My code to Finding the numbers in the Fibonacci sequence is working correct but it is really slooooow. (It is almost impossibleto find number over n=55). Where did I make a mistake?
public static void main(String[] args) {
for (; ; ) {
System.out.println("The value of the fibonacci sequence ");
Scanner numer = new Scanner(System.in);
long n = numer.nextLong();
long result = Fib(n);
System.out.println(result);
}
}
private static long Fib(long n) {
if (n <= 0) { System.out.println("Error"); return 0;}
else if (n == 1 | n == 2) return 1;
else return Fib(n-1)+ Fib(n-2) ;
}
Do you understand how many times you compute fib(k) for each k? If you want your program to run fast, don't use recursion so blindly, use loop instead:
private static long Fib(long n) {
long prev = 1;
long current = 1;
if (n <= 0) { System.out.println("Error"); return 0;}
if (n == 1 | n == 2) return 1;
for (int i = 3; i <= n; i++) {
long next = prev + current;
prev = current;
current = next;
}
return current;
}
If you want to solve this problem only by recursion, then you need to store already computed results. It can be done either by memoization or by using lazy dynamic programming, and it's way harder than just using loop.
Even with the proposed improvements by #Jason, this is still an algorithm with exponential time complexity of O(2^N).
To reduce the time complexity to O(N), also called linear time, I propose the following approach (called "memoization", because we save each value we compute and directly return it for O(1) time whenever it is asked, hence the stack frames will be far fewer and will increase at most linearly with the value of "n").
public static void main(String[] args) {
// receive customer input
HashMap<Long, Long> map = new HashMap<>();
long result = Fib(n, map);
}
private long Fib(long n, HashMap<Long, Long> map) {
if (n <= 0) {
return 0;
}
if (n == 1) {
return 1;
}
if (map.containsKey(n)) {
return map.get(n);
} else {
map.put(n, Fib(n - 1, map) + Fib(n - 2, map));
return map.get(n);
}
}
You can compare the two solutions by submitting them in the Fibonacci exercise in Leetcode.
Also, both approaches (yours and a similar memoization approach to mine) are explained under the "Solution" tab in the link above, where you can learn more about the time complexities.
Given a collection of distinct numbers, return all possible permutations.
For example, [1,2,3] have the following permutations:
[ [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1] ]
My Iterative Solution is :
public List<List<Integer>> permute(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
result.add(new ArrayList<>());
for(int i=0;i<nums.length;i++)
{
List<List<Integer>> temp = new ArrayList<>();
for(List<Integer> a: result)
{
for(int j=0; j<=a.size();j++)
{
a.add(j,nums[i]);
List<Integer> current = new ArrayList<>(a);
temp.add(current);
a.remove(j);
}
}
result = new ArrayList<>(temp);
}
return result;
}
My Recursive Solution is:
public List<List<Integer>> permuteRec(int[] nums) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
if (nums == null || nums.length == 0) {
return result;
}
makePermutations(nums, result, 0);
return result;
}
void makePermutations(int[] nums, List<List<Integer>> result, int start) {
if (start >= nums.length) {
List<Integer> temp = convertArrayToList(nums);
result.add(temp);
}
for (int i = start; i < nums.length; i++) {
swap(nums, start, i);
makePermutations(nums, result, start + 1);
swap(nums, start, i);
}
}
private ArrayList<Integer> convertArrayToList(int[] num) {
ArrayList<Integer> item = new ArrayList<Integer>();
for (int h = 0; h < num.length; h++) {
item.add(num[h]);
}
return item;
}
According to me the time complexity(big-Oh) of my iterative solution is: n * n(n+1)/2~ O(n^3)
I am not able to figure out the time complexity of my recursive solution.
Can anyone explain complexity of both?
The recursive solution has a complexity of O(n!) as it is governed by the equation: T(n) = n * T(n-1) + O(1).
The iterative solution has three nested loops and hence has a complexity of O(n^3).
However, the iterative solution will not produce correct permutations for any number apart from 3.
For n = 3, you can see that n * (n - 1) * (n-2) = n!. The LHS is O(n^3) (or rather O(n^n) since n=3 here) and the RHS is O(n!).
For larger values of the size of the list, say n, you could have n nested loops and that will provide valid permutations. The complexity in that case will be O(n^n), and that is much larger than O(n!), or rather, n! < n^n. There is a rather nice relation called Stirling's approximation which explains this relation.
It's the output (which is huge) matters in this problem, not the routine's implementation. For n distinct items, there are n! permutations to be returned as the answer, and thus we have at least O(n!) complexity.
With a help of Stirling's approximation
O(n!) = O(n^(1/2+n)/exp(n)) = O(sqrt(n) * (n/e)^n)
we can easily see, that O(n!) > O(n^c) for any constant c, that's why it doesn't matter if the implementation itself adds another O(n^3) since
O(n!) + O(n^3) = O(n!)
In terms of number of times the method makePermutations is called, the exact time complexity would be:
O( 1 + n + n(n-1) + n(n-1)(n-2) + ... )
For n = 3:
O( 1 + 3 + (3*2) + (3*2*1) ) = O(16)
This means, for n = 3, the method makePermutations will be called 16 times.
And I think the space complexity for an optimal permutations function would be O(n * n!) because there are n! total arrays to return, and each of those arrays is of size n.
Where ever I see Recursive Fibonacci Series everyone tell that
a[i] = fib(i - 1) + fib( i - 2)
But it can also be solved with
a[i] = fib(i - 1) + a[i-2] // If array 'a' is a global variable.
If array 'a' is a global Variable, then a[i-2] will be calculated when it is calculating for a[i-2];
It can be solved with below program in java..
public class Fibonacci {
public static int maxNumbers = 10;
public static double[] arr = new double[maxNumbers];
public static void main(String args[])
{
arr[0] = 0;
arr[1] = 1;
recur(maxNumbers - 1);
}
public static double recur(int i)
{
if( i > 1)
{
arr[i] = recur(i - 1) + arr[i - 2];
}
return arr[i];
}
}
Further more, complexity is also less when compared with original procedure. Is there any disadvantage of doing this way?
You have done the first step for Dynamic Programming calculation of Fibonacci, idea of DP is to avoid redundant calculations, and your algorithm achieve its goal.
A "classic" Bottom-Up DP Fibonacci implementation is filling the elements from lower to higher:
arr[0] = 0
arr[1] = 1
for (int i = 2; i <= n; i++)
arr[i] = arr[i-1] + arr[i-2]
(Optimization could be storing curr,last alone, and modifying them at each iteration.
Your approach is basically the same in principle.
As a side note, the DP approach to calculate Fibonacci is taking O(n) time, where there is even more efficient solution with exponential of the matrix:
1 1
1 0
The above holds because you use the fact that
1 1 F_{n+1} 1*F{n+1} + 1*F{n} F_{n+2}
* = =
1 0 F_{n} 1*F{n+1} + 0*F{n} F_{n+1}
Using exponent by squaring on the above matrix, this can be solved in O(logn).
If you just want the nth fibonacci number you could do this:
static double fib(double prev, double curr, int n) {
if(n == 0)
return curr;
return fib(curr, prev+curr, n-1);
}
Initial conditions would be prev = 0, curr = 1, n = maxNumbers. This function is tail recursive because you don't need to store the return value of the recursive call for any additional calculations. The initial stack frame gets reused (which saves memory) and once you hit your base case the value that's returned is the same value that would be returned from every other recursive call.
By using an array like you do you only recalculate one of the two branches (the longest one in each iteration) ending up with a O(n) complexity.
If you were to keep track on how large fibonacci number you have caclulated earlier you can use that and produce O(max(n-prevn, 1)). Here is an altered version of your code that fills the array from bottom to i if needed:
public class Fibonacci {
public static final int maxNumbers = 93; // fib(93) > Long.MAX_VALUE
public static long[] arr = new long[maxNumbers];
public static int calculatedN = 0;
public static long fib(int i) throws Exception
{
if( i >= maxNumbers )
throw new Exception("value out of bounds");
if( calculatedN == 0 ) {
arr[0] = 0L;
arr[1] = 1L;
calculatedN = 1;
}
if( i > calculatedN ) {
for( int x=calculatedN+1; x<=i; x++ ){
arr[x] = arr[x-2] + arr[x-1];
}
calculatedN = i;
}
return arr[i];
}
public static void main (String args[]) {
try {
System.out.println(fib(50)); // O(50-2)
System.out.println(fib(30)); // O(1)
System.out.println(fib(92)); // O(92-50)
System.out.println(fib(92)); // O(1)
} catch ( Exception e ) { e.printStackTrace(); }
}
}
I changed double to long. If you need larger fibonacci numbers than fib(92) I would change from long to Biginteger.
You can also code using two recursive function but as the same value is calculating over again and again so all You can do a dynamic programming approach where You can store the value and return it where need.Like this one in C++
#include <bits/stdc++.h>
using namespace std;
int dp[100];
int fib(int n){
if(n <= 1)
return n;
if(dp[n]!= -1)
return dp[n];
dp[n] = fib(n-1) + fib(n-2);
return dp[n];
}
int main(){
memset(dp,-1,sizeof(dp));
for(int i=1 ;i<10 ;i++)
cout<<fib(i)<<endl;
}
This is only step from non recursive version:
https://gist.github.com/vividvilla/4641152
General this partially recursive approach looks incredibly messy
I'm trying to calculate the total, mean and median of an array thats populated by input received by a textfield. I've managed to work out the total and the mean, I just can't get the median to work. I think the array needs to be sorted before I can do this, but I'm not sure how to do this. Is this the problem, or is there another one that I didn't find? Here is my code:
import java.applet.Applet;
import java.awt.Graphics;
import java.awt.*;
import java.awt.event.*;
public class whileloopq extends Applet implements ActionListener
{
Label label;
TextField input;
int num;
int index;
int[] numArray = new int[20];
int sum;
int total;
double avg;
int median;
public void init ()
{
label = new Label("Enter numbers");
input = new TextField(5);
add(label);
add(input);
input.addActionListener(this);
index = 0;
}
public void actionPerformed (ActionEvent ev)
{
int num = Integer.parseInt(input.getText());
numArray[index] = num;
index++;
if (index == 20)
input.setEnabled(false);
input.setText("");
sum = 0;
for (int i = 0; i < numArray.length; i++)
{
sum += numArray[i];
}
total = sum;
avg = total / index;
median = numArray[numArray.length/2];
repaint();
}
public void paint (Graphics graf)
{
graf.drawString("Total = " + Integer.toString(total), 25, 85);
graf.drawString("Average = " + Double.toString(avg), 25, 100);
graf.drawString("Median = " + Integer.toString(median), 25, 115);
}
}
The Arrays class in Java has a static sort function, which you can invoke with Arrays.sort(numArray).
Arrays.sort(numArray);
double median;
if (numArray.length % 2 == 0)
median = ((double)numArray[numArray.length/2] + (double)numArray[numArray.length/2 - 1])/2;
else
median = (double) numArray[numArray.length/2];
Sorting the array is unnecessary and inefficient. There's a variation of the QuickSort (QuickSelect) algorithm which has an average run time of O(n); if you sort first, you're down to O(n log n). It actually finds the nth smallest item in a list; for a median, you just use n = half the list length. Let's call it quickNth (list, n).
The concept is that to find the nth smallest, choose a 'pivot' value. (Exactly how you choose it isn't critical; if you know the data will be thoroughly random, you can take the first item on the list.)
Split the original list into three smaller lists:
One with values smaller than the pivot.
One with values equal to the pivot.
And one with values greater than the pivot.
You then have three cases:
The "smaller" list has >= n items. In that case, you know that the nth smallest is in that list. Return quickNth(smaller, n).
The smaller list has < n items, but the sum of the lengths of the smaller and equal lists have >= n items. In this case, the nth is equal to any item in the "equal" list; you're done.
n is greater than the sum of the lengths of the smaller and equal lists. In that case, you can essentially skip over those two, and adjust n accordingly. Return quickNth(greater, n - length(smaller) - length(equal)).
Done.
If you're not sure that the data is thoroughly random, you need to be more sophisticated about choosing the pivot. Taking the median of the first value in the list, the last value in the list, and the one midway between the two works pretty well.
If you're very unlucky with your choice of pivots, and you always choose the smallest or highest value as your pivot, this takes O(n^2) time; that's bad. But, it's also very unlikely if you choose your pivot with a decent algorithm.
Sample code:
import java.util.*;
public class Utility {
/****************
* #param coll an ArrayList of Comparable objects
* #return the median of coll
*****************/
public static <T extends Number> double median(ArrayList<T> coll, Comparator<T> comp) {
double result;
int n = coll.size()/2;
if (coll.size() % 2 == 0) // even number of items; find the middle two and average them
result = (nth(coll, n-1, comp).doubleValue() + nth(coll, n, comp).doubleValue()) / 2.0;
else // odd number of items; return the one in the middle
result = nth(coll, n, comp).doubleValue();
return result;
} // median(coll)
/*****************
* #param coll a collection of Comparable objects
* #param n the position of the desired object, using the ordering defined on the list elements
* #return the nth smallest object
*******************/
public static <T> T nth(ArrayList<T> coll, int n, Comparator<T> comp) {
T result, pivot;
ArrayList<T> underPivot = new ArrayList<>(), overPivot = new ArrayList<>(), equalPivot = new ArrayList<>();
// choosing a pivot is a whole topic in itself.
// this implementation uses the simple strategy of grabbing something from the middle of the ArrayList.
pivot = coll.get(n/2);
// split coll into 3 lists based on comparison with the pivot
for (T obj : coll) {
int order = comp.compare(obj, pivot);
if (order < 0) // obj < pivot
underPivot.add(obj);
else if (order > 0) // obj > pivot
overPivot.add(obj);
else // obj = pivot
equalPivot.add(obj);
} // for each obj in coll
// recurse on the appropriate list
if (n < underPivot.size())
result = nth(underPivot, n, comp);
else if (n < underPivot.size() + equalPivot.size()) // equal to pivot; just return it
result = pivot;
else // everything in underPivot and equalPivot is too small. Adjust n accordingly in the recursion.
result = nth(overPivot, n - underPivot.size() - equalPivot.size(), comp);
return result;
} // nth(coll, n)
public static void main (String[] args) {
Comparator<Integer> comp = Comparator.naturalOrder();
Random rnd = new Random();
for (int size = 1; size <= 10; size++) {
ArrayList<Integer> coll = new ArrayList<>(size);
for (int i = 0; i < size; i++)
coll.add(rnd.nextInt(100));
System.out.println("Median of " + coll.toString() + " is " + median(coll, comp));
} // for a range of possible input sizes
} // main(args)
} // Utility
If you want to use any external library here is Apache commons math library using you can calculate the Median.
For more methods and use take look at the API documentation
import org.apache.commons.math3.*;
.....
......
........
//calculate median
public double getMedian(double[] values){
Median median = new Median();
double medianValue = median.evaluate(values);
return medianValue;
}
.......
For more on evaluate method AbstractUnivariateStatistic#evaluate
Update
Calculate in program
Generally, median is calculated using the following two formulas given here
If n is odd then Median (M) = value of ((n + 1)/2)th item term.
If n is even then Median (M) = value of [((n)/2)th item term + ((n)/2 + 1)th item term ]/2
In your program you have numArray, first you need to sort array using Arrays#sort
Arrays.sort(numArray);
int middle = numArray.length/2;
int medianValue = 0; //declare variable
if (numArray.length%2 == 1)
medianValue = numArray[middle];
else
medianValue = (numArray[middle-1] + numArray[middle]) / 2;
Arrays.sort(numArray);
return (numArray[size/2] + numArray[(size-1)/2]) / 2;
Arrays.sort(numArray);
int middle = ((numArray.length) / 2);
if(numArray.length % 2 == 0){
int medianA = numArray[middle];
int medianB = numArray[middle-1];
median = (medianA + medianB) / 2;
} else{
median = numArray[middle + 1];
}
EDIT: I initially had medianB setting to middle+1 in the even length arrays, this was wrong due to arrays starting count at 0. I have updated it to use middle-1 which is correct and should work properly for an array with an even length.
You can find good explanation at https://www.youtube.com/watch?time_continue=23&v=VmogG01IjYc
The idea it to use 2 Heaps viz one max heap and mean heap.
class Heap {
private Queue<Integer> low = new PriorityQueue<>(Comparator.reverseOrder());
private Queue<Integer> high = new PriorityQueue<>();
public void add(int number) {
Queue<Integer> target = low.size() <= high.size() ? low : high;
target.add(number);
balance();
}
private void balance() {
while(!low.isEmpty() && !high.isEmpty() && low.peek() > high.peek()) {
Integer lowHead= low.poll();
Integer highHead = high.poll();
low.add(highHead);
high.add(lowHead);
}
}
public double median() {
if(low.isEmpty() && high.isEmpty()) {
throw new IllegalStateException("Heap is empty");
} else {
return low.size() == high.size() ? (low.peek() + high.peek()) / 2.0 : low.peek();
}
}
}
Try sorting the array first. Then after it's sorted, if the array has an even amount of elements the mean of the middle two is the median, if it has a odd number, the middle element is the median.
Use Arrays.sort and then take the middle element (in case the number n of elements in the array is odd) or take the average of the two middle elements (in case n is even).
public static long median(long[] l)
{
Arrays.sort(l);
int middle = l.length / 2;
if (l.length % 2 == 0)
{
long left = l[middle - 1];
long right = l[middle];
return (left + right) / 2;
}
else
{
return l[middle];
}
}
Here are some examples:
#Test
public void evenTest()
{
long[] l = {
5, 6, 1, 3, 2
};
Assert.assertEquals((3 + 4) / 2, median(l));
}
#Test
public oddTest()
{
long[] l = {
5, 1, 3, 2, 4
};
Assert.assertEquals(3, median(l));
}
And in case your input is a Collection, you might use Google Guava to do something like this:
public static long median(Collection<Long> numbers)
{
return median(Longs.toArray(numbers)); // requires import com.google.common.primitives.Longs;
}
I was looking at the same statistics problems. The approach you are thinking it is good and it will work. (Answer to the sorting has been given)
But in case you are interested in algorithm performance, I think there are a couple of algorithms that have better performance than just sorting the array, one (QuickSelect) is indicated by #bruce-feist's answer and is very well explained.
[Java implementation: https://discuss.leetcode.com/topic/14611/java-quick-select ]
But there is a variation of this algorithm named median of medians, you can find a good explanation on this link:
http://austinrochford.com/posts/2013-10-28-median-of-medians.html
Java implementation of this:
- https://stackoverflow.com/a/27719796/957979
I faced a similar problem yesterday.
I wrote a method with Java generics in order to calculate the median value of every collection of Numbers; you can apply my method to collections of Doubles, Integers, Floats and returns a double. Please consider that my method creates another collection in order to not alter the original one.
I provide also a test, have fun. ;-)
public static <T extends Number & Comparable<T>> double median(Collection<T> numbers){
if(numbers.isEmpty()){
throw new IllegalArgumentException("Cannot compute median on empty collection of numbers");
}
List<T> numbersList = new ArrayList<>(numbers);
Collections.sort(numbersList);
int middle = numbersList.size()/2;
if(numbersList.size() % 2 == 0){
return 0.5 * (numbersList.get(middle).doubleValue() + numbersList.get(middle-1).doubleValue());
} else {
return numbersList.get(middle).doubleValue();
}
}
JUnit test code snippet:
/**
* Test of median method, of class Utils.
*/
#Test
public void testMedian() {
System.out.println("median");
Double expResult = 3.0;
Double result = Utils.median(Arrays.asList(3.0,2.0,1.0,9.0,13.0));
assertEquals(expResult, result);
expResult = 3.5;
result = Utils.median(Arrays.asList(3.0,2.0,1.0,9.0,4.0,13.0));
assertEquals(expResult, result);
}
Usage example (consider the class name is Utils):
List<Integer> intValues = ... //omitted init
Set<Float> floatValues = ... //omitted init
.....
double intListMedian = Utils.median(intValues);
double floatSetMedian = Utils.median(floatValues);
Note: my method works on collections, you can convert arrays of numbers to list of numbers as pointed here
And nobody paying attention when list contains only one element (list.size == 1). All your answers will crash with index out of bound exception, because integer division returns zero (1 / 2 = 0). Correct answer (in Kotlin):
MEDIAN("MEDIAN") {
override fun calculate(values: List<BigDecimal>): BigDecimal? {
if (values.size == 1) {
return values.first()
}
if (values.size > 1) {
val valuesSorted = values.sorted()
val mid = valuesSorted.size / 2
return if (valuesSorted.size % 2 != 0) {
valuesSorted[mid]
} else {
AVERAGE.calculate(listOf(valuesSorted[mid - 1], valuesSorted[mid]))
}
}
return null
}
},
As #Bruce-Feist mentions, for a large number of elements, I'd avoid any solution involving sort if performance is something you are concerned about. A different approach than those suggested in the other answers is Hoare's algorithm to find the k-th smallest of element of n items. This algorithm runs in O(n).
public int findKthSmallest(int[] array, int k)
{
if (array.length < 10)
{
Arrays.sort(array);
return array[k];
}
int start = 0;
int end = array.length - 1;
int x, temp;
int i, j;
while (start < end)
{
x = array[k];
i = start;
j = end;
do
{
while (array[i] < x)
i++;
while (x < array[j])
j--;
if (i <= j)
{
temp = array[i];
array[i] = array[j];
array[j] = temp;
i++;
j--;
}
} while (i <= j);
if (j < k)
start = i;
if (k < i)
end = j;
}
return array[k];
}
And to find the median:
public int median(int[] array)
{
int length = array.length;
if ((length & 1) == 0) // even
return (findKthSmallest(array, array.length / 2) + findKthSmallest(array, array.length / 2 + 1)) / 2;
else // odd
return findKthSmallest(array, array.length / 2);
}
public static int median(int[] arr) {
int median = 0;
java.util.Arrays.sort(arr);
for (int i=0;i<arr.length;i++) {
if (arr.length % 2 == 1) {
median = Math.round(arr[arr.length/2]);
} else {
median = (arr[(arr.length/2)] + arr[(arr.length/2)-1])/2;
}
}
return median;
}
Check out the Arrays.sort methods:
http://docs.oracle.com/javase/6/docs/api/java/util/Arrays.html
You should also really abstract finding the median into its own method, and just return the value to the calling method. This will make testing your code much easier.
public int[] data={31, 29, 47, 48, 23, 30, 21
, 40, 23, 39, 47, 47, 42, 44, 23, 26, 44, 32, 20, 40};
public double median()
{
Arrays.sort(this.data);
double result=0;
int size=this.data.length;
if(size%2==1)
{
result=data[((size-1)/2)+1];
System.out.println(" uneven size : "+result);
}
else
{
int middle_pair_first_index =(size-1)/2;
result=(data[middle_pair_first_index+1]+data[middle_pair_first_index])/2;
System.out.println(" Even size : "+result);
}
return result;
}
package arrays;
public class Arraymidleelement {
static public double middleArrayElement(int [] arr)
{
double mid;
if(arr.length%2==0)
{
mid=((double)arr[arr.length/2]+(double)arr[arr.length/2-1])/2;
return mid;
}
return arr[arr.length/2];
}
public static void main(String[] args) {
int arr[]= {1,2,3,4,5,6};
System.out.println( middleArrayElement(arr));
}
}