValue; i++)
{
if ((i % 2) != 0)
{
int root = Convert.ToInt32(Math.Sqrt(i));
for (int j = 1; j < root; j++)
{
for (int k = 2; k < j; k++)
{
if ((j% k) == 0)
{
prime = false;
}
}
if (prime)
{
primeNumbers.Add(j);
}
prime = true;
}
}
foreach (var item in primeNumbers)
{
if ((i%item) == 0)
{
MainPrime = false;
}
}
primeNumbers.Clear();
if (MainPrime)
{
check++;
}
if (check == 10001)
{
Console.WriteLine(i);
break;
}
}
Console.ReadKey();
Several points:
When finding possible prime divisors, you need to check all numbers up to the square root included, so your condition j < root is incorrect.
You don't have to recalculate the primes again for every number. Keep the list as you go and add new primes to it.
As soon as you find a divisor, you can break out of the foreach loop.
Improved code:
List<long> primeNumbers = new List<long>() { 2 };
for (long i = 3; i < long.MaxValue; i += 2)
{
if(!primeNumbers.Any(p => (i % p) == 0))
{
primeNumbers.Add(i);
if (primeNumbers.Count == 10001)
{
Console.WriteLine(i);
break;
}
}
}
Gives 104743 as the 10001st prime.
What we can do is we can use SieveOfEratosthenes to make an bool array in which all the prime numbers value are set to be true than after that;
1.As we found any prime number increment the count with 1;
2.And as count get equal to 10001 we print its value and break through the loop.
Have a Look at code in C++ (I recommend you to learn SieveOfEratosthenes first)
#include <bits/stdc++.h>
using namespace std;
void SieveOfEratosthenes(long long unsigned n)
{
bool prime[n];
memset(prime, true, sizeof(prime)); //This is SieveOfEratosthenes
for (long long p = 2; p * p <= n; p++)
{
if (prime[p] == true)
{
for (long long i = p * p; i <= n; i += p)
prime[i] = false;
}
}
long long count=0; //initializing count as 0;
for (long long p = 2; p <= n; p++) //running the loop form 2 to n
{
if (prime[p]) //we have bool array in which all prime number set to true using sieve
count++; //increment the count because we found a prime number
if(count==10001) // and as count reaches to 10001 we found our number
{
cout<<p;break;} // print the answer and also break form the loop
}
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long unsigned n=999999;
SieveOfEratosthenes(n); //pass the value of n in sieve function
return 0;
}
Try this one out using python
sp=2
cnt = 1
while cnt <= 10001:
primeflag = 0
for j in range(2,sp):
if(sp%j == 0):
primeflag = 1
break;
if(primeflag == 1):
pass
else:
print(cnt ,sp)
cnt = cnt +1
sp =sp+1
#which Gives
#10001 104743
Project Euler challenge 23 states this:
A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.
A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.
As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.
Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.
So I've been trying to get this working, however I keep getting back an incorrect result, I'm not sure where this is going wrong in the code though I have:
static void Main(string[] args)
{
List<int> abundantNums = Enumerable.Range(12, 1000000).Where(i => isAbundant(i)).ToList();
abundantNums = abundantNums.Distinct().ToList();
var boolArr = new bool[28124];
for (int i = 0; i < abundantNums.Count; ++i)
{
for (int j = i; j < abundantNums.Count; ++j)
{
var sum = abundantNums[i] + abundantNums[j];
if (sum < 28124) boolArr[sum] = true;
else break;
}
}
var total = 0;
for (int i = 0; i < boolArr.Length; i++)
{
if (boolArr[i] == false)
{
total += i;
}
}
Console.WriteLine(total);
Console.ReadKey();
}
static bool isAbundant(int num)
{
if (getFactors(num).Sum() > num)
{
return true;
}
else
{
return false;
}
}
And then to find the factors of a number I have:
static List<int> getFactors(int num)
{
List<int> factors = new List<int>();
Stopwatch watch = Stopwatch.StartNew();
for (int i=1; i < Math.Sqrt(num) + 1; i++)
{
if (num % i == 0)
{
factors.Add(i);
if (num / i != i)
{
factors.Add(num / i);
}
}
}
watch.Stop();
factors.Remove(num);
return factors;
}
Now I've been at this for a day or two and as far as I can tell this should be doing the trick, anyone wiser than I able to point out my failings?
The problem is your getFactors loop. Change:
for (int i=1; i < Math.Sqrt(num) + 1; i++)
to
for (int i=1; i <= Math.Sqrt(num); i++)
And it should work. I'll let you try and understand why :-)