CareerCup

array.sort();

public int[] moveSmallNumbersToStartAndBigNumbersToEnd(int[] array) {
    return sort(array);

}

When do I start?

lmfao

public class ArraySortingExample {

	/**
	 * Given an array of numbers, move small numbers to start and big numbers to
	 * end of array
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int array[] = { 25, 3, 14, 1 };
		for (int i = 0; i < array.length; i++) {
			for (int j = i + 1; j < array.length; j++) {
				if (array[i] > array[j]) {
					int temp = array[i];
					array[i] = array[j];
					array[j] = temp;
				}
			}
			System.out.print(array[i] + " ");
		}
	}
}

Пузырьковая сортировка. Сложнааааааа...
The "bubble sort". Too hardddddd XD

//#include<iostream>
//using namespace std;
//
////two numbers m and n(1 <= m <= n <= 1000000000, n - m <= 100000) separated by a space.
////
////p such that m <= p <= n, one number per line, test cases separated by an empty line.
//
//bool isPrime(int number)
//{
//	for (int i = 2; i <= ceil(sqrt(number)); i++)
//	{
//		if (number%i != 0)
//		{
//			return false;
//		}
//	}
//	return true;
//}
//int main()
//{
//	int numberOfCases, m, n;
//	numberOfCases = m = n = 0;
//
//	cin >> numberOfCases;
//
//	while (numberOfCases--)
//	{
//		cin >> m;
//		cin >> n;
//
//		for (int i = m; i<n; i++)
//		{
//			if (isPrime(i) == true)
//				cout << i << " ";
//		}
//		cout << "\n\n";
//	}
//
//	return 0;
//}

#include<iostream>
#include<algorithm>
#include<array>

int main()
{
	std::array<int,10> arr = {1,7,4,62,9,25,54,89,31,3};
	
	std::sort(arr.begin(), arr.end());
	
	for (int i = 0; i < 10; i++)
	{
		std::cout << arr[i] << " ";
	}

	return 0;
}

var arr=[23,58,12,44];


for(var i=0; i<arr.length; i++)
{
for(var j=i+1; j<arr.length; j++)
{
if(arr[j]<arr[i])
{
var tmp=arr[i];

arr[i]=arr[j];
arr[j]=tmp;
}
}
}
console.log(arr)

var arr=[23,58,12,44];


for(var i=0; i<arr.length; i++)
{
for(var j=i+1; j<arr.length; j++)
{
if(arr[j]<arr[i])
{
var tmp=arr[i];

arr[i]=arr[j];
arr[j]=tmp;
}
}
}
console.log(arr)

lol, I like 2 answers from above, array.sort() and LMFAO
array.sort() can have O(n log n) time
But if we just find min and max of the list it can be found in O (n)

remove them from the list and insert them at first and the last position respectively

def doWhatever(arglist):
    
    min = (0, arglist[0])
    max = (0, arglist[0])
    for i in range(1,len(arglist)):
        
        if arglist[i] < min[1]:
            min = (i, arglist[i])

        if arglist[i] > max[1]:
            max = (i, arglist[i])


    del arglist[min[0]]
    del arglist[max[0]]
    
    arglist.insert(0, min[1])
    arglist.append(max[1])
    
    return arglist

#include<stdio.h>
main()
{
int a[10],i,j,n,temp;
printf("\n enter array size ");
scanf("%d",&n);
printf("Enter %d numbers\n",n);
for(i=0;i<n;i++)
{
scanf("%d",&a[i])
}
for(i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
if(a[i]>a[j])
{
temp=a[i];
a[i]=a[j];
a[j]=temp;
}
}
printf("%d\n",a[i]);
}
return 0;
}

def k_least_num():

a1 = input().split(',')
a2 = input().split(',')

k = int(input())
s = []

while len(s)<k:
if (int(a1[0]) < int(a2[0])):
if a1[0] not in s:
s.append(a1[0])
del a1[0]
else:
del a1[0]
else:
if a2[0] not in s:
s.append(a2[0])
del a2[0]
else:
del a2[0]
print("final unique least K numbers list-",s)

def k_least_num():

a1 = input().split(',')
a2 = input().split(',')

k = int(input())
s = []

while len(s)<k:
if (int(a1[0]) < int(a2[0])):
if a1[0] not in s:
s.append(a1[0])
del a1[0]
else:
del a1[0]
else:
if a2[0] not in s:
s.append(a2[0])
del a2[0]
else:
del a2[0]
print("final unique least K numbers list-",s)