This is my insertion sort code. This code works but is this the way to write an insertion sort? I looked few tutorials but they do it in a different way.
int thirdArray[] = { 0, 22, 1, 10, 8, 5,39 };
for (int i = 0; i < thirdArray.length - 1; i++) {
for (int j = i + 1; j > 0; j--) {
if (thirdArray[j] < thirdArray[j - 1]) {
int index = thirdArray[j];
thirdArray[j] = thirdArray[j - 1];
thirdArray[j - 1] = index;
}
}
}
for (int number : thirdArray) {
System.out.print(number + "\t");
}
With Java 7 its so simple to do sorting now days.
Arrays.sort(thirdArray);
This would sort the elements of the array sorted.
In your code you are taking extra space for the swap operation. However, It's a valid approach, but you can optmize it in the following way
int thirdArray[] = { 0, 22, 1, 10, 8, 5,39 };
int key,j;
for(int i = 1 ; i < thirdArray.length ; i++)
{
key = thirdArray[i];
j = i - 1;
while(j >=0 && key < thirdArray[j])
{
thirdArray[j+1] = thirdArray[j];
j--;
}
arr[j+1] = key;
}
}
In the above code, you start comparing from the first index with it's previous, if it's smaller, overwrite it's values (you are already saving the current value in key) in the end place it in it's correct position. We can guarantee that after the sorting that all the element on the left are sorted. Insertion sort has Worst-case performance: О(n^2) Best-case performance: O(n). By swapping we are adding more complexity which you can avoid it.
Related
I came across a Hackerearth coding problem where you have to perform the following tasks over an integer array-
Search for a particular number in the array and replace it's occurrences with 1
Move all the 1s to the first part of the array, maintaining the original order of the array
For example- if we have an integer array {22,1,34,22,16,22,35,1}, here we search for the number "22" (let us assume it is present in the array), replace it with 1 and move all those 1s (including the 1s already present) to the first part of the array and the resultant array should look like {1,1,1,1,1,1,34,16,35} -maintaining the original order of the array, preferably in Java.
I actually have coded a solution and it works fine but is not optimal, can anyone help me find an optimal solution (w.r.t. time-space complexity)?
Below is my solution-
public static void main(String[] args) {
int[] n = rearr(new int[] {22,1,34,22,16,22,1,34,1}, 22);
for(int i=0; i<n.length; i++) {
System.out.print(n[i]+" ");
}
}
static int[] rearr(int[] a, int x) {
int[] temp = new int[a.length];
int j=0, c=0, k=0;
//search and replace
for(int i=0; i<a.length; i++) {
if(a[i] == x) {
a[i] = 1;
}
}
//shift all 1s to first part of array or shift all non-1s to last part of the array
for(int i=0; i<a.length; i++) {
if(a[i] != 1) {
temp[j] = a[i];
j++;
}
if(a[i] == 1) {
c++;
}
}
j=0;
for(int i=0; i<a.length && c>0; i++, c--) {
a[i] = 1;
j++;
}
for(int i=j ;i<a.length; i++) {
a[i] = temp[k];
k++;
}
return a;
}
This can be done in linear time and space complexity, by returning a completely new list instead of modifying the original list.
static int[] rearr(int[] a, int x) {
// allocate the array we'll return
int[] b = new int[a.length];
int fillvalue = 1;
// iterate backwards through the list, and transplant every value OTHER than
// (x or 1) to the last open index in b, which we track with b_idx
int b_idx = b.length - 1;
for (int i = a.length - 1; i >= 0; i--) {
if (a[i] != x && a[i] != fillvalue)) {
b[b_idx] = a[i];
b_idx--;
}
}
// once we've gone through and done that, fill what remains of b with ones
// which are either original or are replacements, we don't care
for (int i = b_idx; i >= 0; i--) {
b[i] = fillvalue;
}
return b;
}
This is linear space complexity because it requires additional space equal to the size of the given list. It's linear time complexity because, in the worst case, it iterates over the size of the list exactly twice.
As a bonus, if we decide we want to leave the original 1s where they were, we can do that without any trouble at all, by simply modifying the if condition. Same if we decide we want to change the fill value to something else.
Doing this with constant space complexity would require O(n^2) list complexity, as it would require swapping elements in a to their proper positions. The easiest way to do that would probably be to do replacements on a first run through the list, and then do something like bubblesort to move all the 1s to the front.
This can be done in a single iteration through the array. We can use 2 pointer approach here where we will use on pointer to iterate through the array and other one to point to the index of 1 in the array.
The code is below:
public static void main(String[] args) {
// input array
int[] arr = { 22, 1, 34, 22, 16, 22, 35, 1, 20, 33, 136 };
// element to be replaced
int x = 22;
int j = -1;
for (int i = arr.length - 1; i >= 0; i--) {
if (arr[i] == 1 || arr[i] == x) {
if (j == -1) {
j = i;
}
// incase arr[i]==x
arr[i] = 1;
} else {
if (j != -1) {
arr[j] = arr[i];
arr[i] = 1;
j--;
}
}
}
for (int i = 0; i < arr.length; i++) {
System.out.println(arr[i]);
}
}
Here we initialise j=-1 since we consider there are no 1's present in the array.
Then we start iterating the array from the end towards the starting of the array as we have to push all the 1's to the starting of the array. Now when we reach to 1 or x (particular number in your case), we check if this is first occurrence of the x or 1, if yes then we initialise the j with this index and change arr[i] = 1 because this could be equal to x then we need to make it 1. If the arr[i] is not 1 or x it means its a number which we need to push at back of the array. We check if we have position of 1 or j=-1. If j=-1 it means this number is already pushed back at end of array else we swap the number at i and j, and decrement j by 1.
At the end of the array we will have the array sorted in a fashion which is required.
Time Complexity: Since we are only iterating the array one, hence the time complexity is O(n).
Space Complexity: Since there are no extra space being used or constant space being used hence the space complexity is O(1)
I need to implement a gnomesort to sort strings on how close they are to the string input. I measure this difference with the Levenshtein-algoritm.
The algoritm works fine but only with if I have two strings in the database. It then sorts it fine, but if there are more than two strings, it just prints them in the order they are in the database. I really can't find the problem
public static void retrieveFromDatabase(String string)
{
String[] sq = new String[database.size()];
database.toArray(sq);
int r = 0, index = 1, y = 2, tmp1 = 0;
String tmp2;
int[] ds = new int[sq.length];
for (int i = 0; i < database.size(); i++) {
ds[i] = sortLevenshtein(string, database.get(i), false);
}
for(index = 1; index < ds.length; index++) // gnomsort
{
if(ds[index - 1] <= ds[index] )
{
++index;
}
else
{
tmp1 = ds[index];
tmp2 = sq[index];
ds[index] = ds[index - 1];
sq[index] = sq[index - 1];
ds[index-1] = tmp1;
sq[index-1] = tmp2;
index--;
if (index == 0)
index++;
}
}
System.out.println("Best matches: ");
for(r=0; r<Math.min(3,sq.length); r++)
{
System.out.println(ds[r] + "\t" + sq[r]);
}
}
The problem
Your gnome sort is not sorting correctly when there are more than two elements to sort.
In your problematic case your ds contains 1, 1, 0 from the outset. In your for loop index is 1. You see that the elements at indices 0 and 1 are in the correct order (both elements are 1), so you increment index to 2 in the if statement. Next your for loop also increments index, so it is now 3. 3 is not less than ds.length (also 3), so the loop terminates.
I don’t know gnome sort, so I can’t tell you the fix. What I can tell you is that manipulating your for loop control variable — index in your code — inside the for loop is the sure way to code that is hard to understand and very hard to find errors in. I never ever do that.
for(index = 1; index < ds.length; index++) // OK: loop control variable is incremented here
{
if(ds[index - 1] <= ds[index] )
{
++index; // No-no: incrementing loop control variable, dangerous
}
else
{
tmp1 = ds[index];
tmp2 = sq[index];
ds[index] = ds[index - 1];
sq[index] = sq[index - 1];
ds[index-1] = tmp1;
sq[index-1] = tmp2;
index--; // No-no: decrementing loop control variable, problematic
if (index == 0)
index++; // No-no: incrementing loop control variable
}
}
I am trying to solve a challenge,
I wrote my solution and it passes all test cases except some hidden test cases. I can't think another case in which my method fails and don't know what to do anymore.
Here it is:
int firstDuplicate(int[] a) {
int[] indexCount;
int duplicate, temp;
boolean check;
duplicate = -1; temp = a.length;
indexCount = new int[a.length];
check = false;
for( int i = 0; i < a.length; i++ ){
if( indexCount[a[i]-1] == 0 ){
indexCount[a[i]-1] = i+1;
check = false;
}else{
indexCount[a[i]-1] = (i+1) - indexCount[a[i]-1];
check = true;
}
if( check && indexCount[a[i]-1] < temp ){
duplicate = a[i];
temp = indexCount[a[i]-1];
}
}
return duplicate;
}
Instructions are:
Write a solution with O(n) time complexity and O(1) additional space complexity.
Given an array a that contains only numbers in the range from 1 to a.length, find the first duplicate number for which the second occurrence has the minimal index.
Example
For a = [2, 3, 3, 1, 5, 2], the output should be
firstDuplicate(a) = 3.
There are 2 duplicates: numbers 2 and 3. The second occurrence of 3 has a smaller index than than second occurrence of 2 does, so the answer is 3.
For a = [2, 4, 3, 5, 1], the output should be
firstDuplicate(a) = -1.
Here is what I have. Runs in O(n) and uses O(1) space. Correct me if I'm wrong here.
Since my input cannot have a value that's more than the length, I can use mod operator for indexing on the same array and add the length to the value in index. As soon as I encounter a value that larger than the length, that means I've already incremented that before, which gives me the duplicate value.
public int firstDuplicate(int[] arr) {
int length = arr.length;
for (int i = 0; i < length; i++) {
int expectedIndex = arr[i] % length;
if (arr[expectedIndex] > length) {
return arr[i] > length ? arr[i] - length : arr[i];
} else {
arr[expectedIndex] += length;
}
}
return -1;
}
This answer is based on #Mehmet-Y's answer and all credit goes to Mehmet-Y. This version addresses the three issues I pointed out in the comments. I will delete this answer if the original gets corrected.
The general approach is to use the original array for storage instead of allocating a new one. The fact that no value may be less than one or greater than the length suggests that you can use the array as a set of indices to flag an element as "already seen" by either negating it or adding/subtracting the array length to/from it.
To achieve O(n) time complexity, you have to solve the problem in a fixed number of passes (not necessarily one pass: the number just can't depend on the size of the array).
But how do you decide which duplicate has the smallest second index? I would suggest using two different flags to indicate an index that is already seen vs. the second item in a duplicate pair. For this example, we can set the index flag by incrementing the elements by the length, and marking duplicates by negating them. You will need a second pass to find the first negagive in the array. You can also use that pass to restore the elements to their original values without sacrificing O(n) time complexity.
Here is a sample implementation:
int firstDuplicate(int[] a)
{
// assume all elements of a are in range [1, a.length]
// An assertion of that would not increase the time complexity from O(n)
int len = a.length;
for(int i = 0; i < len; i++) {
// a[i] may be > len, but not negative.
// Index of bin to check if this element is already seen.
flagIndex = (a[i] - 1) % len;
if(a[flagIndex] > len) {
// If already seen, current element is the second of the pair.
// It doesn't matter if we flag the third duplicate,
// just as long as we don't tag the first be accident.
a[i] = -a[i];
} else {
// Flag the element as "already seen".
// This can be done outside the else, but you might run
// into (more) overflow problems with large arrays.
a[flagIndex] += len;
}
}
// Search and stash index of first negative number
for(int i = 0; i < len; i++) {
if(a[i] < 0) {
return -a[i] % len;
}
}
// Nothing found, oh well
return -1;
}
If you want to take advantage of the second pass to restore the original values of the array, replace
for(int i = 0; i < len; i++) {
if(a[i] < 0) {
return -a[i] % len;
}
}
return -1;
with
int duplicate = -1;
for(int i = 0; i < len; i++) {
if(a[i] < 0) {
a[i] = -a[i];
if(duplicate == -1) {
duplicate = a[i] % len;
}
}
a[i] %= len;
}
return duplicate;
I'm trying to solve the problem below from CodeFights. I left my answer in Java after the question. The code works for all the problems, except the last one. Time limit exception is reported. What could I do to make it run below 3000ms (CodeFights requirement)?
Note: Write a solution with O(n) time complexity and O(1) additional space complexity, since this is what you would be asked to do during a real interview.
Given an array a that contains only numbers in the range from 1 to a.length, find the first duplicate number for which the second occurrence has the minimal index. In other words, if there are more than 1 duplicated numbers, return the number for which the second occurrence has a smaller index than the second occurrence of the other number does. If there are no such elements, return -1.
Example
For a = [2, 3, 3, 1, 5, 2], the output should be
firstDuplicate(a) = 3.
There are 2 duplicates: numbers 2 and 3. The second occurrence of 3 has a smaller index than than second occurrence of 2 does, so the answer is 3.
For a = [2, 4, 3, 5, 1], the output should be
firstDuplicate(a) = -1.
Input/Output
[time limit] 3000ms (java)
[input] array.integer a
Guaranteed constraints:
1 ≤ a.length ≤ 105,
1 ≤ a[i] ≤ a.length.
[output] integer
The element in a that occurs in the array more than once and has the minimal index for its second occurrence. If there are no such elements, return -1.
int storedLeastValue = -1;
int indexDistances = Integer.MAX_VALUE;
int indexPosition = Integer.MAX_VALUE;
for (int i = 0; i < a.length; i++)
{
int tempValue = a[i];
for (int j = i+1; j < a.length; j++) {
if(tempValue == a[j])
{
if(Math.abs(i-j) < indexDistances &&
j < indexPosition)
{
storedLeastValue = tempValue;
indexDistances = Math.abs(i-j);
indexPosition = j;
break;
}
}
}
}
return storedLeastValue;
Your solution has two nested for loops which implies O(n^2) while the question explicitly asks for O(n). Since you also have a space restriction you can't use an additional Set (which can provide a simple solution as well).
This question is good for people that have strong algorithms/graph theory background. The solution is sophisticated and includes finding an entry point for a cycle in a directed graph. If you're not familiar with these terms I'd recommend that you'll leave it and move to other questions.
Check this one, it's also O(n) , but without additional array.
int firstDuplicate(int[] a) {
if (a.length <= 1) return -1;
for (int i = 0; i < a.length; i++) {
int pos = Math.abs(a[i]) - 1;
if (a[pos] < 0) return pos + 1;
else a[pos] = -a[pos];
}
return -1;
}
The accepted answer does not work with the task.
It would work if the input array would indeed contain no bigger value than its length.
But it does, eg.: [5,5].
So, we have to define which number is the biggest.
int firstDuplicate(int[] a) {
int size = 0;
for(int i = 0; i < a.length; i++) {
if(a[i] > size) {
size = a[i];
}
}
int[] t = new int[size+1];
for(int i = 0; i < a.length; i++) {
if(t[a[i]] == 0) {
t[a[i]]++;
} else {
return a[i];
}
}
return -1;
}
What about this:
public static void main(String args[]) {
int [] a = new int[] {2, 3, 3, 1, 5, 2};
// Each element of cntarray will hold the number of occurrences of each potential number in the input (cntarray[n] = occurrences of n)
// Default initialization to zero's
int [] cntarray = new int[a.length + 1]; // need +1 in order to prevent index out of bounds errors, cntarray[0] is just an empty element
int min = -1;
for (int i=0;i < a.length ;i++) {
if (cntarray[a[i]] == 0) {
cntarray[a[i]]++;
} else {
min = a[i];
// no need to go further
break;
}
}
System.out.println(min);
}
You can store array values in hashSet. Check if value is already present in hashSet if not present then add it in hashSet else that will be your answer. Below is code which passes all test cases:-
int firstDuplicate(int[] a) {
HashSet<Integer> hashSet = new HashSet<>();
for(int i=0; i<a.length;i++){
if (! hashSet.contains(a[i])) {
hashSet.add(a[i]);
} else {
return a[i];
}
}
return -1;
}
My simple solution with a HashMap
int solution(int[] a) {
HashMap<Integer, Integer> countMap = new HashMap<Integer, Integer>();
int min = -1;
for (int i=0; i < a.length; i++) {
if (!(countMap.containsKey(a[i]))) {
countMap.put(a[i],1);
}
else {
return a[i];
}
}
return min;
}
Solution is very simple:
Create a hashset
keep iterating over the array
if element is already not in the set, add it.
else element will be in the set, then it mean this is minimal index of first/second the duplicate
int solution(int[] a) {
HashSet<Integer> set = new HashSet<>();
for(int i=0; i<a.length; i++){
if(set.contains(a[i])){
// as soon as minimal index duplicate found where first one was already in the set, return it
return a[i];
}
set.add(a[i]);
}
return -1;
}
A good answer for this exercise can be found here - https://forum.thecoders.org/t/an-interesting-coding-problem-in-codefights/163 - Everything is done in-place, and it has O(1) solution.
Problem H (Longest Natural Successors):
Two consecutive integers are natural successors if the second is the successor of the first in the sequence of natural numbers (1 and 2 are natural successors). Write a program that reads a number N followed by N integers, and then prints the length of the longest sequence of consecutive natural successors.
Example:
Input 7 2 3 5 6 7 9 10 Output 3 this is my code so far and i have no idea why it does not work
import java.util.Scanner;
public class Conse {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int x = scan.nextInt();
int[] array = new int[x];
for (int i = 0; i < array.length; i++) {
array[i] = scan.nextInt();
}
System.out.println(array(array));
}
public static int array(int[] array) {
int count = 0, temp = 0;
for (int i = 0; i < array.length; i++) {
count = 0;
for (int j = i, k = i + 1; j < array.length - 1; j++, k++) {
if (Math.abs(array[j] - array[k]) == 1) {
count++;
} else {
if (temp <= count) {
temp = count;
}
break;
}
}
}
return temp + 1;
}
}
Why two loops? What about
public static int array(final int[] array) {
int lastNo = -100;
int maxConsecutiveNumbers = 0;
int currentConsecutiveNumbers = 0;
for (int i = 0; i < array.length; i++) {
if (array[i] == lastNo + 1) {
currentConsecutiveNumbers++;
maxConsecutiveNumbers = Math.max(maxConsecutiveNumbers,
currentConsecutiveNumbers);
} else {
currentConsecutiveNumbers = 1;
}
lastNo = array[i];
}
return Math.max(maxConsecutiveNumbers, currentConsecutiveNumbers);
}
This seems to work:
public static int longestConsecutive(int[] array) {
int longest = 0;
// For each possible start
for (int i = 0; i < array.length; i++) {
// Count consecutive.
for (int j = i + 1; j < array.length; j++) {
// This one consecutive to last?
if (Math.abs(array[j] - array[j - 1]) == 1) {
// Is it longer?
if (j - i > longest) {
// Yup! Remember it.
longest = j - i;
}
} else {
// Start again.
break;
}
}
}
return longest + 1;
}
public void test() {
int[] a = new int[]{7, 2, 3, 5, 6, 7, 9, 10};
System.out.println("Longest: " + Arrays.toString(a) + "=" + longestConsecutive(a));
}
prints
Longest: [7, 2, 3, 5, 6, 7, 9, 10]=3
Since your question has "Problem H" associated with it, I'm assuming you are just learning. Simpler is always better, so it usually pays to break it down into "what has to be done" before starting on a particular road by writing code that approaches the problem with "how can this be done."
In this case, you may be over-complicating things with arrays. A number is a natural successor if it is one greater than the previous number. If this is true, increment the count of the current sequence. If not, we're starting a new sequence. If the current sequence length is greater than the maximum sequence length we've seen, set the max sequence length to the current sequence length. No arrays needed - you only need to compare two numbers (current and last numbers read).
For example:
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int N = scan.nextInt();
int maxSequenceLen = 0; // longest sequence ever
int curSequenceLen = 0; // when starting new sequence, reset to 1 (count the reset #)
int last = 0;
for(int i = 0; i < N; i++) {
int cur = scan.nextInt();
if ((last+1) == cur){
++curSequenceLen;
}
else{
curSequenceLen = 1;
}
if (curSequenceLen > maxSequenceLen){
maxSequenceLen = curSequenceLen;
}
last = cur;
}
System.out.println(maxSequenceLen);
Caveat: I'm answering this on a computer that does not have my Java development environment on it, so the code is untested.
I'm not sure I understand this question correctly. The answer's written here assumes that the the natural successors occur contiguously. But if this is not the same then the solution here might not give the correct answer.
Suppose instead of [7 2 3 5 6 7 9 10] the input was [7 2 6 3 7 5 6 9 10] then the answer becomes 2 while the natural successor [5 6 7] is present in the array.
If the input is not sorted we'll have to use a different approach. Like using HashSet
Load the entire array into a HashSet which removes duplicates.
Pick the first value from the HashSet and assigned it to start and end and remove it from the set.
Now decrements start and check if it is present in the HashSet and continue till a particular value for start is not present int the HashSetwhile removing the value being searched from the set.
Do the same for end except that you will have to increase the value of end for each iteration.
We now have to continuous range from start to end present in the set and whose range is current_Max = end - start + 1
In each iteration we keep track of this current_Max to arrive at the longest natural successor for the entire array.
And since HashSet supports Add, Remove, Update in O(1) time. This algorithm will run in O(n) time, where n is the length of the input array.
The code for this approach in C# can be found here