Develop Reference

unable to sort or print out array

I got a task to write the methods to 1. take information in and save it into array, 2. to copy the array(only with previously given arguments) and 3. to sort the array to be in descending order. it seems to me like i got the first 2 parts working but i have been stuck on the third method for 2 days without any progress. i am still learning java so i'm pretty sure i have made a dumb mistake somewhere. thanks in advance.
import java.util.*;
public class RevisionExercise {
public static void main(String[] args) {
int[] tempArray = new int[100];
System.out.println("Type in numbers. Type zero to quit.");
int amountOfNumbers = askInfo(tempArray);
int[] realArray = new int[amountOfNumbers];
copyInfo(realArray, tempArray);
setArray(realArray);
printArray(realArray);
}
public static int askInfo(int[] tempArray) {
Scanner reader = new Scanner(System.in);
int i;
for (i = 0; i < tempArray.length; i++) {
System.out.print((i+1) + ". number: ");
tempArray[i] = reader.nextInt();
if (tempArray[i] == 0) {
return tempArray[i];
}
}
return i;
}
private static int[] copyInfo(int[] realArray, int[] tempArray)
{
int targetIndex = 0;
for( int sourceIndex = 0; sourceIndex < tempArray.length; sourceIndex++ )
{
if( tempArray[sourceIndex] != 0 )
tempArray[targetIndex++] = tempArray[sourceIndex];
}
realArray = new int[targetIndex];
System.arraycopy( tempArray, 0, realArray, 0, targetIndex );
return realArray;
}
public static int[] setArray(int[] realArray)
{
int temp = 0;
for (int i = 0; i <realArray.length; i++) {
for (int j = i+1; j <realArray.length; j++) {
if(realArray[i] >realArray[j]) {
temp = realArray[i];
realArray[i] = realArray[j];
realArray[j] = temp;
}
}
}
return realArray;
}
public static void printArray(int[] realArray ) {
System.out.println("\nOrdered array: ");
for(int i = 0; i < realArray .length; i++) {
System.out.println(realArray [i]);
}
}
}
the output i'm getting looks like this.
Type in numbers. Type zero to quit.
1. number: 3
2. number: 8
3. number: 5
4. number: 6
5. number: 9
6. number: 0
Ordered array:
while it should look like this:
Type in numbers. Type zero to quit.
1. number: 3
2. number: 8
3. number: 5
4. number: 6
5. number: 9
6. number: 0
Ordered array:
9
8
6
5
3

You made a mistake in calculating the total numbers entered by the user. Your function always return 0. So change it to something like this:
public static int askInfo(int[] tempArray) {
Scanner reader = new Scanner(System.in);
int totalNumbers = 0;
for (int i = 0; i < tempArray.length; i++) {
System.out.print((i+1) + ". number: ");
int number = reader.nextInt();
if (number != 0) {
totalNumbers++;
tempArray[i] = number;
} else {
break;
}
}
return totalNumbers;
}
Chage the sort function to this:
public static void setArray(int[] realArray) {
int temp = 0;
for (int i = 0; i < realArray.length; i++) {
for (int j = i+1; j < realArray.length; j++) {
// Use < for descending order and > for ascending order
if(realArray[i] < realArray[j]) {
temp = realArray[i];
realArray[i] = realArray[j];
realArray[j] = temp;
}
}
}
}
Your main program
public static void main(String[] args) {
int[] tempArray = new int[100];
System.out.println("Type in numbers. Type zero to quit.");
int amountOfNumbers = askInfo(tempArray);
int[] realArray = new int[amountOfNumbers];
copyInfo(tempArray, realArray, amountOfNumbers);
setArray(realArray);
printArray(realArray);
}
public static int askInfo(int[] tempArray) {
Scanner reader = new Scanner(System.in);
int totalNumbers = 0;
for (int i = 0; i < tempArray.length; i++) {
System.out.print((i+1) + ". number: ");
int number = reader.nextInt();
if (number != 0) {
totalNumbers++;
tempArray[i] = number;
} else {
break;
}
}
return totalNumbers;
}
private static void copyInfo(int[] tempArray, int[] realArray, int size) {
for( int sourceIndex = 0; sourceIndex < size; sourceIndex++ )
{
realArray[sourceIndex] = tempArray[sourceIndex];
}
}
public static void setArray(int[] realArray) {
int temp = 0;
for (int i = 0; i < realArray.length; i++) {
for (int j = i+1; j < realArray.length; j++) {
if(realArray[i] < realArray[j]) {
temp = realArray[i];
realArray[i] = realArray[j];
realArray[j] = temp;
}
}
}
}
public static void printArray(int[] realArray ) {
System.out.println("\nOrdered array: ");
for(int i = 0; i < realArray.length; i++) {
System.out.println(realArray [i]);
}
}

To begin with, in the askInfo method, the size of the realArray will always be 0, since amountOfNumbers is the return of
if (tempArray[i] == 0) {
return tempArray[i];
}
because your condition is tempArray[i] == 0, hence you are always returning 0. Then your copyInfo method is returning an array but you have nothing receiving the returned array. I would suggest changing your copyInfo method to
private static int[] copyInfo(int[] tempArray)
{
// same
int [] realArray = new int[targetIndex];
// same
}
and your main method to
public static void main(String[] args) {
//same
int[] realArray = new int[amountOfNumbers];
realArray = copyInfo(tempArray);
// same
}

General remark: Each of your methods returns an array but you never use this return value.
There are 2 modifications necessary to fix your code:
Keep the value of the method output:
realArray=copyInfo(realArray, tempArray);
So you put the result of the copy in the realArray variable.
You expect the askInfo method to return the number of entered numbers but in fact you return the value of the last entered number.
So instead of return tempArray[i]; you have to return i;

Finding the smallest integer that appears at least k times

You are given an array A of integers and an integer k. Implement an algorithm that determines, in linear time, the smallest integer that appears at least k times in A.
I have been struggling with this problem for awhile, coding in Java, I need to use a HashTable to find the smallest integer that appears at least k times, it also must be in linear time.
This is what I attempted but it does not pass any of the tests
private static int problem1(int[] arr, int k)
{
// Implement me!
HashMap<Integer, Integer> table = new HashMap<Integer, Integer>();
int ans = Integer.MAX_VALUE;
for (int i=0; i < arr.length; i++) {
if(table.containsKey(arr[i])) {
table.put(arr[i], table.get(arr[i]) + 1);
if (k <= table.get(arr[i])) {
ans = Math.min(ans, arr[i]);
}
}else{
table.put(arr[i], 1);
}
}
return ans;
}
Here is the empty code with all of the test cases:
import java.io.*;
import java.util.*;
public class Lab5
{
/**
* Problem 1: Find the smallest integer that appears at least k times.
*/
private static int problem1(int[] arr, int k)
{
// Implement me!
return 0;
}
/**
* Problem 2: Find two distinct indices i and j such that A[i] = A[j] and |i - j| <= k.
*/
private static int[] problem2(int[] arr, int k)
{
// Implement me!
int i = -1;
int j = -1;
return new int[] { i, j };
}
// ---------------------------------------------------------------------
// Do not change any of the code below!
private static final int LabNo = 5;
private static final String quarter = "Fall 2020";
private static final Random rng = new Random(123456);
private static boolean testProblem1(int[][] testCase)
{
int[] arr = testCase[0];
int k = testCase[1][0];
int answer = problem1(arr.clone(), k);
Arrays.sort(arr);
for (int i = 0, j = 0; i < arr.length; i = j)
{
for (; j < arr.length && arr[i] == arr[j]; j++) { }
if (j - i >= k)
{
return answer == arr[i];
}
}
return false; // Will never happen.
}
private static boolean testProblem2(int[][] testCase)
{
int[] arr = testCase[0];
int k = testCase[1][0];
int[] answer = problem2(arr.clone(), k);
if (answer == null || answer.length != 2)
{
return false;
}
Arrays.sort(answer);
// Check answer
int i = answer[0];
int j = answer[1];
return i != j
&& j - i <= k
&& i >= 0
&& j < arr.length
&& arr[i] == arr[j];
}
public static void main(String args[])
{
System.out.println("CS 302 -- " + quarter + " -- Lab " + LabNo);
testProblems(1);
testProblems(2);
}
private static void testProblems(int prob)
{
int noOfLines = prob == 1 ? 100000 : 500000;
System.out.println("-- -- -- -- --");
System.out.println(noOfLines + " test cases for problem " + prob + ".");
boolean passedAll = true;
for (int i = 1; i <= noOfLines; i++)
{
int[][] testCase = null;
boolean passed = false;
boolean exce = false;
try
{
switch (prob)
{
case 1:
testCase = createProblem1(i);
passed = testProblem1(testCase);
break;
case 2:
testCase = createProblem2(i);
passed = testProblem2(testCase);
break;
}
}
catch (Exception ex)
{
passed = false;
exce = true;
}
if (!passed)
{
System.out.println("Test " + i + " failed!" + (exce ? " (Exception)" : ""));
passedAll = false;
break;
}
}
if (passedAll)
{
System.out.println("All test passed.");
}
}
private static int[][] createProblem1(int testNo)
{
int size = rng.nextInt(Math.min(1000, testNo)) + 5;
int[] numbers = getRandomNumbers(size, size);
Arrays.sort(numbers);
int maxK = 0;
for (int i = 0, j = 0; i < size; i = j)
{
for (; j < size && numbers[i] == numbers[j]; j++) { }
maxK = Math.max(maxK, j - i);
}
int k = rng.nextInt(maxK) + 1;
shuffle(numbers);
return new int[][] { numbers, new int[] { k } };
}
private static int[][] createProblem2(int testNo)
{
int size = rng.nextInt(Math.min(1000, testNo)) + 5;
int[] numbers = getRandomNumbers(size, size);
int i = rng.nextInt(size);
int j = rng.nextInt(size - 1);
if (i <= j) j++;
numbers[i] = numbers[j];
return new int[][] { numbers, new int[] { Math.abs(i - j) } };
}
private static void shuffle(int[] arr)
{
for (int i = 0; i < arr.length - 1; i++)
{
int rndInd = rng.nextInt(arr.length - i) + i;
int tmp = arr[i];
arr[i] = arr[rndInd];
arr[rndInd] = tmp;
}
}
private static int[] getRandomNumbers(int range, int size)
{
int numbers[] = new int[size];
for (int i = 0; i < size; i++)
{
numbers[i] = rng.nextInt(2 * range) - range;
}
return numbers;
}
}

private static int problem1(int[] arr, int k) {
// Implement me!
Map<Integer, Integer> table = new TreeMap<Integer, Integer>();
for (int i = 0; i < arr.length; i++) {
if (table.containsKey(arr[i])) {
table.put(arr[i], table.get(arr[i]) + 1);
} else {
table.put(arr[i], 1);
}
}
for (Map.Entry<Integer,Integer> entry : table.entrySet()) {
//As treemap is sorted, we return the first key with value >=k.
if(entry.getValue()>=k)
return entry.getKey();
}
//Not found
return -1;
}

As others have pointed out, there are a few mistakes. First, the line where you initialize ans,
int ans = 0;
You should initialize ans to Integer.MAX_VALUE so that when you find an integer that appears at least k times for the first time that ans gets set to that integer appropriately. Second, in your for loop, there's no reason to skip the first element while iterating the array so i should be initialized to 0 instead of 1. Also, in that same line, you want to iterate through the entire array, and in your loop's condition right now you have i < k when k is not the length of the array. The length of the array is denoted by arr.length so the condition should instead be i < arr.length. Third, in this line,
if (k < table.get(arr[i])){
where you are trying to check if an integer has occurred at least k times in the array so far while iterating through the array, the < operator should be changed to <= since the keyword here is at least k times, not "more than k times". Fourth, k should never change so you can get rid of this line of code,
k = table.get(arr[i]);
After applying all of those changes, your function should look like this:
private static int problem1(int[] arr, int k)
{
// Implement me!
HashMap<Integer, Integer> table = new HashMap<Integer, Integer>();
int ans = Integer.MAX_VALUE;
for (int i=0; i < arr.length; i++) {
if(table.containsKey(arr[i])) {
table.put(arr[i], table.get(arr[i]) + 1);
if (k <= table.get(arr[i])) {
ans = Math.min(ans, arr[i]);
}
}else{
table.put(arr[i], 1);
}
}
return ans;
}

Pseudo code:
collect frequencies of each number in a Map<Integer, Integer> (number and its count)
set least to a large value
iterate over entries
ignore entry if its value is less than k
if entry key is less than current least, store it as least
return least
One line implementation:
private static int problem1(int[] arr, int k) {
return Arrays.stream(arr).boxed()
.collect(groupingBy(identity(), counting()))
.entrySet().stream()
.filter(entry -> entry.getValue() >= k)
.map(Map.Entry::getKey)
.reduce(MAX_VALUE, Math::min);
}

This was able to pass all the cases! Thank you to everyone who helped!!
private static int problem1(int[] arr, int k)
{
// Implement me!
HashMap<Integer, Integer> table = new HashMap<Integer, Integer>();
int ans = Integer.MAX_VALUE;
for (int i=0; i < arr.length; i++) {
if(table.containsKey(arr[i])) {
table.put(arr[i], table.get(arr[i]) + 1);
}else{
table.put(arr[i], 1);
}
}
Set<Integer> keys = table.keySet();
for(int i : keys){
if(table.get(i) >= k){
ans = Math.min(ans,i);
}
}
if(ans != Integer.MAX_VALUE){
return ans;
}else{
return 0;
}
}

Find smallest number K , if exists, such that product of its digits is N. Eg:when N = 6, smallest number is k=16(1*6=6) and not k=23(2*3=6)

I have made this program using array concept in java. I am getting Exception as ArrayIndexOutOfBound while trying to generate product.
I made the function generateFNos(int max) to generate factors of the given number. For example a number 6 will have factors 1,2,3,6. Now,i tried to combine the first and the last digit so that the product becomes equal to 6.
I have not used the logic of finding the smallest number in that array right now. I will do it later.
Question is Why i am getting Exception as ArrayIndexOutOfBound? [i couldn't figure out]
Below is my code
public class SmallestNoProduct {
public static void generateFNos(int max) {
int ar[] = new int[max];
int k = 0;
for (int i = 1; i <= max; i++) {
if (max % i == 0) {
ar[k] = i;
k++;
}
}
smallestNoProduct(ar);
}
public static void smallestNoProduct(int x[]) {
int j[] = new int[x.length];
int p = x.length;
for (int d = 0; d < p / 2;) {
String t = x[d++] + "" + x[p--];
int i = Integer.parseInt(t);
j[d] = i;
}
for (int u = 0; u < j.length; u++) {
System.out.println(j[u]);
}
}
public static void main(String s[]) {
generateFNos(6);
}
}
****OutputShown****
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 6
at SmallestNoProduct.smallestNoProduct(SmallestNoProduct.java:36)
at SmallestNoProduct.generateFNos(SmallestNoProduct.java:27)
at SmallestNoProduct.main(SmallestNoProduct.java:52)
#Edit
The improved Code using array only.
public class SmallestNoProduct {
public static void generateFNos(int max) {
int s = 0;
int ar[] = new int[max];
int k = 0;
for (int i = 1; i <= max; i++) {
if (max % i == 0) {
ar[k] = i;
k++;
s++;
}
}
for (int g = 0; g < s; g++) {
System.out.println(ar[g]);
}
smallestNoProduct(ar, s);
}
public static void smallestNoProduct(int x[], int s) {
int j[] = new int[x.length];
int p = s - 1;
for (int d = 0; d < p;) {
String t = x[d++] + "" + x[p--];
System.out.println(t);
int i = Integer.parseInt(t);
j[d] = i;
}
/*for (int u = 0; u < j.length; u++) {
System.out.println(j[u]);
}*/
}
public static void main(String s[]) {
generateFNos(6);
}
}

Maybe it better:
public class SmallestNoProduct {
public static int smallest(int n) {
int small = n*n;
for(int i = 1; i < Math.sqrt(n); i++) {
if(n%i == 0) {
int temp = Integer.parseInt(""+i+""+n/i);
int temp2 = Integer.parseInt(""+n/i+""+i);
temp = temp2 < temp? temp2: temp;
if(temp < small) {
small = temp;
}
}
}
return small;
}
public static void main(String[] args) {
System.out.println(smallest(6)); //6
System.out.println(smallest(10)); //25
System.out.println(smallest(100)); //205
}
}

Problem lies in this line
String t=x[d++]+""+x[p--];
x[p--] will try to fetch 7th position value, as p is length of array x i.e. 6 which results in ArrayIndexOutOfBound exception. Array index starts from 0, so max position is 5 and not 6.
You can refer this question regarding postfix expression.
Note: I haven't checked your logic, this answer is only to point out the cause of exception.

We are unnecessarily using array here...
below method should work....
public int getSmallerMultiplier(int n)
{
if(n >0 && n <10) // if n is 6
return (1*10+n); // it will be always (1*10+6) - we cannot find smallest number than this
else
{
int number =10;
while(true)
{
//loop throuogh the digits of n and check for their multiplication
number++;
}
}
}

int num = n;
for(i=9;i>1;i--)
{
while(n%d==0)
{
n=n/d;
arr[i++] = d;
}
}
if(num<=9)
arr[i++] = 1;
//printing array in reverse order;
for(j=i-1;j>=0;j--)
system.out.println(arr[j]);

Optimizing N queens puzzle

I'm trying to solve the problem of positioning N queens on NxN board without row, column and diagonal conflicts. I use an algorithm with minimizing the conflicts. Firstly, on each column randomly a queen is positioned. After that, of all conflict queens randomly one is chosen and for her column are calculated the conflicts of each possible position. Then, the queen moves to the best position with min number of conflicts. It works, but it runs extremely slow. My goal is to make it run fast for 10000 queens. Would you, please, suggest me some improvements or maybe notice some mistakes in my logic?
Here is my code:
public class Queen {
int column;
int row;
int d1;
int d2;
public Queen(int column, int row, int d1, int d2) {
super();
this.column = column;
this.row = row;
this.d1 = d1;
this.d2 = d2;
}
#Override
public String toString() {
return "Queen [column=" + column + ", row=" + row + ", d1=" + d1
+ ", d2=" + d2 + "]";
}
#Override
public boolean equals(Object obj) {
return ((Queen)obj).column == this.column && ((Queen)obj).row == this.row;
}
}
And:
import java.util.HashSet;
import java.util.Random;
public class SolveQueens {
public static boolean printBoard = false;
public static int N = 100;
public static int maxSteps = 2000000;
public static int[] queens = new int[N];
public static Random random = new Random();
public static HashSet<Queen> q = new HashSet<Queen>();
public static HashSet rowConfl[] = new HashSet[N];
public static HashSet d1Confl[] = new HashSet[2*N - 1];
public static HashSet d2Confl[] = new HashSet[2*N - 1];
public static void init () {
int r;
rowConfl = new HashSet[N];
d1Confl = new HashSet[2*N - 1];
d2Confl = new HashSet[2*N - 1];
for (int i = 0; i < N; i++) {
r = random.nextInt(N);
queens[i] = r;
Queen k = new Queen(i, r, i + r, N - 1 + i - r);
q.add(k);
if (rowConfl[k.row] == null) {
rowConfl[k.row] = new HashSet<Queen>();
}
if (d1Confl[k.d1] == null) {
d1Confl[k.d1] = new HashSet<Queen>();
}
if (d2Confl[k.d2] == null) {
d2Confl[k.d2] = new HashSet<Queen>();
}
((HashSet<Queen>)rowConfl[k.row]).add(k);
((HashSet<Queen>)d1Confl[k.d1]).add(k);
((HashSet<Queen>)d2Confl[k.d2]).add(k);
}
}
public static void print () {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
System.out.print(queens[i] == j ? "♕ " : "◻◻◻ ");
}
System.out.println();
}
System.out.println();
}
public static boolean checkItLinear() {
Queen r = choseConflictQueen();
if (r == null) {
return true;
}
Queen newQ = findNewBestPosition(r);
q.remove(r);
q.add(newQ);
rowConfl[r.row].remove(r);
d1Confl[r.d1].remove(r);
d2Confl[r.d2].remove(r);
if (rowConfl[newQ.row] == null) {
rowConfl[newQ.row] = new HashSet<Queen>();
}
if (d1Confl[newQ.d1] == null) {
d1Confl[newQ.d1] = new HashSet<Queen>();
}
if (d2Confl[newQ.d2] == null) {
d2Confl[newQ.d2] = new HashSet<Queen>();
}
((HashSet<Queen>)rowConfl[newQ.row]).add(newQ);
((HashSet<Queen>)d1Confl[newQ.d1]).add(newQ);
((HashSet<Queen>)d2Confl[newQ.d2]).add(newQ);
queens[r.column] = newQ.row;
return false;
}
public static Queen choseConflictQueen () {
HashSet<Queen> conflictSet = new HashSet<Queen>();
boolean hasConflicts = false;
for (int i = 0; i < 2*N - 1; i++) {
if (i < N && rowConfl[i] != null) {
hasConflicts = hasConflicts || rowConfl[i].size() > 1;
conflictSet.addAll(rowConfl[i]);
}
if (d1Confl[i] != null) {
hasConflicts = hasConflicts || d1Confl[i].size() > 1;
conflictSet.addAll(d1Confl[i]);
}
if (d2Confl[i] != null) {
hasConflicts = hasConflicts || d2Confl[i].size() > 1;
conflictSet.addAll(d2Confl[i]);
}
}
if (hasConflicts) {
int c = random.nextInt(conflictSet.size());
return (Queen) conflictSet.toArray()[c];
}
return null;
}
public static Queen findNewBestPosition(Queen old) {
int[] row = new int[N];
int min = Integer.MAX_VALUE;
int minInd = old.row;
for (int i = 0; i < N; i++) {
if (rowConfl[i] != null) {
row[i] = rowConfl[i].size();
}
if (d1Confl[old.column + i] != null) {
row[i] += d1Confl[old.column + i].size();
}
if (d2Confl[N - 1 + old.column - i] != null) {
row[i] += d2Confl[N - 1 + old.column - i].size();
}
if (i == old.row) {
row[i] = row[i] - 3;
}
if (row[i] <= min && i != minInd) {
min = row[i];
minInd = i;
}
}
return new Queen(old.column, minInd, old.column + minInd, N - 1 + old.column - minInd);
}
public static void main(String[] args) {
long startTime = System.currentTimeMillis();
init();
int steps = 0;
while(!checkItLinear()) {
if (++steps > maxSteps) {
init();
steps = 0;
}
}
long endTime = System.currentTimeMillis();
System.out.println("Done for " + (endTime - startTime) + "ms\n");
if(printBoard){
print();
}
}
}
Edit:
Here is my a-little-bit-optimized solution with removing some unused objects and putting the queens on diagonal positions when initializing.
import java.util.Random;
import java.util.Vector;
public class SolveQueens {
public static boolean PRINT_BOARD = true;
public static int N = 10;
public static int MAX_STEPS = 5000;
public static int[] queens = new int[N];
public static Random random = new Random();
public static int[] rowConfl = new int[N];
public static int[] d1Confl = new int[2*N - 1];
public static int[] d2Confl = new int[2*N - 1];
public static Vector<Integer> conflicts = new Vector<Integer>();
public static void init () {
random = new Random();
for (int i = 0; i < N; i++) {
queens[i] = i;
}
}
public static int getD1Pos (int col, int row) {
return col + row;
}
public static int getD2Pos (int col, int row) {
return N - 1 + col - row;
}
public static void print () {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
System.out.print(queens[i] == j ? "Q " : "* ");
}
System.out.println();
}
System.out.println();
}
public static boolean hasConflicts() {
generateConflicts();
if (conflicts.isEmpty()) {
return false;
}
int r = random.nextInt(conflicts.size());
int conflQueenCol = conflicts.get(r);
int currentRow = queens[conflQueenCol];
int bestRow = currentRow;
int minConfl = getConflicts(conflQueenCol, queens[conflQueenCol]) - 3;
int tempConflCount;
for (int i = 0; i < N ; i++) {
tempConflCount = getConflicts(conflQueenCol, i);
if (i != currentRow && tempConflCount <= minConfl) {
minConfl = tempConflCount;
bestRow = i;
}
}
queens[conflQueenCol] = bestRow;
return true;
}
public static void generateConflicts () {
conflicts = new Vector<Integer>();
rowConfl = new int[N];
d1Confl = new int[2*N - 1];
d2Confl = new int[2*N - 1];
for (int i = 0; i < N; i++) {
int r = queens[i];
rowConfl[r]++;
d1Confl[getD1Pos(i, r)]++;
d2Confl[getD2Pos(i, r)]++;
}
for (int i = 0; i < N; i++) {
int conflictsCount = getConflicts(i, queens[i]) - 3;
if (conflictsCount > 0) {
conflicts.add(i);
}
}
}
public static int getConflicts(int col, int row) {
return rowConfl[row] + d1Confl[getD1Pos(col, row)] + d2Confl[getD2Pos(col, row)];
}
public static void main(String[] args) {
long startTime = System.currentTimeMillis();
init();
int steps = 0;
while(hasConflicts()) {
if (++steps > MAX_STEPS) {
init();
steps = 0;
}
}
long endTime = System.currentTimeMillis();
System.out.println("Done for " + (endTime - startTime) + "ms\n");
if(PRINT_BOARD){
print();
}
}
}

Comments would have been helpful :)
Rather than recreating your conflict set and your "worst conflict" queen everything, could you create it once, and then just update the changed rows/columns?
EDIT 0:
I tried playing around with your code a bit. Since the code is randomized, it's hard to find out if a change is good or not, since you might start with a good initial state or a crappy one. I tried making 10 runs with 10 queens, and got wildly different answers, but results are below.
I psuedo-profiled to see which statements were being executed the most, and it turns out the inner loop statements in chooseConflictQueen are executed the most. I tried inserting a break to pull the first conflict queen if found, but it didn't seem to help much.
Grouping only runs that took more than a second:
I realize I only have 10 runs, which is not really enough to be statistically valid, but hey.
So adding breaks didn't seem to help. I think a constructive solution will likely be faster, but randomness will again make it harder to check.

Your approach is good : Local search algorithm with minimum-conflicts constraint. I would suggest try improving your initial state. Instead of randomly placing all queens, 1 per column, try to place them so that you minimize the number of conflicts. An example would be to try placing you next queen based on the position of the previous one ... or maybe position of previous two ... Then you local search will have less problematic columns to deal with.

If you randomly select, you could be selecting the same state as a previous state. Theoretically, you might never find a solution even if there is one.
I think you woud be better to iterate normally through the states.
Also, are you sure boards other than 8x8 are solvable?
By inspection, 2x2 is not, 3x3 is not, 4x4 is not.

Checking one dimensional array for duplicate integers Java

hey guys,
right i have an array of 50 random integers and i have to check if any of them are the same,
here is my code so far it only checks ajacent indexs
for (int i =0; i < 50; i++)
{
System.out.print("Student" + i + ": " );
customers[i] = (int)((Math.random()*10000)%10+1);
System.out.print(" " +customers[i]+ "\n");
if( duplicate == customers[i])
{
System.out.println("yup");
}
duplicate = customers[i];
}

Sort the array first. Then you can check just the next index. If it's ever the same, break.
Okay, I hate ridiculous limitations. If you want, you can do it like this without using a sort:
import java.util.ArrayList;
import java.lang.Math;
public class Main {
public static void main(String[] args) {
ArrayList<Integer> list = new ArrayList<Integer>();
Integer currentValue = 0;
int i = 0;
int limit = 20;
for(i = 0; i < limit; i++) {
list.add((int)(Math.random() * 100));
}
for(i = 0; i < limit; i++) {
currentValue = list.get(i);
list.set(i, -1);
if(list.contains(currentValue)) {
System.out.println("yup:" + currentValue);
return;
} else {
list.set(i, currentValue);
}
}
System.out.println("No duplicates!");
return;
}
}
Is it efficient? No.
Does it work? Yes.

I think, if you have to just use if/for functions, you have to make two loops:
for (int i =0; i < 50; i++)
{
customers[i] = (int)((Math.random()*10000)%10+1);
for ( int j = 0; j < i; j++)
{
if( customers[j] == customers[i])
{
// duplicated entry. do what you want
System.out.println("yup");
}
}
}

You could use Arrays.sort (see more at http://www.exampledepot.com/egs/java.util/coll_SortArray.html) to sort your data, and then check for duplicates.

You use value 0-10 so you can save your value in boolean array and it that way verify if this is duplicate or not:
boolean[] checker = new boolean[11];
for (int i =0; i < 50; i++) {
customers[i] = (int)((Math.random()*10000)%10+1);
if (checker[customers[i]]) {
System.out.println("yup"); //duplication
} else {
checker[customers[i]] = true;
}
}

You can use the below code if you are working with Integer array:
public class DuplicateInteger {
private static int countDuplicate;
public static int[] getDuplicateIntegers(int[] integerArray){
int duplicateIntegers[] = new int[integerArray.length];
countDuplicate = 0;
for(int i=0;i<integerArray.length;i++){
for(int j=i+1;j<integerArray.length;j++){
int replicaTest = 0;
if(integerArray[i]==integerArray[j]){
for(int k=0;k<countDuplicate;k++){
if(duplicateIntegers[k]==integerArray[i]){
replicaTest = 1;
}
}
if(replicaTest==0){
duplicateIntegers[countDuplicate] = integerArray[i];
countDuplicate++;
}
}
}
}
return duplicateIntegers;
}
public static void printDuplicateIntegers(int[] duplicateIntegers){
System.out.println("Duplicate Integers:");
System.out.println("-------------------");
for(int i=0;i<countDuplicate;i++){
System.out.println(duplicateIntegers[i]);
}
}
public static void main(String[] args){
int numberArray[] = {1, 2, 3, 4, 5, 6, 7, 1, 3, 5, 7};
printDuplicateIntegers(getDuplicateIntegers(numberArray));
}
}