I have this question:
The method accepts an integer array as its input and returns a new array which is a
permutation of the input array. The method fix34 rearranges the input array such
that every 3 is immediately followed by a 4 (e.g. if there is a 3 at position i, there will
be a 4 at position i+1). The method keeps the original positions of the 3s but may
move any other number, moving the minimal amount of numbers.
Assumptions regarding the input:
The array contains the same number of 3's and 4's (for every 3 there is a 4)
There are no two consecutive 3s in the array
The matching 4 for a 3 at some position i is at position j where j > i
ok, so this is what I wrote:
public class Fix34 {
public static void main(String[] args){
int [] args1 ={3,1,2,3,5,4,4};
int[] args11=fix34(args1);
for (int i = 0; i<=args11.length-1;i++ ){
System.out.print(args11[i]+" ");}}
public static int pos (int[] arr){
int i= arr.length-1;
while (arr[i]!=4){
i=-1;
}
return i;
}
public static int[] fix34(int[] nums){
for(int i = 0; i<=nums.length-1; i++){
if (nums[i] == 3){
nums[pos(nums)]=nums[i+1];
nums[i+1]=4;
}
}
return nums;
}
}
when I insert arrays such {3,2,1,4} it works, but with the array as written in the code, it gives me the error message:
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: -1
at Fix34.pos(Fix34.java:15)
at Fix34.fix34(Fix34.java:25)
at Fix34.main(Fix34.java:6)
how come the arrays gets to -1 position?!
Thanks
you are setting it to -1 here:
i=-1;
Your issue in in this piece of code
public static int pos (int[] arr){
int i= arr.length-1;
while (arr[i]!=4){
i=-1;
}
return i;
}
If the last element in the array is 4 the while loop is never entered so arrays like 3, 1, 2, 4 are fine. Otherwise the loop is entered and i is set to -1. I think that you mean to decrement. In that case replace
i=-1
with
i--
or
i=i-1
and as mentioned in another answer make sure i doesn't go below 0.
I think you ment
i-= 1;
instead of that:
i=-1;
Whoa there that is a little overblown for this problem. Nested loops are they key here…come take a look
public int[] fix34(int[] nums) {
for(int a = 0; a < nums.length; a++){ //we see 4's first
for(int b = 0; b < nums.length - 1; b++){ //then the nested loop finds a 3
//length - 1 to stay in bounds, although not needed here...
if(nums[a] == 4 && nums[b] == 3){
//swap
int tmp = nums[b + 1];
nums[b + 1] = nums[a];
nums[a] = tmp;
}
}
}
return nums;
}
Related
Beginner here. I'm having problems running this series of for loops to find which integers are missing from an array.
public class FunWithArrays{
public static void main(String[] args){
String nString = args[0];
int n = Integer.parseInt(nString);
int inputArray [] = {1,2,4};
System.out.println(" The missing numbers are " );
findMissingNum(n, inputArray);
}
public static void findMissingNum(int n, int[] inputArray){
for (int i = 1; i <= inputArray.length; i++){
int count = 0;
for( int j = 0; j < n; j++){
if(inputArray[j] == i){
count ++;
}
if (count == 0){
System.out.println(i);
}
}
}
}
}
I get the answer I want, namely 3, however it doesn't print but rather shows up in a runtime error:
java.lang.ArrayIndexOutOfBoundsException: 3
at FunWithArrays.findMissingNum(FunWithArrays.java:17)
at FunWithArrays.main(FunWithArrays.java:9)
the method should take an input n from the user (when the program is run) as the largest value of the array and print all the ones missing
The logic is the outer for loop should traverse the array for numbers 1-n, and the inner loop should add to the count variable each time it finds a certain number. At the end of iteration it should print any numbers with a final "count" of 0. THIS IS LITERALLY DRIVING ME CRAZY!!! thanks in advance :)
First of all, you should traverse from 0 to (inputArray.length-1) index of inputArray. This will get rid of the ArrayIndexOutOfBoundsException, because java array indexing starts from 0 not 1.
And for inner loop, run from 0 to n, since n is the max number.
And Thirdly, it should be inputArray[i] == j, not inputArray[j] == i, same for printing the value. In you case I believe you have n>=4, so it was trying to access inputArray[3] via inputArray[j] call. That's why you are getting this out of bound error.
I think your code means like this: nest loop always run through inner loop first before run the outer loop.
public static void findMissingNum(int n, int[] inputArray){
for (int i = 1; i <= n; i++){
int count = 0;
for( int j = 0; j < inputArray.length; j++){
if(inputArray[j] == i){
count ++;
}
if (count == 0){
System.out.println(i);
}
}
}
}
I will just use a while loop instead:
int num =1;
while(num<=n){
for(int i = 0;i<inputArray.length;i++){
if(inputArray[i]!=num){
System.out.println(num);
}
}
num++;
}
The i incrementing to <= ipnutArray.length is not causing the error because i is never used as the index. What is causing the error is when n > length.
Also, you should not be checking n elements starting from the beginning because n is the max value, not the number of elements.
Im currently writing some code that print Pascal's Triangle. I need to use a 2D array for each row but don't know how to get the internal array to have a variable length, as it will also always changed based on what row it is int, for example:
public int[][] pascalTriangle(int n) {
int[][] array = new int[n + 1][]
}
As you can see I know how to get the outer array to have the size of Pascal's Triangle that I need, but I don't know how to get a variable length for the row that corresponds with the line it is currently on.
Also how would I print this 2D array?
Essentially what you want to happen is get the size of each row.
for(int i=0; i<array.size;i++){//this loops through the first part of array
for(int j=0;j<array[i].size;j++){//this loops through the now row
//do something
}
}
You should be able to use this example to also print the triangle now.
This is my first answer on StackOverFlow. I am a freshman and have just studied Java as part of my degree.
To make every step clear, I will put different codes in different methods.
Say n tells us how many rows that we are going to print for the triangle.
public static int[][] createPascalTriangle(int n){
//We first declare a 2D array, we know the number of rows
int[][] triangle = new int[n][];
//Then we specify each row with different lengths
for(int i = 0; i < n; i++){
triangle[i] = new int[i+1]; //Be careful with i+1 here.
}
//Finally we fill each row with numbers
for(int i = 0; i < n; i++){
for(int j = 0; j <= i; j++){
triangle[i][j] = calculateNumber(i, j);
}
}
return triangle;
}
//This method is used to calculate the number of the specific location
//in pascal triangle. For example, if i=0, j=0, we refer to the first row, first number.
public static int calculateNumber(int i, int j){
if(j==0){
return 1;
}
int numerator = computeFactorial(i);
int denominator = (computeFactorial(j)*computeFactorial(i-j));
int result = numerator/denominator;
return result;
}
//This method is used to calculate Factorial of a given integer.
public static int computeFactorial(int num){
int result = 1;
for(int i = 1; i <= num; i++){
result = result * i;
}
return result;
}
Finally, in the main method, we first create a pascalTriangle and then print it out using for loop:
public static void main(String[] args) {
int[][] pascalTriangle = createPascalTriangle(6);
for(int i = 0; i < pascalTriangle.length; i++){
for(int j = 0; j < pascalTriangle[i].length; j++){
System.out.print(pascalTriangle[i][j] + " ");
}
System.out.println();
}
}
This will give an output like this:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
I am very new to Java and I was trying to solve this problem on Hackerrank:
Here's the task:
https://www.hackerrank.com/challenges/cut-the-sticks
You are given N sticks, where the length of each stick is a positive
integer. A cut operation is performed on the sticks such that all of
them are reduced by the length of the smallest stick.
Suppose we have six sticks of the following lengths:
5 4 4 2 2 8
Then, in one cut operation we make a cut of length 2 from each of the six
sticks. For the next cut operation four sticks are left (of non-zero length), > whose lengths are the following:
3 2 2 6
The above step is repeated until no sticks are left.
Given the length of N sticks, print the number of sticks that are left before > each subsequent cut operations.
Note: For each cut operation, you have to recalcuate the length of smallest
sticks (excluding zero-length sticks).
Here is my attempt at it, but it doesnt seem to be working. The output gets stuck in while loop (4 gets printed out infinitely)
import java.io.*;
import java.util.*;
public class Solution {
private static int findMin (int[] A)
{
int min = A[0];
for (int i =0; i<A.length; i++)
{
if (A[i] < min)
{
min = A[i];
}
}
return min;
}
private static int countNonZeros (int[] A)
{
int zeros = 0;
for (int i =0; i<A.length; i++)
{
if (A[i] == 0)
{
zeros++;
}
}
int nonZeros = A.length - zeros;
return nonZeros;
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int[] A = new int[n];
for (int i=0; i<n; i++)
{
A[i] = scanner.nextInt();
}
int nums = countNonZeros(A);
while (nums > 0)
{
int mins = findMin(A);
for (int j = 0; j<A.length; j++)
{
A[j]=A[j]-mins;
}
nums = countNonZeros(A);
System.out.println(nums);
}
}
}
Any help is appreciated
(PS I know I can just look the solution up somewhere, but I want to know why my code isn't working)
The problem that you have is that your findMin is not excluding zero-length elements, so once you have a zero that will be the min, and as a result an iteration of the while loop will be the same as the previous iteration, having subtracted 0 from each of the elements of A.
An integer array stores values 3,2,3,4,5. I am trying to create a program that increments these values by 2 and then saves the result into the same array using a for loop. I tried but something is wrong with my code, here:
public class ArrayClass {
int a[] = {2, 3, 3, 4, 5};
}
public class ArrayObject {
public static void main(String[] Ella) {
int a[] = new int[5];
int i;
for (i = 2; i < a.length; i = i + 2) {
a[i] = i + 2;
System.out.println(a[i]);
}
}
}
This should work:
for (i = 0; i < a.length; i++) {
a[i] += 2;
System.out.println(a[i]);
}
You see, when increasing every single value of an array, the index has to be 0 and max the array's length. By adding one to i, the indexing of the array increases by one, which means the next number will be increased by two. what you did was add two to the "i" variable which means that only 3 of the varialbes would have been changed.
Please make below change to your code.It will work.
for (i = 0; i < a.length; i++) {
a[i] = a[i] + 2;
System.out.println(a[i]);
}
The error is that when you do i = i + 2, you are just incrementing the position index, not the actual value in that position.
you need to do:
a[i] = a[i]+2;
Let me explain what a[i] is:
|3|2|3|4|5|
1 2 3 4 5
The first row are the values. The second row is the index. "Index" means the position number of each of positions in the array.
Another problem is that, when you initialise i, it need to be i=0. That is because i array indices (plural of index) always start from 0. That means that a[0] is the first position in the array That would be number 3 from your data set.
This is more of an self defined programming exercise than a real problem. I have an array of java.lang.Comparable items. I need to maintain two pointers (an index into the array i.e., int values) i,j . i starts at the beginning of array and moves right until it encounters an element which is less than or equal to the previous element. When it does it stops moving right and ends up pointing to the element which is out of order(element which is not greater than the previous). Similarly j starts at the end of the array and moves left until it finds an element which is not less than the previous.
Also, I need to make sure that the indices don't run out of the array i.e., i cannot go below 0 and j cannot go above arraylength-1
lets say we have an array of 5 elements.
i = 0;
j = 4;(which is the arraylength-1 )
if C,D,E,F,G is the array ,the final values of i and j will be
i = 4 and j = 0
if array is J,D,E,F,G ,the final values of i, j will be
i = 0 , j = 0
if array is B,C,A,D,G , final values of i,j will be
i = 2 , j = 1
I tried to code the logic for moving i to the right, using a while loop as below. I was able to get it working for the i pointer in two cases.
public class PointerMovement{
public static void ptrsPointToOutOfOrderElements(Comparable[] a){
int lo = 0;
int hi = a.length-1;
int i = lo;
int t=i+1;
int j = hi;
//only for moving i to the right .
while(less(a[i],a[t])){
if(t == hi){
i=t;
break;
}
i++;
t++;
}
i=t;
for(Comparable x:a){
System.out.print(x+",");
}
System.out.println();
System.out.println("bad element or end of array at i="+i+"==>"+a[i]);
}
private static boolean less(Comparable x,Comparable y){
return x.compareTo(y) < 0;
}
public static void main(String[] args) {
String[] a = new String[]{"C","D","E","F","G"};//works
//String[] a = new String[]{"B","C","A","D","G"};//works
//String[] a = new String[]{"J","D","E","F","G"};//fails!
ptrsPointToOutOfOrderElements(a);
}
}
My line of reasoning given below
I maintain i=0; and another variable t=i+1
when the while loop fails, less(a[i],a[t]) is false .We need to return a pointer to a[t] which is out of order. so i=t and return i.
if we reach right end of array, the test if(t == hi) passes and we assign i=t and now i points to end of array.
However, the code fails when the out of order element is in the 0th position in the array.
J,D,E,F,G
Instead of i (=0) we get i=1 because i=t is assgined.i ends up pointing to D instead of J.
Can someone point me in the right direction?
update:
this seems to work
public static void ptrsPointToOutOfOrderElements(Comparable[] a){
int lo = 0;
int hi = a.length-1;
int i = lo;
while(less(a[i],a[i+1])){
if(i+1 == hi){
break;
}
i++;
}
i++;
int j = hi;
while(less(a[j-1],a[j])){
if(j-1 == lo){
break;
}
j--;
}
j--;
for(Comparable x:a){
System.out.print(x+",");
}
System.out.println();
if(i>=j){
System.out.println("pointers crossed");
}
System.out.println("bad element or end of array at i="+i+"==>"+a[i]);
System.out.println("bad element or end of array at j="+j+"==>"+a[j]);
}
I do not think you have a problem:
String[] a = new String[]{"C","D","E","F","G"};//works, index should be 4 (but should it be so? It would indicate that G is out of order while it is not. I think you should return 5, indicating that none is out of order.
String[] a = new String[]{"B","C","A","D","G"};//works, index should be 2 as A is out of order
String[] a = new String[]{"J","D","E","F","G"};//works since the first out of order element is indeed D, with index 1
I have tried using simple for loop.
var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
for (var i = 0, j = arr.length - 1; i <= j; i++, j--) {
console.log(arr[i] + ' , ' + arr[j]);
}
Output :
1 , 10
2 , 9
3 , 8
4 , 7
5 , 6