I need to get a minimum number that I cant get by adding different numbers of an array. Basically if I have these numbers:1,1,1,5; I can get 1,2,3,5,6... but I cant get 4 so that is the number I am looking for. Now this is my code:
import java.util.Scanner;
public class Broj_6 {
public static void main(String[] args) {
Scanner unos = new Scanner(System.in);
int k;
int n = unos.nextInt();
int niz []= new int [n];
for(int i = 0;i<n;i++){
niz[i]=unos.nextInt();
}
BubbleSort(niz);
for(int i = 0;i<n;i++){
System.out.print(niz[i] + " ");
}
for(int br = 1;br<=10000;br++){
for(k = 1;k<n;k++){
if(niz[k]>br){
break;
}
}
int podniz [] = new int [k];
for(int i=0;i<podniz.length;i++){
niz[i] = podniz[i];
}
//This is where I will need my logic to go
}
}
static void BubbleSort (int [] niz){
int pom;
for(int i = 0;i<niz.length-1;i++){
for(int j = 0;j<niz.length-1-i;j++){
if(niz[j]>niz[j+1]){
pom = niz[j];
niz[j] = niz[j+1];
niz[j+1] = pom;
}
}
}
}
}
So the code goes by testing each number individually from 1 to 100000 and makes a subarray of all numbers given that are less than the number itself. Now here is the problem,I dont know how to mix and match the numbers in the subarray so it can get(or not get) the desired number. When every combination is tested and there is no desired number,I will break; the loop and print i. Just to clarify,I can only use addition,and each number can only go in once
You can achieve this as below:
Use two nested loops, like below to calculate the sum of different numbers:
List<Integer> additionList = new ArrayList<Integer>();
int []inputNumbers = .... // Logic to read inputs
for(int _firstIndex = 0; _firstIndex < totalInputs; _firstIndex++){
for(int _secondIndex = _firstIndex + 1; _secondIndex < totalInputs; _secondIndex++){
additionList.add(inputNumbers[_firstIndex]); // only because you have 1 in the sample output
additionList.add(inputNumbers[_firstIndex] + inputNumbers[_secondIndex ]);
}
}
Then sort additionList and look for any missing entry. The first missing entry will be your answer,
Sorting the whole array and then finding sum of all subarrays does solve the problem, but is costly: O(2n^2) ~ O(n^2).
More efficient way to solve this will be Kadane's Algorithm: http://en.wikipedia.org/wiki/Maximum_subarray_problem
What the algo does:
Start from first element and increase the array size (sub array) till you reach the sum you're desiring.
my_num = 1;
while(true){
if(sum_subarray) > my_num){
current position = new subarray;
}
and this subarray concept is calculated through Kadane's approach:
def sum_subarray(A):
sum_ending_here = sum_so_far = 0
for x in A:
sum_ending_here = max(0, max_ending_here + x)
sum_so_far = max(sum_so_far, sum_ending_here)
return sum_so_far
I couldn't solve the problem completely. 'my_num' mentioned here needs to be incremented from 1, and break when my_num > max_sum. I hope someone can add to it and make it compilable.
Note:
This will also take care if negative elements are present in array.
Related
im trying to find all possible sums from a list.
The list consist of ints that are user inputed, and the number of inputs are decided by the user aswell( see code).
What i want is to check all possible sums without having a target.
The reason i dont want a target is because the program is then later suposed to chose one of these sums which is closest to 1000.
So if i was to pick 1000 as target i wouldnt be able to get say 1001.
Example input from user:
5 // user chose 5 numbers.
500,400,300,50,60 // numbers chosen by user.
Output would then be:
1010 // because 500+400+60+50= 1010 and closest to 1000.
Next example could be:
3 // user chose 3 numbers.
1,2,3 // numbers chosen by user.
Output would then be:
6 // because 1+2+3 = 6.
So back to my original question, how do is this done? Everytime i search "all possible sums of a list of ints" or simular i get with a target and it dosent work in this example.
public static void main(String[] args) throws Exception {
int bästaVikt;
int räknare = 0;
Scanner sc = new Scanner(System.in);
ArrayList<Integer> mylist = new ArrayList<Integer>();
int s;
s = sc.nextInt();
int ans = 0;
for(int i = 1; i <= s; i++) {
mylist.add(sc.nextInt());
}
for(int i = 0; i < mylist.size(); i++) {
Collections.sort(mylist);
System.out.print(mylist.get(i));
System.out.print(" ");
}
}
Half actual code half pseudo code, hope it helps.
input // the array that holds all numbers than the user input
int n = input.length();
ArrayList<Integer> combinations = new ArrayList<Integer>();
int totalSum = input.sum(); // sum of all the values the user input
ArrayList<Integer> allSums = new ArrayList<Ingeter>();
for(int i = 0; i < n; i++){
combinations = combine(i) // find all possible combinations of i integers in the input array; you can surely find code for retrieving combinations in another thread
foreach combination in combinations{
allSums.add(totalSum - combination.sum())
}
combinations = new ArrayList<Integer>(); // clear the combinations arraylist
}
return allSums; //all possible sums
minDistanceToNumber = absolute(allSums.get(0) - 1000); // define a function "int absolute(int value)" which will multiply value by -1 if it is less than 0 and return it or just return it if it is bigger than 0
minNumber = allSums.get(0); // this will hold the number which has the minimum distance to 1000, as in the example above
for each sum in allSums
if(absolute(sum - 1000) < minDistanceToNumber){
minDistanceToNumber = absolute(sum - 1000);
minNumber = sum;
}
}
return minNumber;
I want to find all possible binary permutations with a given number of ones in Java:
x is the desired number of ones in each sequence
n is the desired length of each sequence
For an example:
x=2, n=4
Output: 1100, 0011, 1010, 1001, 0101, 0110
I'm searching for an elegant and fast way to do this. Can you help me?
I've tested eboix solution in Print list of binary permutations but it is unfortunately too slow because the algorithm in this example is searching for all 2^n binary permutations.
I want to find sequences with a length of 50 or 100.
First of all, you're missing 0110 as an output case.
It's fairly intuitive that there are n choose x possibilities. You're finding all valid arrangements of x identical items among n total slots. So you can find the total number of sequences in O(1).
As a hint, try simply finding all permutations of the bitstring consisting of x ones followed n - x zeros.
To specifically address the problem, try creating a recursive algorithm that decides at every ith iteration to either include 1 or 0. If 1 is included, you need to decrement the count of 1's available for the rest of the string.
Actually, there may be an elegant way, but no fast way to do this. The number of string permutations is given by the binomial coefficient (see https://en.wikipedia.org/wiki/Binomial_coefficient). For example, x=10, n= 50 gives over 10 million different strings.
Here is just a basic version that will generate your desired output. Please work on it to make it more accurate/efficient -
This will not generate all the combinations, but you will get the idea of how to do it. Off course, for all the possible combinations generated by this, you will have to generate all the other possible combinations.
public class Test {
static int iter = 0;
public static void main(String args[]){
int n = 50;
int x = 5;
byte[] perms = new byte[n];
for(int i=0; i<x; i++){
perms[i] = 1;
}
print(perms);
for(int j=x-1; j>=0; j--){
for(int i=1; i<(n/2-j); i++){
iter++;
swap(perms, j, i);
}
}
}
public static void swap(byte[] perms, int pos, int by){
byte val = perms[pos+by];
perms[pos+by] = perms[pos];
perms[pos] = val;
print(perms);
val = perms[pos+by];
perms[pos+by] = perms[pos];
perms[pos] = val;
}
public static void print(byte[] perms){
System.out.println("iter = "+iter);
for(int i=0; i<perms.length; i++){
System.out.print(perms[i]);
}
System.out.println();
for(int i=perms.length-1; i>=0; i--){
System.out.print(perms[i]);
}
System.out.println();
}
}
Another inspiration for you. A dirty version which works. It allocates extra array space (you should adjust size) and uses String Set at the end to remove duplicates.
public static void main(String[] args) {
int x = 2;
int n = 4;
Set<BigInteger> result = new LinkedHashSet<>();
for (int j = x; j > 0; j--) {
Set<BigInteger> a = new LinkedHashSet<>();
for (int i = 0; i < n - j + 1; i++) {
if (j == x) {
a.add(BigInteger.ZERO.flipBit(i));
} else {
for (BigInteger num : result) {
if (num != null && !num.testBit(i) && (i >= (n - j) || num.getLowestSetBit() >= i-1))
a.add(num.setBit(i));
}
}
}
result = a;
}
String zeros = new String(new char[n]).replace("\0", "0");
for (BigInteger i : result) {
String binary = i.toString(2);
System.out.println(zeros.substring(0, n - binary.length()) + binary);
}
}
EDIT: changed the primitives version to use BigInteger instead to support larger n,x values.
Java Arrays: Finding Unique Numbers In A Group of 10 Inputted Numbers
I have a problem that I have looked into Doestovsky's question but from his question, I need to know on how to make the part on finding duplicates into a function of it's own:
java.util.Scanner input = new java.util.Scanner(System.in);
int[] numbers = new int[10];
boolean[] usedBefore = new boolean[10];
// Insert all numbers
for (int i = 0; i < numbers.length; i++) {
// Read number from console
numbers[i] = input.nextInt();
// Check if number was inserted before
usedBefore[i] = false;
for(int k = 0; k < i; k++) {
if(numbers[k] == numbers[i]) {
usedBefore[i] = true;
break;
}
}
}
// Print all numbers that were not inserted before
for(int j = 0; j < numbers.length; j++) {
if(!usedBefore[i]) {
System.out.print(String.valueOf(numbers[j])+" ");
}
}
I have tried this part of this code and it worked but I need this to take the part that find duplicates into a function of it's own that is powered by arrays.
Credits to ThimoKl for creating this code.
Let's try something else, using Treeset, and let's get rid of that for for
import java.util.*;
public class duplicate {
private static TreeSet<Integer> numbers = new TreeSet<Integer>();
private static TreeSet<Integer> duplicates = new TreeSet<Integer>();
public static void main (String[] args) {
Scanner input = new Scanner(System.in);
int n=0;
int numberOfIntToRead=10;
for (int i = 0; i < numberOfIntToRead; i++) {
// Read number from console
n=input.nextInt();
// Check if number was inserted before
checkIfDuplicate(n);
}
// Print all numbers that were not inserted more than one time
for (Integer j : numbers) {
System.out.print(j+" ");
}
}
public static void checkIfDuplicate(int n){
if(!numbers.contains(n) && !duplicates.contains(n)){
numbers.add(n);
}else{
numbers.remove(n);
duplicates.add(n);
}
}
}
But if you really want to use arrays an not a Collection of any sort, then you need to declare your arrays as class members, that way you can put the "for" that checks for duplicates in a function a give it your i, and this way you can also put the "for" that does the printing in a function. and give to it numbers.length. That should do the trick.
You can make use of a Set to make finding duplicates easy:
List<Integer> dups = new ArrayList<>();
Set<Integer> set = new HashSet<>();
for (int i : numbers)
if (!set.add(i))
dups.add(i);
This works because Set#add() returns false if it doesn't change as a result if being called, which happens eggnog the set already contains the number - ie it's a dup.
Yes, this is homework, but I need some help with it. I have been able to make it sort through the highest number, but none of the numbers are correct after that. List of numbers: http://pastebin.com/Ss1WFGv1
Right now, we are learning arrays, so is this simply trying to shoot a fly with a cannonball?
package hw2;
import java.io.BufferedReader;
import java.io.FileReader;
import java.util.ArrayList;
public class HW2 {
public static ArrayList<Integer> nums1 = new ArrayList<Integer>();
public static int size = 0;
public static void main(String[] args) throws Exception {
ArrayList<Integer> sortedNums = new ArrayList<Integer>();
readFile();
System.out.println("Name: Jay Bhagat" + "\n" + "Email: xxxxxx");
size = nums1.size();
for(int l = 0; l<=10;l++){
nums1.set(sortThis(nums1, l), 90009);
System.out.println("\n\n");
}
// for (int k = 0; k <= size - 1; k++) {
// System.out.println("Number " + (k + 1) + sortedNums.get(k));
//
// }
}
public static void readFile() throws Exception {
BufferedReader reader = new BufferedReader(new FileReader("L:\\numbers.txt"));
while (reader.readLine() != null) {
nums1.add(Integer.parseInt((reader.readLine())));
}
reader.close();
}
public static int sortThis(ArrayList<Integer> current, int offset) {
int j = 0;
int tempNum = 0;
int curNum = 0;
int finalIndex = 0;
int prevIndex = 0;
int curIndex = 0;
for (j = 0; j < size-offset; j++) {
curIndex = j;
nums1.trimToSize();
curNum = current.get(j);
//Thread.sleep(1000);
if(curNum!=90009){
if (curNum > tempNum) {
tempNum = curNum;
System.out.println(tempNum);
prevIndex = j;
finalIndex = prevIndex;
}
if (curNum < tempNum) {
finalIndex = prevIndex;
}
}
}
return finalIndex;
}
}
An approach that lets you make just one pass through the list and doesn't require sorting:
Declare an array of 5 integers: int[] largest = new int[5];
Put the first 5 elements in the ArrayList into largest.
Starting with the 6th element, look at each element N in the ArrayList, and if N is larger than any element in largest, throw out the smallest number currently in largest and replace it with N.
If you need to exclude duplicates, the algorithm can be modified easily (just skip over any ArrayList element that's already in largest).
Why not use Collections.sort(List list) or Arrays.Sort(arr). This will save much of effort. Or is it part of your task?
Assuming your collection is sorted, and you want the last 5 elements, try this out:
for (int i = sortedNums.size() - 5; i < sortedNums.size(); ++i) {
System.err.println(sortedNums.get(i));
}
How I would go about doing this:
Create a temporary ArrayList, as a copy of the initial one.
After each largest element is found, remove it from the temporary ArrayList and add it to your 5 largest numbers
Repeat until complete
edit* This does not require your elements to be sorted, and has a fairly poor efficiency as a result
I assume you do not have the liberty to use sort and suchlike, considering this is a homework. So here is outline of an algorithm that you can try to implement
create an array of five integers (we will keep this sorted)
for each element in the list
find the index of the element in the array that it is greater than
if no such element exists in the array (i.e. it is smaller than all elements in the array)
continue on to the next element in the list
else
push all elements in the array to one index below, letting them fall off the
edge if need be (e.g. if the number in list is 42 and the array has
45 and 40 at index 3 and 2 respectively then
move arr[1] to arr[0], and arr[2] (40) to arr[1] and set arr[2] = 42)
end if
end for
At the end the array will have the five elements
I will leave one question for you to answer (it is important): what should you set the array to initially?
You only need two lines of code:
Collections.sort(nums1);
List<Integer> high5 = nums1.subList(nums1.size() - 5, nums1.size());
If you must "do it yourself", the simplest way to sort is a bubble sort:
iterate over the list
swap adjacent numbers if they are in the wrong order
repeat n times
Not efficient but very easy to code.
I need help understanding how to sort numbers.
Below is what I have I came up with so far and it didn't work. Can you please point out the mistake and tell me what to do?
I saw some of you guys using java.util.Arrays . Can you describe to me its functions?
import static java.lang.System.*;
import java.util.*;
public class Lab07v2_Task10{
public static void main (String[] args){
Scanner orcho = new Scanner (in);
int quantity = 5;
int[] myArray = new int [quantity];
out.println("Please enter 5 numbers");
for(int count = 0; count<myArray.length; count++){
myArray[count] = orcho.nextInt();
}
int maxSoFar = myArray[0];
for(int count = myArray.length-1; count>=0; count--){
if(myArray[count] > maxSoFar){
maxSoFar = myArray[count];
}
out.println(maxSoFar);
}
}
}
No solution.
The idea is to take several steps, do a for-loop. And assume that you are in the middle. The first part already is sorted, the rest is to-be-done.
Then tackle the current element with respect to what already is sorted.
int maxSoFar = myArray[0];
for (int i = 1; i < myArray.length; i++) {
// The array 0, ..., i-1 is sorted
if (myArray[i] >= maxSoFar) {
// Still sorted
maxSoFar = myArray[i];
} else {
// myArray[i] must be shifted left
...
}
// Now the array 0, ..., i is sorted
}
This is a general trick: assume part is already done, tackle one small step, and let continue.
The java.util.Arrays.sort(int[]) method sorts the specified array of int into ascending numerical order.
try this out..
// sorting array
java.util.Arrays.sort(myArray);
// let us print all the elements available in list
System.out.println("The sorted int array is:");
for (int number : myArray) {
System.out.println("Number = " + number);
}
}
Arrays.sort is a method which is a utility method available in java.util package.
Where Arrays is a system defined Utility class which contains the mehtod sort(int[]) takes int[] (array) as an argument and after sorting this array, It re-assign Array.
For more deep Info Here or Official Java Docs
The way your program runs right now: it will print 5 numbers and the number that it prints is the highest number it finds at that iteration.
The way that you want it to work: sort 5 numbers from lowest to highest. Then print these 5 numbers. This is an implementation of bubble sort in your program:
for(int i = 0; i< myArray.length; i++){
for(int j = 0; j < myArray.length-1; j++){
if(myArray[j] > myArray[j+1]){ //if the current number is less than the one next to it
int temp = myArray[j]; //save the current number
myArray[j] = myArray[j+1]; //put the one next to it in its spot
myArray[j+1] = temp; //put the current number in the next spot
}
}
}
it is probably the easiest sort to understand. Basically, for as many times as the length of your array, comb over the numbers and bring the next highest number as far up as it can go.
When it's done sorting you can then print the numbers.