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Is there a test case scenario in which my program will fail?

Problem : You have L, a list containing some digits (0 to 9). Write a function solution(L) which finds the largest number that can be made from some or all of these digits and is divisible by 3. If it is not possible to make such a number, return 0 as the solution. L will contain anywhere from 1 to 9 digits. The same digit may appear multiple times in the list, but each element in the list may only be used once.
Test Cases :
Input:
Solution.solution({3, 1, 4, 1})
Output: 4311
Input:
Solution.solution({3, 1, 4, 1, 5, 9})
Output: 94311
My Program :
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.stream.IntStream;
public class Solution {
static ArrayList<Integer> al = new ArrayList<Integer>();
static ArrayList<Integer> largest = new ArrayList<Integer>();
static int o = 1;
static int po = 0;
static void combinations(String[] digits, String[] data, int start, int end, int index, int r)
{
if (index == r)
{
String temp = "0";
for (int j = 0; j < r; j++)
{
temp = temp + data[j];
// System.out.print(data[j]);
}
Integer d = Integer.parseInt(temp);
al.add(d);
// System.out.println(al);
}
for (int i = start; i <= end && ((end - i + 1) >= (r - index)); i++)
{
data[index] = digits[i];
combinations(digits, data, i + 1, end, index + 1, r);
}
}
static void printCombinations(String[] sequence, int N)
{
String[] data = new String[N];
for (int r = 0; r < sequence.length; r++)
combinations(sequence, data, 0, N - 1, 0, r);
}
static String[] convert(int[] x)
{
String c[] = new String[x.length];
for(int i=0; i < x.length; i++)
{
Integer k = x[i];
if(k==0)
{
o = o * 10;
continue;
}
c[i] = k.toString();
}
// System.out.println(o);
c = Arrays.stream(c).filter(s -> (s != null && s.length() > 0)).toArray(String[]::new);
po = c.length;
// System.out.println("Come"+ Arrays.asList(c));
return c;
}
public static int solution(int[] l) {
if(l.length==0)
return 0;
if(IntStream.of(l).sum()%3==0)
{
String x = "";
Arrays.sort(l);
for (int i = l.length - 1; i >= 0; i--) {
x = x + l[i];
}
return Integer.parseInt(x);
}
printCombinations(convert(l),po);
al.sort(Comparator.reverseOrder());
al.remove(al.size()-1);
al.removeIf( num -> num%3!=0);
if(al.isEmpty())
return 0;
for(int i=0; i< al.size(); i++)
{
Integer n = al.get(i);
printMaxNum(n);
}
// System.out.println(al);
// System.out.println(largest);
return largest.get(0)*o;
}
static void printMaxNum(int num)
{
// hashed array to store count of digits
int count[] = new int[10];
// Converting given number to string
String str = Integer.toString(num);
// Updating the count array
for(int i=0; i < str.length(); i++)
count[str.charAt(i)-'0']++;
// result is to store the final number
int result = 0, multiplier = 1;
// Traversing the count array
// to calculate the maximum number
for (int i = 0; i <= 9; i++)
{
while (count[i] > 0)
{
result = result + (i * multiplier);
count[i]--;
multiplier = multiplier * 10;
}
}
// return the result
largest.add(result);
}
public static void main(String[] args) {
System.out.println(solution(new int[]{9,8,2,3}));
}
}
My Code passes both given test cases and 3 other hidden test cases except one. I tried all possible input combinations but couldn't get to the exact failure. The return type by default is given as int and therefore they would not pass values which give output that does not fit in int. Any other scenario where my code fails?

Merging two 2D arrays (M + N)

I am a first year programming trying to solve this challenge that was given to us students at uni.
Question image
There's a typo at where it says (N + K) whereas in fact it's actually (M+K) columns.
My attempt for this question goes as follows
public static int[][] mergeArrays(int[][] arrayA, int[][] arrayB){
int rows = 3;
int columns = arrayA[0].length + arrayB[0].length;
int[][] mergedArray = new int[rows][columns];
int k = 0;
for (int i = 0; i < rows; i++)
{
for ( int j = 0 ; j < columns; j++)
{
try
{
mergedArray[i][j] = arrayA[i][j];
}
catch (ArrayIndexOutOfBoundsException e)
{
mergedArray[i][j] = arrayB[i][k];
k += 1;
}
}
}
return mergedArray;
}
public static void main(String [] args)
{
int [][] a1 = { {1,2,3,3,3} , {3,2,1,6,3} , {4,5,6,1,3} };
int [][] a2 = { {1,9,7,2,3} , {0,7,8,3,2} , {3,8,9,7,2} };
int[][] m = mergeArrays(a1,a2);
for (int[] x : m)
{
for (int y : x)
{
System.out.print(y + " ");
}
System.out.println();
}
}
The program doesn't work for some reason. I don't know what's wrong with my approach here. Would really appreciate if someone helps me out.

Without using any libraries, in a manual way, here is my working answer.
I didn't use any of them, since we were not allowed, when I was a student.
public class Main {
private static int[][] mergeArrays(int[][] a1, int[][] a2) {
// Count rows and cols length.
int rows = a1.length;
int cols_a1 = a1[0].length;
int cols_a2 = a2[0].length;
// Total number of cols
int cols = cols_a2 + cols_a1;
int [][] merged = new int[rows][cols];
for (int i = 0; i < rows ; ++i) {
for (int j = 0; j < cols_a1; ++j) {
merged[i][j] = a1[i][j];
}
// To not overwrite values,
// the trick is to add an offset, while assigning,
// which is the amount of elements (cols_a1) used by the previous loop.
// Basically, we are shifting the k-index by this constant,
// as to not overwrite the values assigned from the previous
// inner loop.
for (int k = 0; k < cols_a2; ++k) {
merged[i][cols_a1 + k] = a2[i][k];
}
}
// Return the merged array
return merged;
}
// I refactored your good printing code into a method, for readability.
private static void print2darray(int[][] array2d) {
for (int[] x : array2d) {
for (int y : x) {
System.out.print(y + " ");
}
System.out.println();
}
}
public static void main(String[] args) {
int [][] a1 = {{1,2,3,3,3} , {3,2,1,6,3} , {4,5,6,1,3}};
int [][] a2 = {{1,9,7,2,3} , {0,7,8,3,2} , {3,8,9,7,2}};
int [][] merged = mergeArrays(a1, a2);
print2darray(merged);
}
}
The result is the same, as expected, from your question image:
1 2 3 3 3 1 9 7 2 3
3 2 1 6 3 0 7 8 3 2
4 5 6 1 3 3 8 9 7 2

Since you're a student I think to better if we give a hint, but since the solution is already there you can check this one as well:
public static int[] merge(int[] first,int[] second) {
return ArrayUtils.addAll(first, second);
}
public static void main(String[] args) {
int [][] a1 = { {1,2,3,3,3} , {3,2,1,6,3} , {4,5,6,1,3}};
int [][] a2 = { {1,9,7,2,3} , {0,7,8,3,2} , {3,8,9,7,2}};
int [][] a3 = new int[a1.length][];
for (int i = 0; i < a1.length; i++) {
a3[i] = merge(a1[i],a2[i]);
}
for (int[] ints : a3) {
StringJoiner joiner = new StringJoiner(",","[","]");
for (int i1 : ints) {
joiner.add(i1+"");
}
System.out.println(joiner.toString());
}
}

You are not merging it properly. Your logic is that if arrayA column index is out of bounds, you are adding from arrayB's columns. But what if that is also out of bounds, as in your case. Since you are always incrementing its index k. You could simply iterate over 2 arrays separately and merge into resulting array.
public static int[][] mergeArrays(int[][] arrayA, int[][] arrayB) {
int rows = 3;
int columns = arrayA[0].length + arrayB[0].length;
int[][] mergedArray = new int[rows][columns];
int k = 0;
for (int i = 0; i < rows; i++) {
for (int j = 0; j < arrayA[0].length; j++) {
mergedArray[i][k++] = arrayA[i][j];
}
for (int j = 0; j < arrayB[0].length; j++) {
mergedArray[i][k++] = arrayB[i][j];
}
k=0;
}
return mergedArray;
}

Merge sort 3 way java

I have to make a 3 way merge sort of an array. the array length is a in a power of 3, i.e. 3,9,27 etc. So I can use only one split function and not "left","mid","right".
Would like to get an answer how to repair it and why does not it work.
I have written the code, however don't know how to get it to work.
Here it is:
EDITED THE CODE, STILL DOES NOT WORK
public class Ex3 {
public static void main(String[] args) { //main function
Scanner in = new Scanner(System.in); //scanner
int size = in.nextInt();
int[] arr = new int[size];
for (int i = 0; i<arr.length; i++){
arr[i] = in.nextInt();
}
in.close();
arr = merge3sort (arr); //send to the function to merge
for (int i = 0; i<arr.length; i++){ //printer
System.out.print(arr[i]+ " ");
}
}
static int[] split(int[] m, int thirdNum) { //split function that splits to 3 arrays
int third[] = new int[m.length/3];
int third1[]=new int[m.length/3];
int third2[]=new int[m.length/3];
for(int i = 0; i<=m.length/3; i++)
third[i]=m[i];
for(int i=0; i<=m.length/3;i++)
third1[i]=m[i+thirdNum];
for(int i=0; i<=m.length/3;i++)
third2[i]=m[i+2*thirdNum];
return merge(third,third1,third2);
//return null;
}
static int minOf3(int[] a3) { //function that finds out how what is the index of the smallest number
int num0 = a3[0];
int num1 = a3[1];
int num2 = a3[2];
int idx = 0;
if(num0<num1 && num1<num2)
idx=0;
if(num1<num0 && num0<num2)
idx=1;
else
idx=2;
return idx;
}
static int[] merge(int[] th0, int[] th1, int[] th2) { //function that sorts the numbers between 3 arrays
int len0=th0.length;
int len1=th1.length;
int len2=th2.length;
int[] united = new int[len0+len1+len2];
int ind = 0; int i0=0; int i1=0; int i2=0;
while(i0<len0 && i1<len1 && i2<len2){
if(th0[i0]<th1[i1]){
if(th0[i0]<th2[i2]){
united[ind]=th0[i0];
i0=i0+1;
}//end inner if
else{
united[ind]=th2[i2];
i2=i2+1;
}//end inner else
}//end outer if
else{
united[ind]=th1[i1];
i1=i1+1;
}//end outer else
ind=ind+1;
}//end while
for (int i = i0; i < len0; i = i + 1) {
united[ind] = th0[i];
ind = ind + 1;
}
for (int i = i1; i < len1; i = i + 1) {
united[ind] = th1[i];
ind = ind + 1;
}for (int i = i2; i < len2; i = i + 1) {
united[ind] = th2[i];
ind = ind + 1;
}
return united;
}
static int[] merge3sort(int[] m) { //function that glues all together
if (m.length == 1) {
return m;
}
else{
return merge(merge3sort(split(m,m.length/3)),merge3sort(split(m,m.length/3)),merge3sort(split(m,m.length/3))); }
}
I get the following exception:
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 0
at ololosh1.Ex3.split(Ex3.java:27)
at ololosh1.Ex3.merge3sort(Ex3.java:98)
at ololosh1.Ex3.main(Ex3.java:15)

Look at this part of your code:
for(int i = 0; i<=m.length/3; i++)
third[i]=m[i];
for(int i=0; i<=m.length/3;i++)
third1[i]=m[i+thirdNum];
for(int i=0; i<=m.length/3;i++)
third2[i]=m[i+2*thirdNum];
Arrays are indexed from 0 to length-1. Each third* array has length m.length/3. Therefore their index can only go up to m.length/3 - 1. Yet you are indexing up to and including m.length/3.
Once you get your application working correctly, you really should clean it up. There is a lot of redundancy. For example, you are using the expression m.length/3 multiple times in method split() but you are also passing that same value to it as an argument.

Finding closest number to 0

I have an array of integers, and I need to find the one that's closest to zero (positive integers take priority over negative ones.)
Here is the code I have so far:
public class CloseToZero {
public static void main(String[] args) {
int[] data = {2,3,-2};
int curr = 0;
int near = data[0];
// find the element nearest to zero
for ( int i=0; i < data.length; i++ ){
curr = data[i] * data[i];
if ( curr <= (near * near) ) {
near = data[i];
}
}
System.out.println( near );
}
}
Currently I'm getting a result of -2 but I should be getting 2. What am I doing wrong?

This will do it in O(n) time:
int[] arr = {1,4,5,6,7,-1};
int closestIndex = 0;
int diff = Integer.MAX_VALUE;
for (int i = 0; i < arr.length; ++i) {
int abs = Math.abs(arr[i]);
if (abs < diff) {
closestIndex = i;
diff = abs;
} else if (abs == diff && arr[i] > 0 && arr[closestIndex] < 0) {
//same distance to zero but positive
closestIndex =i;
}
}
System.out.println(arr[closestIndex ]);

If you are using java8:
import static java.lang.Math.abs;
import static java.lang.Math.max;
public class CloseToZero {
public static void main(String[] args) {
int[] str = {2,3,-2};
Arrays.stream(str).filter(i -> i != 0)
.reduce((a, b) -> abs(a) < abs(b) ? a : (abs(a) == abs(b) ? max(a, b) : b))
.ifPresent(System.out::println);
}
}

Sort the array (add one line of code) so the last number you pick up will be positive if the same absolute value is selected for a positive and negative numbers with the same distance.
Source code:
import java.util.Arrays;
public class CloseToZero {
public static void main(String[] args) {
int[] data = {2,3,-2};
int curr = 0;
int near = data[0];
Arrays.sort(data); // add this
System.out.println(Arrays.toString(data));
// find the element nearest to zero
for ( int i=0; i < data.length; i++ ){
System.out.println("dist from " + data[i] + " = " + Math.abs(0 -data[i]));
curr = data[i] * data[i];
if ( curr <= (near * near) ) {
near = data[i];
}
}
System.out.println( near );
}
}

Just add zero to this list.
Then sort the list
Arrays.sort(data);
then grab the number before or after the zero and pick the minimum one greater than zero

Assumption is that the array data has at least 1 value.
int closestToZero = 0;
for ( int i = 1; i < data.length; i++ )
{
if ( Math.abs(data[i]) < Math.abs(data[closestToZero]) ) closestToZero = i;
}
The value in closestToZero is the index of the value closest to zero, not the value itself.

static int Solve(int N, int[] A){
int min = A[0];
for (int i=1; i<N ; i++){
min = min > Math.abs(0- A[i]) ? Math.abs(0- A[i]) : Math.abs(min);
}
return min;
}

As you multiply data[i] with data[i], a value negative and a value positive will have the same impact.
For example, in your example: 2 and -2 will be 4. So, your code is not able to sort as you need.
So, here, it takes -2 as the near value since it has the same "weight" as 2.

I have same answer with different method,Using Collections and abs , we can solved.
static int Solve(int N, int[] A){
List<Integer> mInt=new ArrayList<>();
for ( int i=0; i < A.length; i++ ){
mInt.add(Math.abs(0 -A[i]));
}
return Collections.min(mInt);
}
That all,As simple as that

This is a very easy to read O(n) solution for this problem.
int bigestNegative = Integer.MIN_VALUE;
int smalestpositive = Integer.MAX_VALUE;
int result = 0;
for (int i = 0; i < n; i++) {
//if the zero should be considered as result as well
if ( temperatures[i] == 0 ) {
result = 0;
break;
}
if ( temperatures[i] > 0 && temperatures[i] < smalestpositive ) {
smalestpositive = temperatures[i];
}
if ( temperatures[i] < 0 && temperatures[i] > bigestNegative ) {
bigestNegative = temperatures[i];
}
}
if( (Math.abs(bigestNegative)) < (Math.abs(smalestpositive)) && bigestNegative != Integer.MIN_VALUE)
result = bigestNegative;
else
result = smalestpositive;
System.out.println( result );

First convert the int array into stream. Then sort it with default sorting order. Then filter greater than zero & peek the first element & print it.
Do it in declarative style which describes 'what to do', not 'how to do'. This style is more readable.
int[] data = {2,3,-2};
IntStream.of(data)
.filter(i -> i>0)
.sorted()
.limit(1)
.forEach(System.out::println);

using Set Collection and abs methode to avoid complex algo
public static void main(String[] args) {
int [] temperature={0};
***// will erase double values and order them from small to big***
Set<Integer> s= new HashSet<Integer>();
if (temperature.length!=0) {
for(int i=0; i<temperature.length; i++) {
***// push the abs value to the set***
s.add(Math.abs(temperature[i]));
}
// remove a zero if exists in the set
while(s.contains(0)) {
s.remove(0);
}
***// get first (smallest) element of the set : by default it is sorted***
if (s.size()!=0) {
Iterator iter = s.iterator();
System.out.println(iter.next());
}
else System.out.println(0);
}
else System.out.println(0);
}

static int nearToZero(int[] A){
Arrays.sort(A);
int ans = 0;
List<Integer> list = Arrays.stream(A).boxed().collect(Collectors.toList());
List<Integer> toRemove = new ArrayList<>();
List<Integer> newList = new ArrayList<>();
for(int num: list){
if(newList.contains(num)) toRemove.add(num);
else newList.add(num);
}
list.removeAll(toRemove);
for(int num : list){
if(num == 0 ) return 0;
if(ans == 0 )ans = num;
if(num < 0 && ans < num) ans = num;
if(num < ans) ans = num;
if(num > 0 && Math.abs(ans) >= num) ans = num;
}
return ans;
}

here is a method that gives you the nearest to zero.
use case 1 : {1,3,-2} ==> return 1 : use the Math.abs() for comparison and get the least.
use case 2 : {2,3,-2} ==> return 2 : use the Math.abs() for comparison and get the Math.abs(least)
use case 3 : {-2,3,-2} ==> return -2: use the Math.abs() for comparison and get the least.
public static double getClosestToZero(double[] liste) {
// if the list is empty return 0
if (liste.length != 0) {
double near = liste[0];
for (int i = 0; i < liste.length; i++) {
// here we are using Math.abs to manage the negative and
// positive number
if (Math.abs(liste[i]) <= Math.abs(near)) {
// manage the case when we have two equal neagative numbers
if (liste[i] == -near) {
near = Math.abs(liste[i]);
} else {
near = liste[i];
}
}
}
return near;
} else {
return 0;
}
}

You can do like this:
String res = "";
Arrays.sort(arr);
int num = arr[0];
int ClosestValue = 0;
for (int i = 0; i < arr.length; i++)
{
//for negatives
if (arr[i] < ClosestValue && arr[i] > num)
num = arr[i];
//for positives
if (arr[i] > ClosestValue && num < ClosestValue)
num = arr[i];
}
res = num;
System.out.println(res);
First of all you need to store all your numbers into an array. After that sort the array --> that's the trick who will make you don't use Math.abs(). Now is time to make a loop that iterates through the array. Knowing that array is sorted is important that you start to make first an IF statement for negatives numbers then for the positives (in this way if you will have two values closest to zero, let suppose -1 and 1 --> will print the positive one).
Hope this will help you.

The easiest way to deal with this is split the array into positive and negative sort and push the first two items from both the arrays into another array. Have fun!
function closeToZeroTwo(arr){
let arrNeg = arr.filter(x => x < 0).sort();
let arrPos = arr.filter(x => x > 0).sort();
let retArr = [];
retArr.push(arrNeg[0], arrPos[0]);
console.log(retArr)
}

Easiest way to just sort that array in ascending order suppose input is like :
int[] array = {10,-5,5,2,7,-4,28,65,95,85,12,45};
then after sorting it will gives output like:
{-5,-4,2,5,7,10,12,28,45,65,85,95,}
and for positive integer number, the Closest Positive number is: 2
Logic :
public class Closest {
public static int getClosestToZero(int[] a) {
int temp=0;
//following for is used for sorting an array in ascending nubmer
for (int i = 0; i < a.length-1; i++) {
for (int j = 0; j < a.length-i-1; j++) {
if (a[j]>a[j+1]) {
temp = a[j];
a[j]=a[j+1];
a[j+1]=temp;
}
}
}
//to check sorted array with negative & positive values
System.out.print("{");
for(int number:a)
System.out.print(number + ",");
System.out.print("}\n");
//logic for check closest positive and Integer
for (int i = 0; i < a.length; i++) {
if (a[i]<0 && a[i+1]>0) {
temp = a[i+1];
}
}
return temp;
}
public static void main(String[] args) {
int[] array = {10,-5,5,2,7,-4,28,65,95,85,12,45};
int closets =getClosestToZero(array);
System.out.println("The Closest Positive number is : "+closets);
}
}

static void closestToZero(){
int[] arr = {45,-4,-12,-2,7,4};
int max = Integer.MAX_VALUE;
int closest = 0;
for (int i = 0; i < arr.length; i++){
int value = arr[i];
int abs = Math.abs(value);
if (abs < max){
max = abs;
closest = value;
}else if (abs == max){
if (value > closest){
closest = value;
}
}
}
Return a positive integer if two absolute values are the same.

package solution;
import java.util.Scanner;
public class Solution {
public static void trier(int tab[]) {
int tmp = 0;
for(int i = 0; i < (tab.length - 1); i++) {
for(int j = (i+1); j< tab.length; j++) {
if(tab[i] > tab[j]) {
tmp = tab[i];
tab[i] = tab[j];
tab[j] = tmp;
}
}
}
int prochePositif = TableauPositif(tab);
int procheNegatif = TableauNegatif(tab);
System.out.println(distanceDeZero(procheNegatif,prochePositif));
}
public static int TableauNegatif(int tab[]) {
int taille = TailleNegatif(tab);
int tabNegatif[] = new int[taille];
for(int i = 0; i< tabNegatif.length; i++) {
tabNegatif[i] = tab[i];
}
int max = tabNegatif[0];
for(int i = 0; i <tabNegatif.length; i++) {
if(max < tabNegatif[i])
max = tabNegatif[i];
}
return max;
}
public static int TableauPositif(int tab[]) {
int taille = TailleNegatif(tab);
if(tab[taille] ==0)
taille+=1;
int taillepositif = TaillePositif(tab);
int tabPositif[] = new int[taillepositif];
for(int i = 0; i < tabPositif.length; i++) {
tabPositif[i] = tab[i + taille];
}
int min = tabPositif[0];
for(int i = 0; i< tabPositif.length; i++) {
if(min > tabPositif[i])
min = tabPositif[i];
}
return min;
}
public static int TailleNegatif(int tab[]) {
int cpt = 0;
for(int i = 0; i < tab.length; i++) {
if(tab[i] < 0) {
cpt +=1;
}
}
return cpt;
}
public static int TaillePositif(int tab[]) {
int cpt = 0;
for(int i = 0; i < tab.length; i++) {
if(tab[i] > 0) {
cpt +=1;
}
}
return cpt;
}
public static int distanceDeZero(int v1, int v2) {
int absv1 = v1 * (-1);
if(absv1 < v2)
return v1;
else if(absv1 > v2)
return v2;
else
return v2;
}
public static void main(String[] args) {
int t[] = {6,5,8,8,-2,-5,0,-3,-5,9,7,4};
Solution.trier(t);
}
}

To maintain O(n) time complexity and getting the desired results we have to add another variable called 'num' and assign to it 'near' before changing it's value. And finally make necessary checks. The improvements in the code are are:
public class CloseToZero {
public static void main(String[] args) {
int[] data = {2,3,-2};
int curr = 0;
int near = data[0];
int num=near;
// find the element nearest to zero
for ( int i=0; i < data.length; i++ ){
curr = data[i] * data[i];
if ( curr <= (near * near) ) {
num=near;
near = data[i];
}
}
if(near<0 && near*(-1)==num)
near=num;
System.out.println( near );
}
}

We have to find the Closest number to zero.
The given array can have negative values also.
So the easiest approach would append the '0' in the given array and sort it and return the element next to '0'
append the 0
Sort the Array
Return the element next to 0.
`
N = int(input())
arr = list(map(int, input().split()))
arr.append(0)
arr.sort()
zeroIndex = arr.index(0)
print(arr[zeroIndex + 1])
--> If this solution leaves corner cases please let me know also.
`

if you don't wanna use the inbuilt library function use the below code (just an and condition with your existing code)-
public class CloseToZero {
public static void main(String[] args) {
int[] data = {2,3,-2,-1,1};
int curr = 0;
int near = data[0];
// find the element nearest to zero
for ( int i=0; i < data.length; i++ ){
curr = data[i] * data[i];
if ( curr <= (near * near) && !((curr - (near * near) == 0) && data[i] < 0)) {
near = data[i];
}
}
System.out.println( near );
}
}
!((curr - (near * near) == 0) && data[i] < 0) : skip asignment if if near and curr is just opposit in sign and the curr is negative

public static int find(int[] ints) {
if (ints==null) return 0;
int min= ints[0]; //a random value initialisation
for (int k=0;k<ints.length;k++) {
// if a positive value is matched it is prioritized
if (ints[k]==Math.abs(min) || Math.abs(ints[k])<Math.abs(min))
min=ints[k];
}
return min;
}

public int check() {
int target = 0;
int[] myArray = { 40, 20, 100, 30, -1, 70, -10, 500 };
int result = myArray[0];
for (int i = 0; i < myArray.length; i++) {
if (myArray[i] == target) {
result = myArray[i];
return result;
}
if (myArray[i] > 0 && result >= (myArray[i] - target)) {
result = myArray[i];
}
}
return result;
}
I have added a check for the positive number itself.
Please share your views folks!!

public class ClosesttoZero {
static int closZero(int[] ints) {
int result=ints[0];
for(int i=1;i<ints.length;i++) {
if(Math.abs(result)>=Math.abs(ints[i])) {
result=Math.abs(ints[i]);
}
}
return result;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] ints= {1,1,5,8,4,-9,0,6,7,1};
int result=ClosesttoZero.closZero(ints);
System.out.println(result);
}
}

It can be done simply by making all numbers positive using absolute value then sort the Array:
int[] arr = {9, 1, 4, 5, 6, 7, -1, -2};
for (int i = 0; i < arr.length; ++i)
{
arr[i] = Math.abs(arr[i]);
}
Arrays.sort(arr);
System.out.println("Closest value to 0 = " + arr[0]);

import java.math.*;
class Solution {
static double closestToZero(double[] ts) {
if (ts.length == 0)
return 0;
double closestToZero = ts[0];
double absClosest = Math.abs(closestToZero);
for (int i = 0; i < ts.length; i++) {
double absValue = Math.abs(ts[i]);
if (absValue < absClosest || absValue == absClosest && ts[i] > 0) {
closestToZero = ts[i];
absClosest = absValue;
}
}
return closestToZero;
}
}

//My solution priorizing positive numbers contraint
int closestToZero = Integer.MAX_VALUE;//or we
for(int i = 0 ; i < arrayInt.length; i++) {
if (Math.abs(arrayInt[i]) < closestToZero
|| Math.abs(closestToZero) == Math.abs(arrayInt[i]) && arrayInt[i] > 0 ) {
closestToZero = arrayInt[i];
}
}

Write the integers of an array vertically

I know this is probably very simple but I really can't seem to understand how to write the integers vertically. For instance, there is an array that has 4 integers which are 9, 21, 63, and 501, the outcome would be the following
9 2 5 6
1 0 3
1
This is a small step of my program and probably the easiest but I can't understand how to do it :(
Can someone please help me or guide me so I can finish my program

Try this pseudo code
int[] list = new int[] {9,21,63,501};
bool finished = false;
if (list.Count > 0) {
for (var j=0;!finshed; j++) {
finished = true;
for (var i = 0; i<list.Count;i++) {
String val = list[i].ToString();
if (val.length>j) {
write(val.charAt(j));
finished = false;
}
}
}
}

I have created a very modular and easy to follow solution.
Edit: Converted digitAtIndex() to a purely numerical calculation.
Kept the original and called it digitAtStrIndex().
public class IntegerColumns {
public IntegerColumns() {
int[] arr = new int[] {9, 21, 501, 63};
printColumnMajorOrder(arr);
}
public static void main(String[] args) {
new IntegerColumns();
}
// --------------------- Primary Functions --------------------------
// Prints out an Array of Integers, each in a vertical column
public void printColumnMajorOrder(int[] arr) {
int cols = arr.length;
int rows = maxDigits(arr);
for (int r = 0; r < rows; r++) {
for (int c = 0; c < cols; c++) {
int d = digitAtIndex(arr[c], r);
System.out.printf("%s\t", d >= 0 ? Integer.toString(d) : " ");
}
System.out.println();
}
}
// Returns the length of an Integer
public int numDigits(int i) {
if (i <= 0) return 0;
return (int)Math.floor(Math.log10(i))+1;
}
// Numeric calculation to find a digit at a specified index
public int digitAtIndex(int num, int index) {
int digits = numDigits(num);
int deg = digits - index - 1;
int pow = (int)Math.pow(10, deg);
return pow > 0 ? (int)(num/pow)%10 : -1;
}
// Returns the number of digits for the longest Integer in an Array
public int maxDigits(int[] arr) {
int max = 0;
for (int i : arr) {
int size = numDigits(i);
if (size > max) max = size;
}
return max;
}
// ---------------------- Extra Functions ---------------------------
// Hybrid of Integer and Substrings - String manipulation = slow
public int digitAtStrIndex(int number, int i) {
String n = Integer.toString(number);
return n.length() > i ? Integer.parseInt(n.substring(i, i+1)) : -1;
}
// Prints the digits of a number vertically
public void printNumberVertical(int num) {
for (int i = 0; i < numDigits(num); i++)
System.out.println(digitAtIndex(num, i));
}
}

`public class VerticalPrintService {
private int[] data;
public VerticalPrintService( int[] intArray ) {
this.data = intArray;
}
public void printVertically(){
int cols = data.length; // # of columns
int rows = getRows(); // # of rows
System.out.println("cols: " + cols);
System.out.println("rows: " + rows);
String[][] matrix = new String[rows][cols];
int rowIndex = 0;
int colIndex = 0;
// populate 2d array
for ( int i : data ) {
String str = String.valueOf(i);
for ( int j = 0; j < str.length(); j++ ) {
matrix[rowIndex][colIndex] = String.valueOf(str.charAt(j));
rowIndex++;
}
colIndex++;
rowIndex = 0;
}
// print
for ( int i = 0; i < rows; i++ ) {
for ( int j = 0; j < cols; j++ ) {
if ( null == matrix[i][j] ) {
System.out.print("\t");
} else {
System.out.print( matrix[i][j] + "\t" );
}
}
System.out.println();
}
}
private int getRows(){
int max = 0;
for ( int i : data ) {
int len = String.valueOf(i).length();
if ( len > max ) {
max = len;
}
}
return max;
}
}`
and in your main method
`public static void main(String[] args) {
int[] array = { 9, 53, 501, 90 };
VerticalPrintService vps = new VerticalPrintService(array);
vps.printVertically();
}`