Let's say I have an array
int[] array={1,2,3,1,6,3,1};
and I want to replace all the 1 with 4. Are there any ways to do that besides changing it to a string array and use replace, then change it back?
Just loop through the array and swap out any 1's with 4's.
for (int i = 0; i < array.length; i++) {
if (array[i]==1) {
array[i]=4;
}
}
Loop through the array and check each element for the value you are looking for.
for (int i = 0; i < array.length; i++) {
if (array[i] == 1)
array[i] = 4;
}
Related
I'm trying to remove duplicates from an int array in Java. This method is void and should only alter the array I'm given. I'm also supposed to replace each duplicate with 0.
//array's duplicates are replaced with zero
//array is returned
public static void removeDuplicates(int[] array) {
//create a temp variable to compare all other other int values in result array
int temp;
//double for loop iterates through array and compares each value to temp.
for(int i = 0; i < array.length; i++) {
temp = array[i];
for(int j = 0; j < array.length; j++) {
if ((array[j] == temp) && (j+1 < array.length)) {
array[j] = 0;
}
}
}
}
This is what I've tried so far. It runs, however, I'm not able to replace all of my duplicates with zero. For example, if I have an array {2,2}, then I'll get an array {0,2} which is not what I want. Rather, I should be getting {0,0}. I've tried drawing this out and still can't find the problem.
Temp holds the value(s) to be deleted but only if it is a duplicate. The inner loop must go from i+1 to the end to see if other values of temp exist. If so, zero them out. But since the first one is also equal to temp it must be deleted too if it was a duplicate.
Change your loops to the following:
for(int i = 0; i < array.length-1; i++) {
temp = array[i];
for(int j = i+1; j < array.length; j++) {
if (array[j] == temp) {
array[j] = 0;
array[i] = 0; // remove the first one too
}
}
}
The first 4 values are set properly in the new array. It has to do with something with my variable 'count' which is not being set properly. The goal of the program is to simply grab the even numbers, and put them in a new array.
I have added 4 to count as a test, and that seems to work perfectly but I dont think that is the issue here.
int[] list = {8,5,4,11,12,2,1,3,10,6,7};
int count = 0;
int gr = 0;
for(int n=0; n<list.length; n++)
{
if(list[n] % 2 == 0)
{
count++;
}
}
int[] evn = new int[count];
for(int k = 0; k<=count; k++)
{
if(list[k] % 2 == 0)
evn[gr++] = list[k];
}
return evn;
Currently, the array prints "8,4,12,2,0,0" when it should print "8,4,12,2,10,6"
This happens because count is always less than the size of the array(list.length), so in the second for-loop you are never iterating till the end of the array.
Change your second for-loop to iterate till the end of the array as shown below :
for(int k = 0; k < list.length; k++)
You're only traversing part of list, as stated in the for condition:
for(int k = 0; k<=count; k++)
^--here--^
This is because count has a lower value than the length of the original array. Change this condition to traverse the whole array:
for(int k = 0; k<list.length; k++)
To traverse the whole list change the following:
for(int k = 0; k<=count; k++)
To
for(int k = 0; k<list.length; k++)
I can't wrap my head around this. Need to find duplicates and I did. All now that is left is to print how many times a duplicate appears in the array. I just started with Java,so this needs to be hard coded for me to understand. Spend last two days trying to figure it out but with no luck.. Any help will be great! Talk is cheap,here is the code..
import java.util.Arrays;
public class LoopTest {
public static void main(String[] args) {
int[] array = {12,23,-22,0,43,545,-4,-55,43,12,0,-999,-87};
int positive_counter = 0;
int negative_counter = 0;
for (int i = 0; i < array.length; i++) {
if(array[i] > 0) {
positive_counter++;
} else if(array[i] < 0) {
negative_counter++;
}
}
int[] positive_array = new int[positive_counter];
int[] negative_array = new int[negative_counter];
positive_counter = 0;
negative_counter = 0;
for (int i = 0; i < array.length; i++) {
if(array[i] > 0) {
positive_array[positive_counter++] = array[i];
} else if(array[i] < 0) {
negative_array[negative_counter++] = array[i];
}
}
System.out.println("Positive array: " + (Arrays.toString(positive_array)));
System.out.println("Negative array: " + (Arrays.toString(negative_array)));
Arrays.sort(array);
System.out.println("Array duplicates: ");
for (int i = 0; i < array.length; i++) {
for (int j = i + 1; j < array.length; j++) {
if(array[i] == array[j]) {
System.out.println(array[j]);
}
}
}
}
}
Since you are already sorting the array you can find the duplicates with just one loop (they will be next to each other right?). So you can do something like:
Arrays.sort(array);
System.out.println("Array duplicates: ");
int lastValueCount=1; //How many times we met the current value (at least 1 - this time)
for (int i = 1; i < array.length; i++){
if(array[i] == array[i-1])
lastValueCount++; //If it is the same as the previous increase the count
else {
if(lastValueCount>1) //If it is duplicate print it
System.out.println(array[i-1]+" was found "+lastValueCount+" times");
lastValueCount=1; //reset the counter
}
}
Result for your array is:
Array duplicates:
0 was found 2 times
12 was found 2 times
43 was found 2 times
Also you can use some of the Java bells and whistles like inserting the values into Map or something like that but I guess you are looking from an algorithmic point of view so the above is the simple answer with just one loop
Just go through your solution, first you separate positive and negative numbers in two different arrays, then you never use them, so what's the purpose of this separation ?
I am giving you just an idea related to your problem, it's better to solve it by your self so that you can get hands on Java.
Solution: you can use Dictionary-key value pair. Go through your array, put element in dictionary as a key and value as zero, on every iteration check if that key already exist in Dictionary, just increment its value. In the end, all of the values are duplicates that occurs in your array.
Hope it helps you.
From the algorithmic point of view, Veselin Davidov's answer is good (the most efficient).
In a production code, you would rather write it like this :
Map<Integer, Long> result =
Arrays.stream(array)
.boxed() //converts IntStream to Stream<Int>
.collect(Collectors.groupingBy(i -> i, Collectors.counting()));
The result is this Map :
System.out.println(result);
{0=2, 545=1, -4=1, -22=1, -87=1, -999=1, -55=1, 23=1, 43=2, 12=2}
An easy way would be using Maps. Without changing code too much:
for (int i = 0; i < array.length; i++) {
int count = 0;
for (int j = i + 1; j < array.length; j++) {
if(array[i] == array[j]) {
System.out.println(array[j]);
count++;
}
}
map.put(array[i], count);
}
Docs:
https://docs.oracle.com/javase/7/docs/api/java/util/Map.html
Edit: As a recommendation, after you are done with the example, you should analize your code and find what isn´t neccesary, what could be done better, etc.
Are all your auxiliary arrays neccesary? Are all loops necessary?
You can do it by creating an array list for duplicate values:-
Arrays.sort(array);
System.out.println("Array duplicates: ");
ArrayList<Integer> duplicates = new ArrayList<Integer>();
for (int i = 0; i < array.length; i++) {
for (int j = 0; j < array.length; j++) {
if(j != i && array[i] == array[j] && !duplicates.contains(array[i])){
duplicates.add(array[i]);
System.Out.println(duplicates[duplicates.size()-1]);
}
}
}
public static void findDuplicate(String s){
char[] charArray=s.toCharArray();
ArrayList<Character> duplicateList = new ArrayList<>();
System.out.println(Arrays.toString(charArray));
for(int i=0 ; i<=charArray.length-1; i++){
if(duplicateList.contains(charArray[i]))
continue;
for(int j=0 ; j<=charArray.length-1; j++){
if(i==j)
continue;
if(charArray[i] == charArray[j]){
duplicateList.add(charArray[j]);
System.out.println("Dupliate at "+i+" and "+j);
}
}
}
}
I am having trouble creating multiple arrays with a loop in Java. What I am trying to do is create a set of arrays, so that each following array has 3 more numbers in it, and all numbers are consecutive. Just to clarify, what I need to get is a set of, let's say 30 arrays, so that it looks like this:
[1,2,3]
[4,5,6,7,8,9]
[10,11,12,13,14,15,16,17,18]
[19,20,21,22,23,24,25,26,27,28,29,30]
....
And so on. Any help much appreciated!
Do you need something like this?
int size = 3;
int values = 1;
for (int i = 0; i < size; i = i + 3) {
int[] arr = new int[size];
for (int j = 0; j < size; j++) {
arr[j] = values;
values++;
}
size += 3;
int count = 0;
for (int j : arr) { // for display
++count;
System.out.print(j);
if (count != arr.length) {
System.out.print(" , ");
}
}
System.out.println();
if (i > 6) { // to put an end to endless creation of arrays
break;
}
}
To do this, you need to keep track of three things: (1) how many arrays you've already created (so you can stop at 30); (2) what length of array you're on (so you can create the next array with the right length); and (3) what integer-value you're up to (so you can populate the next array with the right values).
Here's one way:
private Set<int[]> createArrays() {
final Set<int[]> arrays = new HashSet<int[]>();
int arrayLength = 3;
int value = 1;
for (int arrayNum = 0; arrayNum < 30; ++arrayNum) {
final int[] array = new int[arrayLength];
for (int j = 0; j < array.length; ++j) {
array[j] = value;
++value;
}
arrays.add(array);
arrayLength += 3;
}
return arrays;
}
I don't think that you can "create" arrays in java, but you can create an array of arrays, so the output will look something like this:
[[1,2,3],[4,5,6,7,8,9],[10,11,12,13...]...]
you can do this very succinctly by using two for-loops
Quick Answer
==================
int arrays[][] = new int[30][];
for (int j = 0; j < 30; j++){
for (int i = 0; i < (j++)*3; i++){
arrays[j][i] = (i++)+j*3;
}
}
the first for-loop tells us, via the variable j, which array we are currently adding items to. The second for-loop tells us which item we are adding, and adds the correct item to that position.
All you have to remember is that j++ means j + 1.
Now, the super long-winded explanation:
I've used some simple (well, I say simple, but...) maths to generate the correct item each time:
[1,2,3]
here, j is 0, and we see that the first item is one. At the first item, i is also equal to 0, so we can say that, here, each item is equal to i + 1, or i++.
However, in the next array,
[4,5,6,7,8,9]
each item is not equal to i++, because i has been reset to 0. However, j=1, so we can use this to our advantage to generate the correct elements this time: each item is equal to (i++)+j*3.
Does this rule hold up?
Well, we can look at the next one, where j is 2:
[10,11,12,13,14...]
i = 0, j = 2 and 10 = (0+1)+2*3, so it still follows our rule.
That's how I was able to generate each element correctly.
tl;dr
int arrays[][] = new int[30][];
for (int j = 0; j < 30; j++){
for (int i = 0; i < (j++)*3; i++){
arrays[j][i] = (i++)+j*3;
}
}
It works.
You have to use a double for loop. First loop will iterate for your arrays, second for their contents.
Sor the first for has to iterate from 0 to 30. The second one is a little less easy to write. You have to remember where you last stop and how many items you had in the last one. At the end, it will look like that:
int base = 1;
int size = 3;
int arrays[][] = new int[30][];
for(int i = 0; i < 30; i++) {
arrays[i] = new int[size];
for(int j = 0; j < size; j++) {
arrays[i][j] = base;
base++;
}
size += 3;
}
Rather trivial piece of Java code here.. But getting an OutOfBoundsException and I'm not quite sure why? Any help would be great!
Before anyone asks if this is homework, no it isn't it's for exam prep.
import java.util.Scanner;
public class exampractice {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int[] a = {1,2,3};
int min = a[0];
int max = a[0];
for (int i = 0; i <= a.length; i++){
if(a[i] < min)
min = a[i];
else
if(a[i] > max)
max = a[i];
}
System.out.println("Min is"+min+ "\nMax is: " + max);
}
}
It should be < not <=. Zero based indices on an array vary from 0 to length - 1.
Update
Someone in grad school taught me to build up a collection of mental patterns like this and use them; you'll avoid bugs. This one is that any array in a C-like language is always searched as
for(index = 0; index < array.length ; index++)
Having trained myself over the years with this, I saw this bug at, literally, first glance, which is how I got in at the head of the line.
Here's are some more examples, for C:
char aString[MAXLENTH]; // declare a string array
char * aString ;
aString = (char *) malloc(MAXLENGTH]; // or malloc it
// constructing a string with catenation
aString[0] = '\0'; // now either way this is guaranteed an empty string
But of course I should have used another pattern:
if((aString = (char *) malloc(MAXLENGTH) == NULL){
// report out of memory
exit(BAD);
}
Think about what other patterns might be good.
Arrays use 0 based counting and in your for loop you go i<= a.length which means you'll be accessing a element that doesn't exists. just remove the = and your good.
for (int i = 0; i < a.length; i++)
for (int i = 0; i <= a.length; i++)
should be
for (int i = 0; i < a.length; i++)
While a is an array and arrays are zero-based indexed, you have to loop it from 0 to a.length -1. It's enough to change i <= a.length to i < a.length.
for (int i = 0; i < a.length; i++)
i should be between 0 and a.length - 1, since the first element of an array starts with key 0.
So the for condition should be strictly less than instead of less than or equal to a.length.
Your for loop should be < a.length and not <= a.length:
for (int i = 0; i < a.length; i++) {