I was able to write a code that draws different circles on a canvas and i need to find a way i could delete the leftmost circle when any key is pressed. i've been at this for hours and i feel like i am close to the answer. i am most klikely going to look for the array whenever a key is pressed and delete the array position.
float colour = random(256);
final int DIAM = 20;
final int MAX_NUM = 1000;
int numPointsX = 0;
int numPointsY = 0;
int [] xPos = new int[MAX_NUM];
int [] yPos = new int [MAX_NUM];
boolean start = false;
void setup() {
size (500, 500);
}
void draw() {
background(150);
fill(random(256), random(256), random(256));
for (int i=0; i<numPointsX; i++) {
circle(xPos[i], yPos[i], DIAM);
}
println(xPos[0]);
}
void mouseClicked() {
insertXandY();
}
void insertXandY() {
int x = mouseX;
int y = mouseY;
xPos[numPointsX] = x;
yPos[numPointsY] = y;
numPointsX += 1;
numPointsY += 1;
start = true;
}
void printArrays() {
println("X Positions");
for (int i = 0; i < 20; i++) {
println("\t" + xPos[i]);
}
}
void keyPressed() {
if (key == 'p') {
printArrays();
}
}
You are on the right track.
In broad terms you'd need two steps:
find the smallest X
delete the data associated with the smallest X
The 1st part is trivial:
use a variable to keep track of the currently smallest value (initialised with a bigger than than your data has)
iterate through each value
compare each value with the current smallest:
if it's bigger ignore
if it's smallest: update the currently smallest value (and remember the index)
at the end of the iteration the currently smallest value is the smallest possible value and index can be used to associate between x,y arrays (which are incremented in sync)
Here's a slightly modified version of your code to illustrate this:
float colour = random(256);
final int DIAM = 20;
final int MAX_NUM = 1000;
int numPoints = 0;
int [] xPos = new int[MAX_NUM];
int [] yPos = new int [MAX_NUM];
void setup() {
size (500, 500);
}
void draw() {
background(150);
fill(random(256), random(256), random(256));
for (int i=0; i < numPoints; i++) {
circle(xPos[i], yPos[i], DIAM);
}
}
void mouseClicked() {
insertXandY();
}
void insertXandY() {
int x = mouseX;
int y = mouseY;
xPos[numPoints] = x;
yPos[numPoints] = y;
numPoints++;
}
void deleteLeftMost(){
// find leftmost index
// start with a large X value
int smallestX = width;
int smallestXIndex = -1;
// iterate through each X
for(int i = 0 ; i < numPoints; i++){
// if xPos[i] is smaller than the smallest value so far...
if (xPos[i] < smallestX){
// ...remember it's value and index
smallestX = xPos[i];
smallestXIndex = i;
}
}
// delete the item at this index: fake it for now: move coordinates offscreen (to the right so left search still works)
xPos[smallestXIndex] = width * 2;
}
void printArrays() {
println("X Positions");
for (int i = 0; i < 20; i++) {
println("\t" + xPos[i]);
}
}
void keyPressed() {
if (key == 'p') {
printArrays();
}
if (keyCode == DELETE || keyCode == BACKSPACE){
deleteLeftMost();
}
}
I've made a few of other minor adjustments:
deleted start since it was assigned but not used (when debugging delete anything that isn't necessary)
renamed numPointsX to numPoints and deleted numPointsY: you are using two arrays indeed, however there is only one index for each point that could be re-used to access each array
numPoints++ is shorthand for numPoints = numPoints + 1;
Also, I've used a hacky placeholder for the remove a point just visually.
This means in terms of memory the xPos/yPos for deleted points will still be allocated.
To actually delete the array is a bit tricker since the array datatype does not change size, however you could manually put something together using subset() and concat(). You can achieve a similar effect to deleting an element by concatenating two subset array: from the start to the index to delete and from the index next to the one to delete to the end of the array.
Something like this:
void setup(){
println(deleteIndex(new int[]{1,2,3,4,5,6},-1));
println(deleteIndex(new int[]{1,2,3,4,5,6},2));
println(deleteIndex(new int[]{1,2,3,4,5,6},6));
}
int[] deleteIndex(int[] sourceArray, int indexToDelete){
if(sourceArray == null){
System.err.println("can't process null array");
return null;
}
if(indexToDelete < 0){
System.err.println("invalid index " + indexToDelete + "\nit's < 0");
return null;
}
if(indexToDelete >= sourceArray.length){
System.err.println("invalid index " + indexToDelete + "\nmax index = " + sourceArray.length);
return null;
}
return concat(subset(sourceArray, 0, indexToDelete),
subset(sourceArray, indexToDelete + 1, sourceArray.length - indexToDelete - 1));
}
It's a good idea to check arguments to a method to ensure they are valid and test with at least a few edge cases.
Here's a version of the above sketch using this delete method:
float colour = random(256);
final int DIAM = 20;
final int MAX_NUM = 1000;
int numPoints = 0;
int [] xPos = new int[MAX_NUM];
int [] yPos = new int [MAX_NUM];
void setup() {
size (500, 500);
}
void draw() {
background(150);
fill(random(256), random(256), random(256));
for (int i=0; i < numPoints; i++) {
circle(xPos[i], yPos[i], DIAM);
}
}
void mouseClicked() {
insertXandY();
}
void insertXandY() {
int x = mouseX;
int y = mouseY;
xPos[numPoints] = x;
yPos[numPoints] = y;
numPoints++;
}
void deleteLeftMost(){
// find leftmost index
// start with a large X value
int smallestX = width;
int smallestXIndex = -1;
// iterate through each X
for(int i = 0 ; i < numPoints; i++){
// if xPos[i] is smaller than the smallest value so far...
if (xPos[i] < smallestX){
// ...remember it's value and index
smallestX = xPos[i];
smallestXIndex = i;
}
}
// delete xPos item at this index
xPos = deleteIndex(xPos, smallestXIndex);
// delete yPos as well
yPos = deleteIndex(yPos, smallestXIndex);
// update size counter
numPoints--;
}
int[] deleteIndex(int[] sourceArray, int indexToDelete){
if(sourceArray == null){
System.err.println("can't process null array");
return null;
}
if(indexToDelete < 0){
System.err.println("invalid index " + indexToDelete + "\nit's < 0");
return null;
}
if(indexToDelete >= sourceArray.length){
System.err.println("invalid index " + indexToDelete + "\nmax index = " + sourceArray.length);
return null;
}
return concat(subset(sourceArray, 0, indexToDelete),
subset(sourceArray, indexToDelete + 1, sourceArray.length - indexToDelete - 1));
}
void printArrays() {
println("X Positions");
for (int i = 0; i < xPos.length; i++) {
println("\t" + xPos[i]);
}
}
void keyPressed() {
if (key == 'p') {
printArrays();
}
if (keyCode == DELETE || keyCode == BACKSPACE){
deleteLeftMost();
}
}
If manually deleting an item from an array looks tedious it's because it is :)
Array is meant to be fixed size: deleting an item actually allocates 3 arrays: two subset arrays and one for concatenation.
A better option is to use a dynamic sized array data structure like ArrayList. Speaking of data structures, to represent a point you can use the PVector class (which has x,y properties, but can also do much more).
You might have not encountered ArrayList and PVector yet, but there are plenty of resources out there (including CodingTrain/NatureOfCode videos).
Here's an example using these:
final int DIAM = 20;
final int MAX_NUM = 1000;
ArrayList<PVector> points = new ArrayList<PVector>();
void setup() {
size (500, 500);
}
void draw() {
background(150);
fill(random(256), random(256), random(256));
for (PVector point : points) {
circle(point.x, point.y, DIAM);
}
}
void mouseClicked() {
insertXandY();
}
void insertXandY() {
if(points.size() < MAX_NUM){
points.add(new PVector(mouseX, mouseY));
}
}
void deleteLeftMost(){
// find leftmost index
// start with a large X value
float smallestX = Float.MAX_VALUE;
int smallestXIndex = -1;
// iterate through each X
for(int i = 0 ; i < points.size(); i++){
PVector point = points.get(i);
// if xPos[i] is smaller than the smallest value so far...
if (point.x < smallestX){
// ...remember it's value and index
smallestX = point.x;
smallestXIndex = i;
}
}
// remove item from list
points.remove(smallestXIndex);
}
void keyPressed() {
if (key == 'p') {
println(points);
}
if (keyCode == DELETE || keyCode == BACKSPACE){
deleteLeftMost();
}
}
Hopefully this step by step approach is easy to follow.
Have fun learning !
I'm trying to solve this problem: https://www.interviewbit.com/problems/knight-on-chess-board/#
Basically, you're given a board, a start point and an end point and have to find the shortest path. I'm trying to do BFS on the the board using the 8 possible moves a knight can make and returning the number of moves it took, or -1 if there was no solution. I'm getting a run time out of memory error. I'm not sure where the error (or potential errors) are occurring.
Edit: Previously I was getting an error because I forgot got to mark nodes as visited. I've added that in but I'm still not getting the right answer.
public class Solution {
private class Node {
int row;
int col;
int count;
public Node() {
this.row = 0;
this.col = 0;
this.count = 0;
}
public Node(int row, int col, int count) {
this.row = row;
this.col = col;
this.count = count;
}
}
public int knight(int A, int B, int sr, int sc, int er, int ec) {
int[][] matrix = new int[A][B];
Queue<Node> q = new LinkedList<>(); //linkedlist??
Node n = new Node(sr, sc, 0);
q.add(n);
matrix[sr][sc] = -1;
final int[][] SHIFTS = {
{-2,1},
{-2,-1},
{2,1},
{2,-1},
{-1,2},
{-1,-2},
{1,2},
{1,-2}
};
int count = 0;
while(!q.isEmpty()) {
Node cur = q.remove();
if(cur.row == er && cur.col == ec) {
return cur.count;
}
for(int[] i : SHIFTS) {
if(canTraverse(matrix, cur.row + i[0], cur.col + i[1])) {
matrix[cur.row + i[0]][cur.col + i[1]] = -1;
q.add(new Node(cur.row + i[0], cur.col + i[1], cur.count + 1));
}
}
}
return -1;
}
public static boolean canTraverse(int[][] matrix, int sr, int sc) {
if(sr < 0 || sr >= matrix.length || sc < 0 || sc >= matrix[sr].length || matrix[sr][sc] == -1) {
return false;
}
return true;
}
}
BFS algorithm needs to mark every visited position (node) to work properly. Else, such code could cause (almost certainly) runtime error or memory limit exceded (in short terms: A calls B and B calls A).
Solution: Create a boolean array and mark the nodes at the time they enter to the queue and you are done.
There are 10 balloons and each balloon has some point written onto it. If a customer shoots a balloon, he will get points equal to points on left balloon multiplied by points on the right balloon. A Customer has to collect maximum points in order to win this game. What will be maximum points and in which order should he shoot balloons to get maximum points ?
Please note that if there is only one balloon then you return the points on that balloon.
I am trying to check all 10! permutations in order to find out maximum points. Is there any other way to solve this in efficient way ?
As i said in the comments a Dynamic programming solution with bitmasking is possible, what we can do is keep a bitmask where a 1 at a bit indexed at i means that the ith baloon has been shot, and a 0 tells that it has not been shot.
So a Dynamic Programming state of only mask is required, where at each state we can transition to the next state by iterating over all the ballons that have not been shot and try to shoot them to find the maximum.
The time complexity of such a solution would be : O((2^n) * n * n) and the space complexity would be O(2^n).
Code in c++, it is not debugged you may need to debug it :
int n = 10, val[10], dp[1024]; //set all the values of dp table to -1 initially
int solve(int mask){
if(__builtin_popcount(mask) == n){
return 0;
}
if(dp[mask] != -1) return dp[mask];
int prev = 1, ans = 0;
for(int i = 0;i < n;i++){
if(((mask >> i) & 1) == 0){ //bit is not set
//try to shoot current baloon
int newMask = mask | (1 << i);
int fwd = 1;
for(int j = i+1;j < n;j++){
if(((mask >> j) & 1) == 0){
fwd = val[j];
break;
}
}
ans = max(ans, solve(newMask) + (prev * fwd));
prev = val[i];
}
}
return dp[mask] = ans;
}
#include<iostream>
using namespace std;
int findleft(int arr[],int n,int j ,bool isBurst[],bool &found)
{
if(j<=0)
{
found=false;
return 1;
}
for(int i=j-1;i>=0;i--)
{
if(!isBurst[i])
{
return arr[i];
}
}
found = false;
return 1;
}
int findright(int arr[],int n,int j,bool isBurst[],bool &found)
{
if(j>=n)
{
found = false;
return 1;
}
for(int i= j+1;i<=n;i++)
{
if(!isBurst[i])
{
return arr[i];
}
}
found=false;
return 1;
}
int calc(int arr[],int n,int j,bool isBurst[])
{
int points =0;
bool leftfound=true;
bool rightfound=true;
int left= findleft( arr, n-1, j,isBurst , leftfound);
int right = findright( arr,n-1, j,isBurst, rightfound);
if(!leftfound && !rightfound)
{
points+=arr[j];
}
else
{
points+=left*right*arr[j];
}
return points;
}
void maxpoints(int arr[],int n,int cp,int curr_ans,int &ans,int count,bool isBurst[])
{
if(count==n)
{
if(curr_ans>ans)
{
ans=curr_ans;
return;
}
}
for(int i=0;i<n;i++)
{
if(!isBurst[i])
{
isBurst[i]=true;
maxpoints(arr,n,i,curr_ans+calc(arr,n,i,isBurst),ans,count+1,isBurst);
isBurst[i]=false;
}
}
}
int main()
{
int n;
cin>>n;
int ans=0;
int arr[n];
bool isBurst[n];
for(int i=0;i<n;i++)
{
cin>>arr[i];
isBurst[i]=false;
}
maxpoints(arr,n,0,0,ans,0,isBurst);
cout<<ans;
return 0;
}
Saw this question on Careercup.com:
Given heights of n persons standing in a line and a list of numbers corresponding to each person (p) that gives the number of persons who are taller than p and standing in front of p. For example,
Heights: 5 3 2 6 1 4
InFronts:0 1 2 0 3 2
Means that the actual actual order is: 5 3 2 1 6 4
The question gets the two lists of Heights and InFronts, and should generate the order standing in line.
My solution:
It could be solved by first sorting the list in descending order. Obviously, to sort, we need to define an object Person (with two attributes of Height and InFront) and then sort Persons based on their height. Then, I would use two stacks, a main stack and a temp one, to build up the order.
Starting from the tallest, put it in the main stack. If the next person had an InFront value of greater than the person on top of the stack, that means the new person should be added before the person on top. Therefore, we need to pop persons from the main stack, insert the new person, and then return the persons popped out in the first step (back to the main stack from temp one). I would use a temp stack to keep the order of the popped out persons. But how many should be popped out? Since the list is sorted, we need to pop exactly the number of persons in front of the new person, i.e. corresponding InFront.
I think this solution works. But the worst case order would be O(n^2) -- when putting a person in place needs popping out all previous ones.
Is there any other solutions? possibly in O(n)?
The O(nlogn) algoritm is possible.
First assume that all heights are different.
Sort people by heights. Then iterate from shortest to tallest. In each step you need an efficient way to put the next person to the correct position. Notice that people we've already placed are not taller that the current person. And the people we place after are taller than the current. So we have to find a place such that the number of empty positions in the front is equal to the inFronts value of this person. This task can be done using a data structure called interval tree in O(logn) time. So the total time of an algorithm is O(nlogn).
This algorithm works well in case where there's no ties. As it may be safely assumed that empty places up to front will be filled by taller people.
In case when ties are possible, we need to assure that people of the same height are placed in increasing order of their positions. It can be achieved if we will process people by non-decreasing inFronts value. So, in case of possible ties we should also consider inFronts values when sorting people.
And if at some step we can't find a position for next person then the answer it "it's impossible to satisfy problem constraints".
There exists an algorithm with O(nlogn) average complexity, however worst case complexity is still O(n²).
To achieve this you can use a variation of a binary tree. The idea is, in this tree, each node corresponds to a person and each node keeps track of how many people are in front of him (which is the size of the left subtree) as nodes are inserted.
Start iterating the persons array in decreasing height order and insert each person into the tree starting from the root. Insertion is as follows:
Compare the frontCount of the person with the current node's (root at the beginning) value.
If it is smaller than it insert the node to the left with value 1. Increase the current node's value by 1.
Else, descend to the right by decreasing the person's frontCount by current node's value. This enables the node to be placed in the correct location.
After all nodes finished, an inorder traversal gives the correct order of people.
Let the code speak for itself:
public static void arrange(int[] heights, int[] frontCounts) {
Person[] persons = new Person[heights.length];
for (int i = 0; i < persons.length; i++)
persons[i] = new Person(heights[i], frontCounts[i]);
Arrays.sort(persons, (p1, p2) -> {
return Integer.compare(p2.height, p1.height);
});
Node root = new Node(persons[0]);
for (int i = 1; i < persons.length; i++) {
insert(root, persons[i]);
}
inOrderPrint(root);
}
private static void insert(Node root, Person p) {
insert(root, p, p.frontCount);
}
private static void insert(Node root, Person p, int value) {
if (value < root.value) { // should insert to the left
if (root.left == null) {
root.left = new Node(p);
} else {
insert(root.left, p, value);
}
root.value++; // Increase the current node value while descending left!
} else { // insert to the right
if (root.right == null) {
root.right = new Node(p);
} else {
insert(root.right, p, value - root.value);
}
}
}
private static void inOrderPrint(Node root) {
if (root == null)
return;
inOrderPrint(root.left);
System.out.print(root.person.height);
inOrderPrint(root.right);
}
private static class Node {
Node left, right;
int value;
public final Person person;
public Node(Person person) {
this.value = 1;
this.person = person;
}
}
private static class Person {
public final int height;
public final int frontCount;
Person(int height, int frontCount) {
this.height = height;
this.frontCount = frontCount;
}
}
public static void main(String[] args) {
int[] heights = {5, 3, 2, 6, 1, 4};
int[] frontCounts = {0, 1, 2, 0, 3, 2};
arrange(heights, frontCounts);
}
I think one approach can be the following. Although it again seems to be O(n^2) at present.
Sort the Height array and corresponding 'p' array in ascending order of heights (in O(nlogn)). Pick the first element in the list. Put that element in the final array in the position given by the p index.
For example after sorting,
H - 1, 2, 3, 4, 5, 6
p - 3, 2, 1, 2, 0, 0.
1st element should go in position 3. Hence final array becomes:
---1--
2nd element shall go in position 2. Hence final array becomes:
--21--
3rd element should go in position 1. Hence final array becomes:
-321--
4th element shall go in position 2. This is the position among the empty ones. Hence final array becomes:
-321-4
5th element shall go in position 0. Hence final array becomes:
5321-4
6th element should go in position 0. Hence final array becomes:
532164
I think the approach indicated above is correct. However a critical piece missing in the solutions above are.
Infronts is the number of taller candidate before the current person. So after sorting the persons based on height(Ascending), when placing person 3 with infront=2, if person 1 and 2 was in front placed at 0, 1 position respectively, you need to discount their position and place 3 at position 4, I.E 2 taller candidates will take position 2,3.
As some indicated interval tree is the right structure. However a dynamic sized container, with available position will do the job.(code below)
struct Person{
int h, ct;
Person(int ht, int c){
h = ht;
ct = c;
}
};
struct comp{
bool operator()(const Person& lhs, const Person& rhs){
return (lhs.h < rhs.h);
}
};
vector<int> heightOrder(vector<int> &heights, vector<int> &infronts) {
if(heights.size() != infronts.size()){
return {};
}
vector<int> result(infronts.size(), -1);
vector<Person> persons;
vector<int> countSet;
for(int i= 0; i< heights.size(); i++){
persons.emplace_back(Person(heights[i], infronts[i]));
countSet.emplace_back(i);
}
sort(persons.begin(), persons.end(), comp());
for(size_t i=0; i<persons.size(); i++){
Person p = persons[i];
if(countSet.size() > p.ct){
int curr = countSet[p.ct];
//cout << "the index to place height=" << p.h << " , is at pos=" << curr << endl;
result[curr] = p.h;
countSet.erase(countSet.begin() + p.ct);
}
}
return result;
}
I'm using LinkedList for the this. Sort the tallCount[] in ascending order and accordingly re-position the items in heights[]. This is capable of handling the duplicate elements also.
public class FindHeightOrder {
public int[] findOrder(final int[] heights, final int[] tallCount) {
if (heights == null || heights.length == 0 || tallCount == null
|| tallCount.length == 0 || tallCount.length != heights.length) {
return null;
}
LinkedList list = new LinkedList();
list.insertAtStart(heights[0]);
for (int i = 1; i < heights.length; i++) {
if (tallCount[i] == 0) {
Link temp = list.getHead();
while (temp != null && temp.getData() <= heights[i]) {
temp = temp.getLink();
}
if (temp != null) {
if (temp.getData() <= heights[i]) {
list.insertAfterElement(temp.getData(), heights[i]);
} else {
list.insertAtStart(heights[i]);
}
} else {
list.insertAtEnd(heights[i]);
}
} else {
Link temp = list.getHead();
int pos = tallCount[i];
while (temp != null
&& (temp.getData() <= heights[i] || pos-- > 0)) {
temp = temp.getLink();
}
if (temp != null) {
if (temp.getData() <= heights[i]) {
list.insertAfterElement(temp.getData(), heights[i]);
} else {
list.insertBeforeElement(temp.getData(), heights[i]);
}
} else {
list.insertAtEnd(heights[i]);
}
}
}
Link fin = list.getHead();
int i = 0;
while (fin != null) {
heights[i++] = fin.getData();
fin = fin.getLink();
}
return heights;
}
public class Link {
private int data;
private Link link;
public Link(int data) {
this.data = data;
}
public int getData() {
return data;
}
public void setData(int data) {
this.data = data;
}
public Link getLink() {
return link;
}
public void setLink(Link link) {
this.link = link;
}
#Override
public String toString() {
return this.data + " -> "
+ (this.link != null ? this.link : "null");
}
}
public class LinkedList {
private Link head;
public Link getHead() {
return head;
}
public void insertAtStart(int data) {
if (head == null) {
head = new Link(data);
head.setLink(null);
} else {
Link link = new Link(data);
link.setLink(head);
head = link;
}
}
public void insertAtEnd(int data) {
if (head != null) {
Link temp = head;
while (temp != null && temp.getLink() != null) {
temp = temp.getLink();
}
temp.setLink(new Link(data));
} else {
head = new Link(data);
}
}
public void insertAfterElement(int after, int data) {
if (head != null) {
Link temp = head;
while (temp != null) {
if (temp.getData() == after) {
Link link = new Link(data);
link.setLink(temp.getLink());
temp.setLink(link);
break;
} else {
temp = temp.getLink();
}
}
}
}
public void insertBeforeElement(int before, int data) {
if (head != null) {
Link current = head;
Link previous = null;
Link ins = new Link(data);
while (current != null) {
if (current.getData() == before) {
ins.setLink(current);
break;
} else {
previous = current;
current = current.getLink();
if (current != null && current.getData() == before) {
previous.setLink(ins);
ins.setLink(current);
break;
}
}
}
}
}
#Override
public String toString() {
return "LinkedList [head=" + this.head + "]";
}
}
}
As people already corrected for original input:
Heights : A[] = { 5 3 2 6 1 4 }
InFronts: B[] = { 0 1 2 0 3 2 }
Output should look like: X[] = { 5 3 1 6 2 4 }
Here is the O(N*logN) way to approach solution (with assumption that there are no ties).
Iterate over array B and build chain of inequalities (by placing items into a right spot on each iteration, here we can use hashtable for O(1) lookups):
b0 > b1
b0 > b1 > b2
b3 > b0 > b1 > b2
b3 > b0 > b1 > b4 > b2
b3 > b0 > b5 > b1 > b4 > b2
Sort array A and reverse it
Initialize output array X, iterate over chain from #1 and fill array X by placing items from A into a position defined in a chain
Steps #1 and #3 are O(N), step #2 is the most expensive O(N*logN).
And obviously reversing sorted array A (in step #2) is not required.
This is the implementation for the idea provided by user1990169. Complexity being O(N^2).
public class Solution {
class Person implements Comparator<Person>{
int height;
int infront;
public Person(){
}
public Person(int height, int infront){
this.height = height;
this.infront = infront;
}
public int compare(Person p1, Person p2){
return p1.height - p2.height;
}
}
public ArrayList<Integer> order(ArrayList<Integer> heights, ArrayList<Integer> infronts) {
int n = heights.size();
Person[] people = new Person[n];
for(int i = 0; i < n; i++){
people[i] = new Person(heights.get(i), infronts.get(i));
}
Arrays.sort(people, new Person());
Person[] rst = new Person[n];
for(Person p : people){
int count = 0;
for(int i = 0; i < n ; i++){
if(count == p.infront){
while(rst[i] != null && i < n - 1){
i++;
}
rst[i] = p;
break;
}
if(rst[i] == null) count++;
}
}
ArrayList<Integer> heightrst = new ArrayList<Integer>();
for(int i = 0; i < n; i++){
heightrst.add(rst[i].height);
}
return heightrst;
}
}
Was solving this problem today, here is what I came up with:
The idea is to sort the heights array in descending order. Once, we have this sorted array - pick up an element from this element and place it in the resultant array at the corresponding index (I am using an ArrayList for the same, it would be nice to use LinkedList) :
public class Solution {
public ArrayList<Integer> order(ArrayList<Integer> heights, ArrayList<Integer> infronts) {
Person[] persons = new Person[heights.size()];
ArrayList<Integer> res = new ArrayList<>();
for (int i = 0; i < persons.length; i++) {
persons[i] = new Person(heights.get(i), infronts.get(i));
}
Arrays.sort(persons, (p1, p2) -> {
return Integer.compare(p2.height, p1.height);
});
for (int i = 0; i < persons.length; i++) {
//System.out.println("adding "+persons[i].height+" "+persons[i].count);
res.add(persons[i].count, persons[i].height);
}
return res;
}
private static class Person {
public final int height;
public final int count;
public Person(int h, int c) {
height = h;
count = c;
}
}
}
I found this kind of problem on SPOJ. I created a binary tree with little variation. When a new height is inserted, if the front is smaller than the root's front then it goes to the left otherwise right.
Here is the C++ implementation:
#include<bits/stdc++.h>
using namespace std;
struct TreeNode1
{
int val;
int _front;
TreeNode1* left;
TreeNode1*right;
};
TreeNode1* Add(int x, int v)
{
TreeNode1* p= (TreeNode1*) malloc(sizeof(TreeNode1));
p->left=NULL;
p->right=NULL;
p->val=x;
p->_front=v;
return p;
}
TreeNode1* _insert(TreeNode1* root, int x, int _front)
{
if(root==NULL) return Add(x,_front);
if(root->_front >=_front)
{
root->left=_insert(root->left,x,_front);
root->_front+=1;
}
else
{
root->right=_insert(root->right,x,_front-root->_front);
}
return root;
}
bool comp(pair<int,int> a, pair<int,int> b)
{
return a.first>b.first;
}
void in_order(TreeNode1 * root, vector<int>&v)
{
if(root==NULL) return ;
in_order(root->left,v);
v.push_back(root->val);
in_order(root->right,v);
}
vector<int>soln(vector<int>h, vector<int>in )
{
vector<pair<int , int> >vc;
for(int i=0;i<h.size();i++) vc.push_back( make_pair( h[i],in[i] ) );
sort(vc.begin(),vc.end(),comp);
TreeNode1* root=NULL;
for(int i=0;i<vc.size();i++)
root=_insert(root,vc[i].first,vc[i].second);
vector<int>v;
in_order(root,v);
return v;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
vector<int>h;
vector<int>in;
for(int i=0;i<n;i++) {int x;
cin>>x;
h.push_back(x);}
for(int i=0;i<n;i++) {int x; cin>>x;
in.push_back(x);}
vector<int>v=soln(h,in);
for(int i=0;i<n-1;i++) cout<<v[i]<<" ";
cout<<v[n-1]<<endl;
h.clear();
in.clear();
}
}
Here is a Python solution that uses only elementary list functions and takes care of ties.
def solution(heights, infronts):
person = list(zip(heights, infronts))
person.sort(key=lambda x: (x[0] == 0, x[1], -x[0]))
output = []
for p in person:
extended_output = output + [p]
extended_output.sort(key=lambda x: (x[0], -x[1]))
output_position = [p for p in extended_output].index(p) + p[1]
output.insert(output_position, p)
for c, p in enumerate(output):
taller_infronts = [infront for infront in output[0:c] if infront[0] >= p[0]]
assert len(taller_infronts) == p[1]
return output
Simple O(n^2) solution for this in Java:
Algorith:
If the position of the shortest person is i, i-1 taller people will be in front of him.
We fix the position of shortest person and then move to second shortest person.
Sort people by heights. Then iterate from shortest to tallest. In each step you need an efficient way to put the next person to the correct position.
We can optimise this solution even more by using segment tree. See this link.
class Person implements Comparable<Person>{
int height;
int pos;
Person(int height, int pos) {
this.height = height;
this.pos = pos;
}
#Override
public int compareTo(Person person) {
return this.height - person.height;
}
}
public class Solution {
public int[] order(int[] heights, int[] positions) {
int n = heights.length;
int[] ans = new int[n];
PriorityQueue<Person> pq = new PriorityQueue<Person>();
for( int i=0; i<n; i++) {
pq.offer(new Person(heights[i], positions[i]) );
}
for(int i=0; i<n; i++) {
Person person = pq.poll();
int vacantTillNow = 0;
int index = 0;
while(index < n) {
if( ans[index] == 0) vacantTillNow++;
if( vacantTillNow > person.pos) break;
index++;
}
ans[index] = person.height;
}
return ans;
}
}
Segment tree can be used to solve this in O(nlog n) if there are no ties in heights.
Please look for approach 3 in this link for a clear explanation of this method.
https://www.codingninjas.com/codestudio/problem-details/order-of-people-heights_1170764
Below is my code for the same approach in python
def findEmptySlot(tree, root, left, right, K, result):
tree[root]-=1
if left==right:
return left
if tree[2*root+1] >= K:
return findEmptySlot(tree, 2*root+1, left, (left+right)//2, K, result)
else:
return findEmptySlot(tree, 2*root+2, (left+right)//2+1, right, K-tree[2*root+1], result)
def buildsegtree(tree, pos, start, end):
if start==end:
tree[pos]=1
return tree[pos]
mid=(start+end)//2
left = buildsegtree(tree, 2*pos+1,start, mid)
right = buildsegtree(tree,2*pos+2,mid+1, end)
tree[pos]=left+right
return tree[pos]
class Solution:
# #param A : list of integers
# #param B : list of integers
# #return a list of integers
def order(self, A, B):
n=len(A)
people=[(A[i],B[i]) for i in range(len(A))]
people.sort(key=lambda x: (x[0], x[1]))
result=[0]*n
tree=[0]*(4*n)
buildsegtree(tree,0, 0, n-1)
for i in range(n):
idx=findEmptySlot(tree, 0, 0, n-1, people[i][1]+1, result)
result[idx]=people[i][0]
return result