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How can I get N smallest number in an array?

I'm trying to get the N smallest numbers (given by the user) in an array without using methods like sort()... in the last step, I keep getting only the smallest values and 0 for the rest.. where's the problem?
//1- Scanner to take inputs
Scanner input = new Scanner(System.in);
//2- Take the array size as input and store it in "sizeOfArr" var
System.out.print("Enter the array size: ");
int sizeOfArr = input.nextInt();
//3- Assign the input as an array size
int array[] = new int[sizeOfArr];
//4- Looping on the array and update its values by inputs taken from the user
for(int i = 0; i < array.length; i++) {
System.out.print("Enter "+ (i+1) + "-st element: ");
array[i] = input.nextInt();
}
//5- Print out the array after convert it to String
System.out.println(Arrays.toString(array));
//6- Find the smallest element in the array and print it
int minVal = array[0];
for(int i = 0; i < array.length; i++) {
if (array[i] < minVal) {
minVal = array[i];
}
}
// System.out.println(minVal);
//7- Find the (n) smallest of number defined by the user
System.out.print("Enter the number of smallest numbers do you want: ");
int n = input.nextInt();
//8- new array to store n smallest numbers
int smallestNums[] = new int[n];
//9- trying to loop on the original array n times
int counter;
for(int i = 0; i < n ; i++) {
//10- trying to loop on the original array to store the smallest values in smallestNum[] array.
for(int j = 0; j < array.length; j++) {
smallestNums[i] = minVal;
}
if(smallestNums[i] == smallestNums[i]) {
break;
}
}
System.out.println(Arrays.toString(smallestNums));

Here is one way. Just do a partial sort with the outer loop limit equal to the number of items required. This is variant of the selection sort. This example, varies n in the outer list for demo purposes.
int[] array = { 10, 1, 5, 8, 7, 6, 3 };
for (int n = 1; n <= array.length; n++) {
int[] smallest = getNSmallest(n, array);
System.out.printf("smallest %2d = %s%n", n,
Arrays.toString(smallest));
}
prints
smallest 1 = [1]
smallest 2 = [1, 3]
smallest 3 = [1, 3, 5]
smallest 4 = [1, 3, 5, 6]
smallest 5 = [1, 3, 5, 6, 7]
smallest 6 = [1, 3, 5, 6, 7, 8]
smallest 7 = [1, 3, 5, 6, 7, 8, 10]
Here is the method. The first thing to do is copy the array so the
original is preserved. Then just do the sort and return array of smallest elements.
public static int[] getNSmallest(int n, int[] arr) {
int[] ar = Arrays.copyOf(arr, arr.length);
int[] smallest = new int[n];
for (int i = 0; i < n; i++) {
for (int k = i + 1; k < ar.length; k++) {
if (ar[i] > ar[k]) {
int t = ar[i];
ar[i] = ar[k];
ar[k] = t;
}
}
smallest[i] = ar[i];
}
return smallest;
}

For this task, you don't have to sort the whole array. Only a group of N elements has to be sorted. I.e. only a partial sorting is required.
Below, I've provided two implementations for this problem. The first utilizes only plane arrays and loops, the second makes use of the PriorytyQueue.
The first solution maintains a variable pos which denotes the position in the result array which isn't assigned yet. Note that the default value for an element of the int[] is 0. It's important to be able to distinguish between the default value and a zero-element from the given array. Hence we can't rely on the values and have to track the number of elements that are assigned.
Every element of the source array gets compared with all the elements of the result array that are already assigned. The new element will be added to the result array in two cases:
nested loop has reached an unoccupied position pos in the result array;
an element in the result array that is greater than the next element from the given array has been found.
In the first case, a new element gets assigned the position denoted by pos. In the second case, a new element has to be inserted
nested loop iterates over the given array at the current position i and all elements must be shifted to the right. That's what the method shiftElements() does.
The First solution - Arrays & Loops
public static int[] getSmallest(int[] arr, int limit) {
int[] result = new int[Math.min(limit, arr.length)];
int pos = 0;
for (int next: arr) {
for (int i = 0; i < Math.min(pos + 1, result.length); i++) {
if (i == pos) result[i] = next;
else if (result[i] > next) {
shiftElements(result, next, i, Math.min(pos + 1, result.length));
break;
}
}
pos++;
}
return result;
}
private static void shiftElements(int[] arr, int val, int start, int end) {
int move = arr[start];
arr[start] = val;
for (int i = start + 1; i < end; i++) {
int temp = arr[i];
arr[i] = move;
move = temp;
}
}
Maybe you'll be more comfortable with the first version, but if you are somehow familiar with the Collections framework, then it's a good time to get acquainted with PriorytyQueue. In the nutshell, this collection is backed by an array and maintains its element in the same order as they were added, but when an element is being deleted collection retrieves the smallest one according to the natural order or based on the Comparator, which can be provided while instantiating the PriorytyQueue. It uses a sorting algorithm that is called a heapsort which allows removing a single element in O(log N) time.
The Second solution - PriorytyQueue
public static int[] getSmallestWithPriorityQueue(int[] arr, int limit) {
Queue<Integer> queue = new PriorityQueue<>();
populateQueue(queue, arr);
int[] result = new int[Math.min(limit, arr.length)];
for (int i = 0; i < result.length; i++) {
result[i] = queue.remove();
}
return result;
}
private static void populateQueue(Queue<Integer> queue, int[] arr) {
for (int next: arr) {
queue.add(next);
}
}
main & utility-method to generate an array
public static void main(String[] args) {
int[] source = generateArr(100, 10);
System.out.println("source : " + Arrays.toString(source));
int[] result1 = getSmallest(source, 3);
System.out.println("result(Arrays & Loops) : " + Arrays.toString(result1));
int[] result2 = getSmallestWithPriorityQueue(source, 3);
System.out.println("result(PriorityQueue) : " + Arrays.toString(result2));
}
public static int[] generateArr(int maxVal, int limit) {
Random random = new Random();
return IntStream.generate(() -> random.nextInt(maxVal + 1))
.limit(limit)
.toArray();
}
output
source : [61, 67, 78, 53, 74, 51, 50, 83, 59, 21]
result(Arrays & Loops) : [21, 50, 51]
result(PriorityQueue) : [21, 50, 51]

Randomized select allows to find k-th ranked element in linear time on average.
It alters the input order, so practically, it makes sense to just sort and return k-th element of the sorted array. Especially if there are several such calls on the given input array.

How to find the even numbers of a list in Java [duplicate]

Code written below is correct, but I want to shorten this code.
Write a program in java to enter 10 numbers in Single dimensional array and arrange them in such a way that all even numbers are followed by all odd numbers.
int a[] = new int[6];
int b[] = new int[6];
int i, j;
int k = 0;
System.out.println("enter array");
for (i = 0; i < 6; i++) {
a[i] = sc.nextInt();
}
for (j = 0; j < 6; j++) {
if (a[j] % 2 == 0) {
b[k] = a[j];
k++;
}
}
for (j = 0; j < 6; j++) {
if (a[j] % 2 != 0) {
b[k] = a[j];
k++;
}
}
System.out.println("out-put");
for (i = 0; i < 6; i++) {
System.out.println(b[i]);
}
Can I arrange the even numbers and the odd numbers in a single for loop instead of two for loop? I am using two for loop to transfer the even and the odd numbers into b[] array. Please shorten code. One for loop traverse for checking even number and second for odd numbers.

Here is a simple program for you.
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
import java.util.Scanner;
/**
*
* #author Momir Sarac
*/
public class GroupByEvenAndOddNumbers {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
// create a collection
List<Integer> listOfNumbers = new ArrayList<>();
// do code within a loop for 10 times
for(int i=0;i<10;i++)
{
//print to screen this text
System.out.println("Input your number:");
//get next input integer
int number = scanner.nextInt();
// add it to collection
listOfNumbers.add(number);
}
// sort this collection, list of numbers
// convert all numbers(positive and negative ) within to 0 or 1 depending whether or not they are even or odd and sort them accordignaly.
Collections.sort(listOfNumbers, Comparator.comparingInt(n -> Math.floorMod(n, 2)));
//print sorted collection
System.out.println("Ordered list ..." + listOfNumbers);
}
}

In this version, it copies the even to the start, and the odd to the end.
static int[] sortEvenOdd(int... nums) {
int even = 0, odd = nums.length, ret[] = new int[nums.length];
for (int num : nums)
if (num % 2 == 0)
ret[even++] = num;
else
ret[--odd] = num;
return ret;
}
public static void main(String[] args) {
int[] arr = {1, 3, 2, 4, 7, 6, 9, 10};
int[] sorted = sortEvenOdd(arr);
System.out.println(Arrays.toString(sorted));
}
prints
[2, 4, 6, 10, 9, 7, 3, 1]

This Code will help you to segregate Even and Odd numbers.
// java code to segregate even odd
// numbers in an array
public class GFG {
// Function to segregate even
// odd numbers
static void arrayEvenAndOdd(
int arr[], int n)
{
int i = -1, j = 0;
while (j != n) {
if (arr[j] % 2 == 0)
{
i++;
// Swapping even and
// odd numbers
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
j++;
}
// Printing segregated array
for (int k = 0; k < n; k++)
System.out.print(arr[k] + " ");
}
// Driver code
public static void main(String args[])
{
int arr[] = { 1, 3, 2, 4, 7,
6, 9, 10 };
int n = arr.length;
arrayEvenAndOdd(arr, n);
}
}

As you don't have any requirements that the even and odd numbers itself have to be ordered in their respectively half of the array you can just assign them to their associated array part while entering them.
Therefore you just have to use two "counter" variables one for the left which starts at zero and is incremented and one for the right which starts at your array length minus one and is decremented. Then you can add your numbers, checking if one is even add assign it with your left counter post incremented and if one is odd assign it with your right counter post decremented. Do this within a loop, until your left counter is bigger than your right counter.
I created a simple example where I did not check for NumberFormatException when parsing the String to an int:
import java.util.Arrays;
import java.util.Scanner;
public class SortedArrayInput {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.print("Enter length of array: ");
final int arrayLength = Integer.parseInt(scanner.nextLine());
int intArray[] = new int[arrayLength];
for (int l = 0, r = arrayLength - 1; l <= r; ) {
System.out.print("Enter new array value: ");
int v = Integer.parseInt(scanner.nextLine());
intArray[v % 2 == 0 ? l++ : r--] = v;
}
System.out.println("Output: " + Arrays.toString(intArray));
}
}
Sample input/output:
Enter length of array: 6
Enter new array value: 1
Enter new array value: 2
Enter new array value: 3
Enter new array value: 4
Enter new array value: 5
Enter new array value: 6
Output: [2, 4, 6, 5, 3, 1]

I recommend reading up on streams, they will make collection processing a lot easier for you
List<Integer> numbers = new ArrayList<>();
numbers.add(1);
numbers.add(2);
numbers.add(3);
numbers.add(4);
numbers.add(5);
numbers.add(6);
numbers.add(7);
numbers.add(8);
numbers.add(9);
numbers.add(0);
//this way you simply traverse the numbers twice and output the needed ones
System.out.println(numbers.stream()
.filter(x->x%2==0)
.collect(Collectors.toList()));
System.out.println(numbers.stream()
.filter(x->x%2==1)
.collect(Collectors.toList()));
//this way you can have the numbers in two collections
numbers.forEach(x-> x%2==0? addItToEvenCollection : addItToOddCollection);
//this way you will have a map at the end. The boolean will tell you if the numbers are odd or even,
// and the list contains the numbers, in order of apparition in the initial list
numbers.stream().collect(Collectors.groupingBy(x->x%2==0));

A performant way to check if a number is even, is to use
if ( (x & 1) == 0 )

How to print out even-numbered indexes for arrays in Java?

I'm supposed to write a program using for loops that print out the even indexes of my array. For example, if I create an array that has 10 numbers, it will have indexes from 0-9 so in that case I would print out the numbers at index 2, 4, 6 and 8. This is what I wrote so far but it doesn't work. Please note that I am not trying to print out the even numbers of the array. All I want are the even indexes.
Example I enter the following array: 3,7,5,5,5,7,7,9,9,3
Program output:
5 // (the number at index 2)
5 // (the number at index 4)
7 // (the number at index 6)
9 // (the number at index 8)
My Code:
public class Arrayevenindex
{
public static void main(String[] args)
{
int number; // variable that will represent how many elements the user wants the array to have
Scanner key = new Scanner(System.in);
System.out.println(" How many elements would you like your array to have");
number = key.nextInt();
int [] array = new int [number];
// let the user enter the values of the array.
for (int index = 0; index < number; index ++)
{
System.out.print(" Value" + (index+1) + " :");
array[index] = key.nextInt();
}
// Print out the even indexes
System.out.println("/nI am now going to print out the even indexes");
for (int index = 0; index < array.length; index ++)
{
if (array[number+1]%2==0)
System.out.print(array[number]);
}
}
}

You can just change your for loop and get rid of the inner IF...
for( int index = 0; index < array.length; index += 2) {
System.out.println(array[index]);
}

Just absolutely same thing using java 8 Stream API
Integer[] ints = {0,1,2,3,4,5,6,7,8,9};
IntStream.range(0, ints.length).filter(i -> i % 2 == 0).forEach(i -> System.out.println(ints[i]));

I assume this would be sufficient
// For loop to search array
for (int i = 0; i < array.length; i++) {
// If to validate that the index is divisible by 2
if (i % 2 == 0) {
System.out.print(array[i]);
}
}

This is what I did and it works:also I am not printing out index[0] because technically its not even thats why I started the for loop at 2. Your post did help me a lot. I also thank everyone else as well that took the time to post an answer.
import java.util.Scanner;
public class Arrayevenindex
{
public static void main(String[] args)
{
int number; // variable that will represent how many elements the user wants the array to have
Scanner key = new Scanner(System.in);
System.out.println(" How many elements would you like your array to have");
number = key.nextInt();
int [] array = new int [number];
// let the user enter the values of the array.
for ( int index = 0; index < number; index ++)
{
System.out.print(" Value" + (index+1) + " :");
array[index] = key.nextInt();
}
// Print out the even indexes
System.out.println("/nI am now going to print out the even indexes");
for ( int index = 2; index < array.length; index +=2)
{
System.out.print(array[index] + " ");
}
}
}

Java cut the sticks

I am very new to Java and I was trying to solve this problem on Hackerrank:
Here's the task:
https://www.hackerrank.com/challenges/cut-the-sticks
You are given N sticks, where the length of each stick is a positive
integer. A cut operation is performed on the sticks such that all of
them are reduced by the length of the smallest stick.
Suppose we have six sticks of the following lengths:
5 4 4 2 2 8
Then, in one cut operation we make a cut of length 2 from each of the six
sticks. For the next cut operation four sticks are left (of non-zero length), > whose lengths are the following:
3 2 2 6
The above step is repeated until no sticks are left.
Given the length of N sticks, print the number of sticks that are left before > each subsequent cut operations.
Note: For each cut operation, you have to recalcuate the length of smallest
sticks (excluding zero-length sticks).
Here is my attempt at it, but it doesnt seem to be working. The output gets stuck in while loop (4 gets printed out infinitely)
import java.io.*;
import java.util.*;
public class Solution {
private static int findMin (int[] A)
{
int min = A[0];
for (int i =0; i<A.length; i++)
{
if (A[i] < min)
{
min = A[i];
}
}
return min;
}
private static int countNonZeros (int[] A)
{
int zeros = 0;
for (int i =0; i<A.length; i++)
{
if (A[i] == 0)
{
zeros++;
}
}
int nonZeros = A.length - zeros;
return nonZeros;
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int[] A = new int[n];
for (int i=0; i<n; i++)
{
A[i] = scanner.nextInt();
}
int nums = countNonZeros(A);
while (nums > 0)
{
int mins = findMin(A);
for (int j = 0; j<A.length; j++)
{
A[j]=A[j]-mins;
}
nums = countNonZeros(A);
System.out.println(nums);
}
}
}
Any help is appreciated
(PS I know I can just look the solution up somewhere, but I want to know why my code isn't working)

The problem that you have is that your findMin is not excluding zero-length elements, so once you have a zero that will be the min, and as a result an iteration of the while loop will be the same as the previous iteration, having subtracted 0 from each of the elements of A.

finding longest sequence of consecutive numbers

Problem H (Longest Natural Successors):
Two consecutive integers are natural successors if the second is the successor of the first in the sequence of natural numbers (1 and 2 are natural successors). Write a program that reads a number N followed by N integers, and then prints the length of the longest sequence of consecutive natural successors.
Example:
Input 7 2 3 5 6 7 9 10 Output 3 this is my code so far and i have no idea why it does not work
import java.util.Scanner;
public class Conse {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int x = scan.nextInt();
int[] array = new int[x];
for (int i = 0; i < array.length; i++) {
array[i] = scan.nextInt();
}
System.out.println(array(array));
}
public static int array(int[] array) {
int count = 0, temp = 0;
for (int i = 0; i < array.length; i++) {
count = 0;
for (int j = i, k = i + 1; j < array.length - 1; j++, k++) {
if (Math.abs(array[j] - array[k]) == 1) {
count++;
} else {
if (temp <= count) {
temp = count;
}
break;
}
}
}
return temp + 1;
}
}

Why two loops? What about
public static int array(final int[] array) {
int lastNo = -100;
int maxConsecutiveNumbers = 0;
int currentConsecutiveNumbers = 0;
for (int i = 0; i < array.length; i++) {
if (array[i] == lastNo + 1) {
currentConsecutiveNumbers++;
maxConsecutiveNumbers = Math.max(maxConsecutiveNumbers,
currentConsecutiveNumbers);
} else {
currentConsecutiveNumbers = 1;
}
lastNo = array[i];
}
return Math.max(maxConsecutiveNumbers, currentConsecutiveNumbers);
}

This seems to work:
public static int longestConsecutive(int[] array) {
int longest = 0;
// For each possible start
for (int i = 0; i < array.length; i++) {
// Count consecutive.
for (int j = i + 1; j < array.length; j++) {
// This one consecutive to last?
if (Math.abs(array[j] - array[j - 1]) == 1) {
// Is it longer?
if (j - i > longest) {
// Yup! Remember it.
longest = j - i;
}
} else {
// Start again.
break;
}
}
}
return longest + 1;
}
public void test() {
int[] a = new int[]{7, 2, 3, 5, 6, 7, 9, 10};
System.out.println("Longest: " + Arrays.toString(a) + "=" + longestConsecutive(a));
}
prints
Longest: [7, 2, 3, 5, 6, 7, 9, 10]=3

Since your question has "Problem H" associated with it, I'm assuming you are just learning. Simpler is always better, so it usually pays to break it down into "what has to be done" before starting on a particular road by writing code that approaches the problem with "how can this be done."
In this case, you may be over-complicating things with arrays. A number is a natural successor if it is one greater than the previous number. If this is true, increment the count of the current sequence. If not, we're starting a new sequence. If the current sequence length is greater than the maximum sequence length we've seen, set the max sequence length to the current sequence length. No arrays needed - you only need to compare two numbers (current and last numbers read).
For example:
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int N = scan.nextInt();
int maxSequenceLen = 0; // longest sequence ever
int curSequenceLen = 0; // when starting new sequence, reset to 1 (count the reset #)
int last = 0;
for(int i = 0; i < N; i++) {
int cur = scan.nextInt();
if ((last+1) == cur){
++curSequenceLen;
}
else{
curSequenceLen = 1;
}
if (curSequenceLen > maxSequenceLen){
maxSequenceLen = curSequenceLen;
}
last = cur;
}
System.out.println(maxSequenceLen);
Caveat: I'm answering this on a computer that does not have my Java development environment on it, so the code is untested.

I'm not sure I understand this question correctly. The answer's written here assumes that the the natural successors occur contiguously. But if this is not the same then the solution here might not give the correct answer.
Suppose instead of [7 2 3 5 6 7 9 10] the input was [7 2 6 3 7 5 6 9 10] then the answer becomes 2 while the natural successor [5 6 7] is present in the array.
If the input is not sorted we'll have to use a different approach. Like using HashSet
Load the entire array into a HashSet which removes duplicates.
Pick the first value from the HashSet and assigned it to start and end and remove it from the set.
Now decrements start and check if it is present in the HashSet and continue till a particular value for start is not present int the HashSetwhile removing the value being searched from the set.
Do the same for end except that you will have to increase the value of end for each iteration.
We now have to continuous range from start to end present in the set and whose range is current_Max = end - start + 1
In each iteration we keep track of this current_Max to arrive at the longest natural successor for the entire array.
And since HashSet supports Add, Remove, Update in O(1) time. This algorithm will run in O(n) time, where n is the length of the input array.
The code for this approach in C# can be found here