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I've been grinding leetcode recently and am perplexed on why my solution is timing out when I submit it to Leetcode.
Here is the question:
https://leetcode.com/explore/learn/card/data-structure-tree/133/conclusion/942/
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
For example, given
inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]
Return the following binary tree:
3
/ \
9 20
/ \
15 7
Here is my solution that times out in one of the test cases:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode buildTree(int[] inorder, int[] postorder) {
if (inorder == null || inorder.length == 0) {
return null; // input error
}
if (postorder == null || postorder.length == 0) {
return null; // input error
}
if (postorder.length != inorder.length) {
return null; // input error
}
List<Integer> inOrder = new ArrayList<Integer>();
List<Integer> postOrder = new ArrayList<Integer>();
for (int i = 0; i < inorder.length; i++) {
inOrder.add(inorder[i]);
postOrder.add(postorder[i]);
}
return buildBinaryTree(inOrder, postOrder);
}
public TreeNode buildBinaryTree(List<Integer> inOrder, List<Integer> postOrder) {
boolean found = false;
int root = 0;
int rootIndex = 0;
// for given in-order scan the post-order right to left to find the root
for (int j = postOrder.size() - 1; j >= 0 && !found; j--) {
root = postOrder.get(j);
if (inOrder.contains(root)) {
rootIndex = inOrder.indexOf(root);
root = inOrder.get(rootIndex);
found = true;
break;
}
}
if (found) {
List<Integer> leftOfRoot = new ArrayList<Integer>();
List<Integer> rightOfRoot = new ArrayList<Integer>();
if (rootIndex > 0) {
leftOfRoot.addAll(inOrder.subList(0, rootIndex));
}
if ((rootIndex + 1) < inOrder.size()) {
rightOfRoot.addAll(inOrder.subList(rootIndex + 1, inOrder.size()));
}
TreeNode node = new TreeNode(root);
node.left = buildBinaryTree(leftOfRoot, postOrder);
node.right = buildBinaryTree(rightOfRoot, postOrder);
return node;
}
return null;
}
}
Can anyone help determine why this is happening? I'm thinking it is the Leetcode judge at fault here and my code is fine.
Leetcode's judge is probably OK. This code is too casual about nested linear array operations and heap allocations. Creating ArrayLists and calling contains, addAll, subList and indexOf may appear innocuous, but they should all be thought of as extremely expensive operations when inside a recursive function that spawns two child calls in every frame.
Let's unpack the code a bit:
List<Integer> inOrder = new ArrayList<Integer>();
List<Integer> postOrder = new ArrayList<Integer>();
for (int i = 0; i < inorder.length; i++) {
inOrder.add(inorder[i]);
postOrder.add(postorder[i]);
}
This is a minor up-front cost but it's an omen of things to come. We've done 2 heap allocations that weren't necessary and walked n. I'd stick to primitive arrays here--no need to allocate objects other than the result nodes. A lookup map for inOrder with value -> index pairs might be useful to allocate if you feel compelled to create a supporting data structure here.
Next, we step into buildBinaryTree. Its structure is basically:
function buildBinaryTree(root) {
// do some stuff
if (not base case reached) {
buildBinaryTree(root.left)
buildBinaryTree(root.right)
}
}
This is linear on the number of nodes in the tree, so it's important that // do some stuff is efficient, hopefully constant time. Walking n in this function would give us quadratic complexity.
Next there's
for (int j = postOrder.size() - 1; j >= 0 && !found; j--) {
root = postOrder.get(j);
if (inOrder.contains(root)) {
rootIndex = inOrder.indexOf(root);
This looks bad, but by definition the root is always the last element in a postorder traversal array, so if we keep a pointer to it, we can remove this outer loop. You can use indexOf directly and avoid the contains call since indexOf returns -1 to indicate a failed search.
The code:
if (found) {
List<Integer> leftOfRoot = new ArrayList<Integer>();
List<Integer> rightOfRoot = new ArrayList<Integer>();
does more unnecessary heap allocations for every call frame.
Here,
leftOfRoot.addAll(inOrder.subList(0, rootIndex));
Walks the list twice, once to create the sublist and again to add the entire sublist to the ArrayList. Repeat for the right subtree for two full walks on n per frame. Using start and end indices per call frame means you never need to allocate heap memory or copy anything to prepare the next call. Adjust the indices and pass a reference to the same two arrays along the entire time.
I recommend running your code with a profiler to see exactly how much time is spent copying and scanning your ArrayLists. The correct implementation should do at most one walk through one of the lists per call frame to locate root in inOrder. No array copying should be done at all.
With these modifications, you should be able to pass, although wrangling the pointers for this problem is not obvious. A hint that may help is this: recursively process the right subtree before the left.
Yes, it would be much faster with arrays. Try this:
public static TreeNode buildTree(int[] inorder, int[] postorder, int start,
int end) {
for (int i = postorder.length-1; i >= 0; --i) {
int root = postorder[i];
int index = indexOf(inorder, start, end, root);
if (index >= 0) {
TreeNode left = index == start
? null
: buildTree(inorder, postorder, start, index);
TreeNode right = index+1 == end
? null
: buildTree(inorder, postorder, index+1, end);
return new TreeNode(root, left, right);
}
}
return null;
}
private static int indexOf(int[] array, int start, int end, int value) {
for (int i = start; i < end; ++i) {
if (array[i] == value) {
return i;
}
}
return -1;
}
Write a function:
class Solution{
public int solution(int[] A);
}
that, given an array A of N integers, returns the smallest positive integer(greater than 0)
that does not occur in A.
For example, given A = [1,3,6,4,1,2], the function should return 5.
Given A = [1,2,3], the function should return 4.
Given A = [-1, -3], the function should return 1.
Write an efficient algorithm for the following assumptions.
N is an integer within the range [1..100,000];
each element of array A is an integer within the range [-1,000,000..1,000,000].
I wrote the following algorithm in Java:
public class TestCodility {
public static void main(String args[]){
int a[] = {1,3,6,4,1,2};
//int a[] = {1,2,3};
//int b[] = {-1,-3};
int element = 0;
//checks if the array "a" was traversed until the last position
int countArrayLenght = 0;
loopExtern:
for(int i = 0; i < 1_000_000; i++){
element = i + 1;
countArrayLenght = 0;
loopIntern:
for(int j = 0; j < a.length; j++){
if(element == a[j]){
break loopIntern;
}
countArrayLenght++;
}
if(countArrayLenght == a.length && element > 0){
System.out.println("Smallest possible " + element);
break loopExtern;
}
}
}
}
It does the job but I am pretty sure that it is not efficient. So my question is, how to improve this algorithm so that it becomes efficient?
You should get a grasp on Big O, and runtime complexities.
Its a universal construct for better understanding the implementation of efficiency in code.
Check this website out, it shows the graph for runtime complexities in terms of Big O which can aid you in your search for more efficient programming.
http://bigocheatsheet.com/
However, long story short...
The least amount of operations and memory consumed by an arbitrary program is the most efficient way to achieve something you set out to do with your code.
You can make something more efficient by reducing redundancy in your algorithms and getting rid of any operation that does not need to occur to achieve what you are trying to do
Point is to sort your array and then iterate over it. With sorted array you can simply skip all negative numbers and then find minimal posible element that you need.
Here more general solution for your task:
import java.util.Arrays;
public class Main {
public static int solution(int[] A) {
int result = 1;
Arrays.sort(A);
for(int a: A) {
if(a > 0) {
if(result == a) {
result++;
} else if (result < a){
return result;
}
}
}
return result;
}
public static void main(String args[]){
int a[] = {1,3,6,4,1,2};
int b[] = {1,2,3};
int c[] = {-1,-3};
System.out.println("a) Smallest possible " + solution(a)); //prints 5
System.out.println("b) Smallest possible " + solution(b)); //prints 4
System.out.println("c) Smallest possible " + solution(c)); //prints 1
}
}
Complexity of that algorithm should be O(n*log(n))
The main idea is the same as Denis.
First sort, then process but using java8 feature.
There are few methods that may increase timings.(not very sure how efficient java 8 process them:filter,distinct and even take-while ... in the worst case you have here something similar with 3 full loops. One additional loop is for transforming array into stream). Overall you should get the same run-time complexity.
One advantage could be on verbosity, but also need some additional knowledge compared with Denis solution.
import java.util.function.Supplier;
import java.util.stream.IntStream;
public class AMin
{
public static void main(String args[])
{
int a[] = {-2,-3,1,2,3,-7,5,6};
int[] i = {1} ;
// get next integer starting from 1
Supplier<Integer> supplier = () -> i[0]++;
//1. transform array into specialized int-stream
//2. keep only positive numbers : filter
//3. keep no duplicates : distinct
//4. sort by natural order (ascending)
//5. get the maximum stream based on criteria(predicate) : longest consecutive numbers starting from 1
//6. get the number of elements from the longest "sub-stream" : count
long count = IntStream.of(a).filter(t->t>0).distinct().sorted().takeWhile(t->t== supplier.get()).count();
count = (count==0) ? 1 : ++count;
//print 4
System.out.println(count);
}
}
There are many solutions with O(n) space complexity and O(n) type complexity. You can convert array to;
set: array to set and for loop (1...N) check contains number or not. If not return number.
hashmap: array to map and for loop (1...N) check contains number or not. If not return number.
count array: convert given array to positive array count array like if arr[i] == 5, countArr[5]++, if arr[i] == 1, countArr[1]++ then check each item in countArr with for loop (1...N) whether greate than 1 or not. If not return it.
For now, looking more effective algoritm like #Ricola mentioned. Java solution with O(n) time complexity and O(1) space complexity:
static void swap(final int arr[], final int i,final int j){
final int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static boolean isIndexInSafeArea(final int arr[], final int i){
return arr[i] > 0 && arr[i] - 1 < arr.length && arr[i] != i + 1 ;
}
static int solution(final int arr[]){
for (int i = 0; i < arr.length; i++) {
while (isIndexInSafeArea(arr,i) && arr[i] != arr[arr[i] - 1]) {
swap(arr, i, arr[i] - 1);
}
}
for (int i = 0; i < arr.length; i++) {
if (arr[i] != i + 1) {
return i+1;
}
}
return arr.length + 1;
}
I've been looking for a recursive selection sort, using only 2 parameters:
The array that has to be sorted
a value k, which indicates till which
element it has to be sorted.
Example: SelectionSort(array[] a, int k) with a being {6,3,5,7,2} and k being 2 will sort the first 3 elements, and will keep the last elements untouched.
I was thinking about starting with an if-statement for k being 0, and if that was the case, it would just return the array as it is, since you cant sort an array of size 1.
Something like:
public int[] sort(int[] a){
a = selectionSort(a, n-1);
return a;
}
public int[] selectionSort(int[] a, int k){
if (k = 0){
return a;
}
else{
selectionSort(a, k-1 );
... (part i really don't know)
}
I have no clue how to do the 'else' part since I only know that it has to call the method again.
I'm not allowed to create other methods. I also need to make sure I use exactly 2 parameters, nothing more, nothing less.
I have to work it out in pseudocode, but I understand some Java, so if someone could help me by using either pseudo, or Java, it would be so helpful
First some remarks to your code:
Your methods sort and selectionSort don't need to return an int[] array,
since the array object a stays the same all the time.
It is only the content within this array which changes.
Hence, you can use void as return-type.
In your if use (k == 0) instead of (k = 0)
You already figured out the first part.
Here it is how you can do the second part in pseudo code:
public void selectionSort(int[] a, int k) {
if (k == 0) {
return;
}
else {
selectionSort(a, k-1 );
find x such that a[x] is the smallest of a[k] ... a[a.length - 1]
if (a[k-1] > a[x]) {
swap a[k-1] and a[x]
}
}
}
I'm sure you are able to refine the pseudo code to real Java code.
By doing a simple google search, I found the biggest part of the code below on this site. I added the selectionSort method myself to suit your parameters.
public void selectionSort(int a[], int n)
{
recurSelectionSort(a, n, 0);
}
// Recursive selection sort. n is size of a[] and index
// is index of starting element.
static void recurSelectionSort(int a[], int n, int index)
{
// Return when starting and size are same
if (index == n)
return;
// calling minimum index function for minimum index
int k = minIndex(a, index, n-1);
// Swapping when index nd minimum index are not same
if (k != index){
// swap
int temp = a[k];
a[k] = a[index];
a[index] = temp;
}
// Recursively calling selection sort function
recurSelectionSort(a, n, index + 1);
}
// Return minimum index
static int minIndex(int a[], int i, int j)
{
if (i == j)
return i;
// Find minimum of remaining elements
int k = minIndex(a, i + 1, j);
// Return minimum of current and remaining.
return (a[i] < a[k])? i : k;
}
Like a Max-heap and Min-heap, I want to implement a Median-heap to keep track of the median of a given set of integers. The API should have the following three functions:
insert(int) // should take O(logN)
int median() // will be the topmost element of the heap. O(1)
int delmedian() // should take O(logN)
I want to use an array (a) implementation to implement the heap where the children of array index k are stored in array indices 2*k and 2*k + 1. For convenience, the array starts populating elements from index 1.
This is what I have so far:
The Median-heap will have two integers to keep track of number of integers inserted so far that are > current median (gcm) and < current median (lcm).
if abs(gcm-lcm) >= 2 and gcm > lcm we need to swap a[1] with one of its children.
The child chosen should be greater than a[1]. If both are greater,
choose the smaller of two.
Similarly for the other case. I can't come up with an algorithm for how to sink and swim elements. I think it should take into consideration how close the number is to the median, so something like:
private void swim(int k) {
while (k > 1 && absless(k, k/2)) {
exch(k, k/2);
k = k/2;
}
}
I can't come up with the entire solution though.
You need two heaps: one min-heap and one max-heap. Each heap contains about one half of the data. Every element in the min-heap is greater or equal to the median, and every element in the max-heap is less or equal to the median.
When the min-heap contains one more element than the max-heap, the median is in the top of the min-heap. And when the max-heap contains one more element than the min-heap, the median is in the top of the max-heap.
When both heaps contain the same number of elements, the total number of elements is even.
In this case you have to choose according your definition of median: a) the mean of the two middle elements; b) the greater of the two; c) the lesser; d) choose at random any of the two...
Every time you insert, compare the new element with those at the top of the heaps in order to decide where to insert it. If the new element is greater than the current median, it goes to the min-heap. If it is less than the current median, it goes to the max heap. Then you might need to rebalance. If the sizes of the heaps differ by more than one element, extract the min/max from the heap with more elements and insert it into the other heap.
In order to construct the median heap for a list of elements, we should first use a linear time algorithm and find the median. Once the median is known, we can simply add elements to the Min-heap and Max-heap based on the median value. Balancing the heaps isn't required because the median will split the input list of elements into equal halves.
If you extract an element you might need to compensate the size change by moving one element from one heap to another. This way you ensure that, at all times, both heaps have the same size or differ by just one element.
Here is a java implementaion of a MedianHeap, developed with the help of above comocomocomocomo 's explanation .
import java.util.Arrays;
import java.util.Comparator;
import java.util.PriorityQueue;
import java.util.Scanner;
/**
*
* #author BatmanLost
*/
public class MedianHeap {
//stores all the numbers less than the current median in a maxheap, i.e median is the maximum, at the root
private PriorityQueue<Integer> maxheap;
//stores all the numbers greater than the current median in a minheap, i.e median is the minimum, at the root
private PriorityQueue<Integer> minheap;
//comparators for PriorityQueue
private static final maxHeapComparator myMaxHeapComparator = new maxHeapComparator();
private static final minHeapComparator myMinHeapComparator = new minHeapComparator();
/**
* Comparator for the minHeap, smallest number has the highest priority, natural ordering
*/
private static class minHeapComparator implements Comparator<Integer>{
#Override
public int compare(Integer i, Integer j) {
return i>j ? 1 : i==j ? 0 : -1 ;
}
}
/**
* Comparator for the maxHeap, largest number has the highest priority
*/
private static class maxHeapComparator implements Comparator<Integer>{
// opposite to minHeapComparator, invert the return values
#Override
public int compare(Integer i, Integer j) {
return i>j ? -1 : i==j ? 0 : 1 ;
}
}
/**
* Constructor for a MedianHeap, to dynamically generate median.
*/
public MedianHeap(){
// initialize maxheap and minheap with appropriate comparators
maxheap = new PriorityQueue<Integer>(11,myMaxHeapComparator);
minheap = new PriorityQueue<Integer>(11,myMinHeapComparator);
}
/**
* Returns empty if no median i.e, no input
* #return
*/
private boolean isEmpty(){
return maxheap.size() == 0 && minheap.size() == 0 ;
}
/**
* Inserts into MedianHeap to update the median accordingly
* #param n
*/
public void insert(int n){
// initialize if empty
if(isEmpty()){ minheap.add(n);}
else{
//add to the appropriate heap
// if n is less than or equal to current median, add to maxheap
if(Double.compare(n, median()) <= 0){maxheap.add(n);}
// if n is greater than current median, add to min heap
else{minheap.add(n);}
}
// fix the chaos, if any imbalance occurs in the heap sizes
//i.e, absolute difference of sizes is greater than one.
fixChaos();
}
/**
* Re-balances the heap sizes
*/
private void fixChaos(){
//if sizes of heaps differ by 2, then it's a chaos, since median must be the middle element
if( Math.abs( maxheap.size() - minheap.size()) > 1){
//check which one is the culprit and take action by kicking out the root from culprit into victim
if(maxheap.size() > minheap.size()){
minheap.add(maxheap.poll());
}
else{ maxheap.add(minheap.poll());}
}
}
/**
* returns the median of the numbers encountered so far
* #return
*/
public double median(){
//if total size(no. of elements entered) is even, then median iss the average of the 2 middle elements
//i.e, average of the root's of the heaps.
if( maxheap.size() == minheap.size()) {
return ((double)maxheap.peek() + (double)minheap.peek())/2 ;
}
//else median is middle element, i.e, root of the heap with one element more
else if (maxheap.size() > minheap.size()){ return (double)maxheap.peek();}
else{ return (double)minheap.peek();}
}
/**
* String representation of the numbers and median
* #return
*/
public String toString(){
StringBuilder sb = new StringBuilder();
sb.append("\n Median for the numbers : " );
for(int i: maxheap){sb.append(" "+i); }
for(int i: minheap){sb.append(" "+i); }
sb.append(" is " + median()+"\n");
return sb.toString();
}
/**
* Adds all the array elements and returns the median.
* #param array
* #return
*/
public double addArray(int[] array){
for(int i=0; i<array.length ;i++){
insert(array[i]);
}
return median();
}
/**
* Just a test
* #param N
*/
public void test(int N){
int[] array = InputGenerator.randomArray(N);
System.out.println("Input array: \n"+Arrays.toString(array));
addArray(array);
System.out.println("Computed Median is :" + median());
Arrays.sort(array);
System.out.println("Sorted array: \n"+Arrays.toString(array));
if(N%2==0){ System.out.println("Calculated Median is :" + (array[N/2] + array[(N/2)-1])/2.0);}
else{System.out.println("Calculated Median is :" + array[N/2] +"\n");}
}
/**
* Another testing utility
*/
public void printInternal(){
System.out.println("Less than median, max heap:" + maxheap);
System.out.println("Greater than median, min heap:" + minheap);
}
//Inner class to generate input for basic testing
private static class InputGenerator {
public static int[] orderedArray(int N){
int[] array = new int[N];
for(int i=0; i<N; i++){
array[i] = i;
}
return array;
}
public static int[] randomArray(int N){
int[] array = new int[N];
for(int i=0; i<N; i++){
array[i] = (int)(Math.random()*N*N);
}
return array;
}
public static int readInt(String s){
System.out.println(s);
Scanner sc = new Scanner(System.in);
return sc.nextInt();
}
}
public static void main(String[] args){
System.out.println("You got to stop the program MANUALLY!!");
while(true){
MedianHeap testObj = new MedianHeap();
testObj.test(InputGenerator.readInt("Enter size of the array:"));
System.out.println(testObj);
}
}
}
Here my code based on the answer provided by comocomocomocomo :
import java.util.PriorityQueue;
public class Median {
private PriorityQueue<Integer> minHeap =
new PriorityQueue<Integer>();
private PriorityQueue<Integer> maxHeap =
new PriorityQueue<Integer>((o1,o2)-> o2-o1);
public float median() {
int minSize = minHeap.size();
int maxSize = maxHeap.size();
if (minSize == 0 && maxSize == 0) {
return 0;
}
if (minSize > maxSize) {
return minHeap.peek();
}if (minSize < maxSize) {
return maxHeap.peek();
}
return (minHeap.peek()+maxHeap.peek())/2F;
}
public void insert(int element) {
float median = median();
if (element > median) {
minHeap.offer(element);
} else {
maxHeap.offer(element);
}
balanceHeap();
}
private void balanceHeap() {
int minSize = minHeap.size();
int maxSize = maxHeap.size();
int tmp = 0;
if (minSize > maxSize + 1) {
tmp = minHeap.poll();
maxHeap.offer(tmp);
}
if (maxSize > minSize + 1) {
tmp = maxHeap.poll();
minHeap.offer(tmp);
}
}
}
Isn't a perfectly balanced binary search tree (BST) a median heap? It is true that even red-black BSTs aren't always perfectly balanced, but it might be close enough for your purposes. And log(n) performance is guaranteed!
AVL trees are more tighly balanced than red-black BSTs so they come even closer to being a true median heap.
Here is a Scala implementation, following the comocomocomocomo's idea above.
class MedianHeap(val capacity:Int) {
private val minHeap = new PriorityQueue[Int](capacity / 2)
private val maxHeap = new PriorityQueue[Int](capacity / 2, new Comparator[Int] {
override def compare(o1: Int, o2: Int): Int = Integer.compare(o2, o1)
})
def add(x: Int): Unit = {
if (x > median) {
minHeap.add(x)
} else {
maxHeap.add(x)
}
// Re-balance the heaps.
if (minHeap.size - maxHeap.size > 1) {
maxHeap.add(minHeap.poll())
}
if (maxHeap.size - minHeap.size > 1) {
minHeap.add(maxHeap.poll)
}
}
def median: Double = {
if (minHeap.isEmpty && maxHeap.isEmpty)
return Int.MinValue
if (minHeap.size == maxHeap.size) {
return (minHeap.peek+ maxHeap.peek) / 2.0
}
if (minHeap.size > maxHeap.size) {
return minHeap.peek()
}
maxHeap.peek
}
}
Another way to do it without using a max-heap and a min-heap would be to use a median-heap right away.
In a max-heap, the parent is greater than the children.
We can have a new type of heap where the parent is in the 'middle' of the children - the left child is smaller than the parent and the right child is greater than the parent. All even entries are left children and all odd entries are right children.
The same swim and sink operations which can be performed in a max-heap, can also be performed in this median-heap - with slight modifications. In a typical swim operation in a max-heap, the inserted entry swims up till it is smaller than a parent entry, here in a median-heap, it will swim up till it is lesser than a parent (if it is an odd entry) or greater than a parent (if it is an even entry).
Here's my implementation for this median-heap. I have used an array of Integers for simplicity.
package priorityQueues;
import edu.princeton.cs.algs4.StdOut;
public class MedianInsertDelete {
private Integer[] a;
private int N;
public MedianInsertDelete(int capacity){
// accounts for '0' not being used
this.a = new Integer[capacity+1];
this.N = 0;
}
public void insert(int k){
a[++N] = k;
swim(N);
}
public int delMedian(){
int median = findMedian();
exch(1, N--);
sink(1);
a[N+1] = null;
return median;
}
public int findMedian(){
return a[1];
}
// entry swims up so that its left child is smaller and right is greater
private void swim(int k){
while(even(k) && k>1 && less(k/2,k)){
exch(k, k/2);
if ((N > k) && less (k+1, k/2)) exch(k+1, k/2);
k = k/2;
}
while(!even(k) && (k>1 && !less(k/2,k))){
exch(k, k/2);
if (!less (k-1, k/2)) exch(k-1, k/2);
k = k/2;
}
}
// if the left child is larger or if the right child is smaller, the entry sinks down
private void sink (int k){
while(2*k <= N){
int j = 2*k;
if (j < N && less (j, k)) j++;
if (less(k,j)) break;
exch(k, j);
k = j;
}
}
private boolean even(int i){
if ((i%2) == 0) return true;
else return false;
}
private void exch(int i, int j){
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
private boolean less(int i, int j){
if (a[i] <= a[j]) return true;
else return false;
}
public static void main(String[] args) {
MedianInsertDelete medianInsertDelete = new MedianInsertDelete(10);
for(int i = 1; i <=10; i++){
medianInsertDelete.insert(i);
}
StdOut.println("The median is: " + medianInsertDelete.findMedian());
medianInsertDelete.delMedian();
StdOut.println("Original median deleted. The new median is " + medianInsertDelete.findMedian());
}
}
I have array with constant size (size = 20 in real life), duplicates are allowed For example:
1 2 2 3 3 4 5 6 7 8 9
Now exactly one element updates:
1 5 2 3 3 4 5 6 7 8 9
I need to resort this array. Should I just use bubblesort?
update I don't know how to call what I wrote. But i suppose it is not possible to sort faster. comments are welcome!
// array is already almost sorted and INCREASING, element at pos need to be inserted to the right place
private void SortQuotes(List<Quote> quoteList, int pos)
{
var quoteToMove = quoteList[pos];
if (pos == 0 || quoteList[pos - 1].Price < quoteToMove.Price)
{
MoveElementsDown(quoteList, pos);
} else if (pos == quoteList.Count - 1 || quoteList[pos + 1].Price > quoteToMove.Price)
{
MoveElementsUp(quoteList, pos);
}
}
private void MoveElementsDown(List<Quote> quoteList, int pos)
{
var quoteToInsert = quoteList[pos];
var price = quoteToInsert.Price;
for (int i = pos - 1; i >= 0; i--)
{
var nextQuote = quoteList[i];
if (nextQuote.Price > price)
{
quoteList[i + 1] = quoteList[i];
if (i == 0) // last element
{
quoteList[i] = quoteToInsert;
}
}
else
{
quoteList[i + 1] = quoteToInsert;
break;
}
}
}
private void MoveElementsUp(List<Quote> quoteList, int pos)
{
var quoteToInsert = quoteList[pos];
var price = quoteToInsert.Price;
for (int i = pos + 1; i < quoteList.Count; i++)
{
var nextQuote = quoteList[i];
if (nextQuote.Price < price)
{
quoteList[i - 1] = quoteList[i];
if (i == quoteList.Count - 1) // last element
{
quoteList[i] = quoteToInsert;
}
}
else
{
quoteList[i - 1] = quoteToInsert;
break;
}
}
}
updated i do know which element is odd, i.e. it's position is known!
This solution shifts each element by one until the right position for the odd element is found. As it has been overwritten already in the first step, it is saved in a temporary variable 'h' and then written to the final position. It requires the minimum of comparisions and shift operations:
static void MoveOddElementToRightPosition(int[] a, int oddPosition)
{
int h = a[oddPosition];
int i;
if (h > a[oddPosition + 1])
for (i = oddPosition; i < a.Count()-1 && a[i+1] <= h; i++)
a[i] = a[i+1];
else
for (i = oddPosition; i > 0 && a[i-1] >= h; i--)
a[i] = a[i - 1];
a[i] = h;
}
Bubblesort will use n^2 time if the last element needs to get to the front. Use insertion sort.
As the array is small, insertion sort takes roughly ~O(n) time for small arrays and if you are just updating 1 value, insertion sort should fulfil your purpose in the best possible way.
It can be done in O(n). If you don't know the element then search for the element in O(n) and then You just need to compare and swap for each element and that would take O(n). So total 2n which means O(n).If you know the element which has been modified then compare and swap for each element.
If you're interested in replacing an element quickly, then you can also use a structure where deletion and insertion is fast, like for example a TreeSet in Java. That means O(log(n)) theoretically, but if you just manipulate arrays of 20 elements it may not be worth it
If the set of all different elements is finite, like in your example where you just use numbers for 1 to 9, then there is a solution in O(1). Instead of having a sorted list you just keep an array with the number of occurrences of your elements.
If you still want to keep everything in an array, then the fastest way is this
find the position A of of the element you're going to remove by bisection in O(log(n)).
find the position B of where your new element is going to end up in the same way. More precisely B is the smallest index where new_element < a[k] for every k > B
if A < B, move all elements between A + 1 and B to their left, then set the new element to position B. if B > A, you do the same thing but to the right. Now this step is in O(n), but there's no logic, it's just moving memory around. In C you'd use memmove for this and it's heavily optimized, but I don't know any Java equivalent.
You don't need to sort it again.
Only one element changes. So you just need to go through the list and put the changed number to appropriate place. This will be of O(n) complexity.
int a[] = {1, 5, 2, 3, 3, 4, 5, 6, 7, 8, 9};
int count = 0;
//find the odd element
for(int jj=1; jj< a.length; jj++){
if(a[jj] < a[count])
break;
else count++;
}
System.out.println(" Odd position " + count);
//put odd element to proper position
for(int k= count+1; k<a.length; k++){
if(a[count] > a[k]){
int t = a[count];
a[count] = a[k];
a[k] = t;
count++;
}
}
Above is the working code tested for given input.
Enjoy.
Bubblesort is quite OK in this case with 20 compares max.
But finding the new position with binary search is O(log(n)), that is 5 compares in this case.
Somewhat faster, if you need the last bit odd speed use the binary search otherwise you can stick with bubble sort.
Here is a naive implementation in plain C. Remove the fprintf(stderr, ... after testing. The ITEM can be anything, as long as a comparison function is possible. Otherwise: use pointers to ITEM, (and maybe add an extra sizeofelem argument, ala qsort)
#include <stdio.h>
#include <string.h>
typedef int ITEM;
int item_cmp(ITEM one, ITEM two);
unsigned one_bubble( ITEM *arr, unsigned cnt, unsigned hot , int (*cmp)(ITEM,ITEM) );
int item_cmp(ITEM one, ITEM two)
{
fprintf(stderr,"Cmp= %u to %u: %d\n", one, two, one-two);
if (one > two) return 1;
else if (one < two) return -1;
else return 0;
}
unsigned one_bubble( ITEM *arr, unsigned cnt, unsigned hot , int (*cmp)(ITEM,ITEM) )
{
unsigned goal = cnt;
int diff;
ITEM temp;
/* hot element should move to the left */
if (hot > 0 && (diff=cmp( arr[hot-1], arr[hot])) > 0) {
/* Find place to insert (this could be a binary search) */
for (goal= hot; goal-- > 0; ) {
diff=cmp( arr[goal], arr[hot]);
if (diff <= 0) break;
}
goal++;
fprintf(stderr,"Move %u LEFT to %u\n", hot, goal);
if (goal==hot) return hot;
temp = arr[hot];
/* shift right */
fprintf(stderr,"memmove(%u,%u,%u)\n", goal+1, goal, (hot-goal) );
memmove(arr+goal+1, arr+goal, (hot-goal) *sizeof temp);
arr[goal] = temp;
return goal; /* new position */
}
/* hot element should move to the right */
else if (hot < cnt-1 && (diff=cmp( arr[hot], arr[hot+1])) > 0) {
/* Find place to insert (this could be a binary search) */
for (goal= hot+1; goal < cnt; goal++ ) {
diff=cmp( arr[hot], arr[goal]);
if (diff <= 0) break;
}
goal--;
fprintf(stderr,"Move %u RIGHT to %u\n", hot, goal);
if (goal==hot) return hot;
temp = arr[hot];
/* shift left */
fprintf(stderr,"memmove(%u,%u,%u)\n", hot, hot+1, (goal-hot) );
memmove(arr+hot, arr+hot+1, (goal-hot) *sizeof temp);
arr[goal] = temp;
return goal; /* new position */
}
fprintf(stderr,"Diff=%d Move %u Not to %u\n", diff, hot, goal);
return hot;
}
ITEM array[10] = { 1,10,2,3,4,5,6,7,8,9,};
#define HOT_POS 1
int main(void)
{
unsigned idx;
idx = one_bubble(array, 10, HOT_POS, item_cmp);
printf("%u-> %u\n", HOT_POS, idx );
for (idx = 0; idx < 10; idx++) {
printf("%u: %u\n", idx, array[idx] );
}
return 0;
}