the problem is there is two types of codes for same result(triangular number)
1.
for (int i = 1; i <= 10; i++) {
int triangular = 0;
for (int j = 1; j <= i; j++) {
triangular = triangular + j;
}
System.out.println(i + " = " + triangular);
2.
int x =1;
int triangular = 1;
while(x<=10){
System.out.println(x+ "=" +triangular);
x++;
triangular= triangular+x;
}
why for (1) "int triangualr" is 0 and for (2) its 1 ??? idont understand
In the method 1, int triangular is initialized everytime inside the for loop. Nested for loop is calculating the value for triangular and then you are printing the value
for (int i = 1; i <= 10; i++) {
int triangular = 0;
for (int j = 1; j <= i; j++) {
triangular = triangular + j;
}
System.out.println(i + " = " + triangular);
}
But in Method 2, Value for x=1 is printed in the first line of the while loop and then the value of x is incremented
int x =1;
int triangular = 1;
while(x<=10){
System.out.println(x+ "=" +triangular);
x++;
triangular= triangular+x;
}
So in first method, value for triangular is calculated starting from 1 and in second method value for triangular is calculated is not calculated for 1st Iteration
Because in the second code the "triangular" first is printed with value=1
while in the first code the "triangular" first is increased (goes from 0 to 1) and then is printed with value=1.
So in both cases what you see first is triangular=1.
Related
I have got this algorithm
int count = 0;
for(int i = n; i >= 1; i = i/2) {
for ( int j = 1; j <= i; j++) {
count++;
}
}
Am I right in saying that the Big-O for this would be n/2?
TL;DR The time complexity is O(n).
More details
Am I right in saying that the BigO for this would be n/2?
No that is accurate, in big-O notation you drop the constant part so (1/2)n simplifies to O(n).
I am not sure where that n/2 comes from because only the outer loop
for(int i = n; i >= 1; i = i/2) {
...
}
is log2n not n/2.
And with both loops together:
int count = 0;
for(int i = n; i >= 1; i = i/2) {
for ( int j = 1; j <= i; j++) {
count++;
}
}
the count would vary between N and 2N.
Let us go through the calculations:
int count = 0;
for(int i = n; i >= 1; i = i/2) {
for ( int j = 1; j <= i; j++) {
count++;
}
}
The inner loop will execute N iterations then N/2, then N/4 ... until N/N.
In another words we have (N + N/2 + N/4 ... N/N) which can be simplified to N * (1/2 + 1/4 + .. + 1/2^L)), with L = Log2 N.
This (1/2 + 1/4 + .. + ) series is well-known for being 1. Therefore, we can simplified N * (1/2 + 1/4 + .. + 1/2^L)) to O(N).
You are correct! This is basically a geometric progression with a quotient of 2 and the number of elements is lg(n) as we divide i by 2 each iteration of the outer loop.
1, 2, 4, ..., n
Using a known formula to calculate the sum, we get:
The reason we have lg (n) elements, is because we divide i each iteration by 2, thus we need to solve for the number of iterations k:
I have an assignment of which a part is to generate n random numbers between 0-99 inclusive in a 1d array, where the user enters n. Now, I have to print out those numbers formatted like this:
What is your number? 22 //user entered
1 2 3 4 5 6 7 8 9 10
----random numbers here---------
11 12 13 14 15 16 17 18 19 20
-----random numbers here--------
21 22
---two random numbers here---
Using those numbers, I have find lots of other things, (like min, max, median, outliers, etc.) and I was able to do so. However, I wasn't able to actually print it out in the format shown above, with no more than 10 numbers in one row.
Edit: Hello, I managed to figure it out, here's how I did it:
int counter = 0;
int count2 = 0;
int count3 = 0;
int add = 0;
int idx = 1;
int idx2 = 0;
if (nums > 10)
{
count3 = 10;
count2 = 10;
}
else
{
count3 = nums;
count2 = nums;
}
if (nums%10 == 0) add = 0;
else add = 1;
for (int i = 0; i < nums/10 + add; i++)
{
for (int j = 0; j < count3; j++)
{
System.out.print(idx + "\t");
idx++;
}
System.out.println();
for (int k = 0; k < count2; k++)
{
System.out.print(numbers[idx2] + "\t");
idx2++;
counter++;
}
System.out.println("\n");
if (nums-counter > 10)
{
count3 = 10;
count2 = 10;
}
else
{
count3 = nums-counter;
count2 = nums-counter;
}
}
Thank you to everyone who helped! Also, please let me know if you find a way to shorten what I have done above.
*above, nums was the number of numbers the user entered
I'd use a for-loop to make an array of arrays: and then formatting the lines using those values:
var arr_random_n = [1,2,3,4,5,6,7,8,9,0,1,2,3,6,4,6,7,4,7,3,1,5,7,9,5,3,2,54,6,8,5,2];
var organized_arr = [];
var idx = 0
for(var i = 0; i < arr.length; i+=10){
organized_arr[idx] = arr.slice(i, i+10); //getting values in groups of 10
idx+=1 //this variable represents the idx of the larger array
}
Now organized_arr has an array of arrays, where each array in index i contains the values to be printed in line i.
There's probably more concise ways of doing this. but this is very intuitive.
Let me know of any improvements.
Something like this might be what you're looking for.
private static void printLine(String msg)
{
System.out.println("\r\n== " + msg + " ==\r\n");
}
private static void printLine(int numDisplayed)
{
printLine(numDisplayed + " above");
}
public static void test(int total)
{
int[] arr = new int[total];
// Fill our array with random values
for (int i = 0; i < total; i++)
arr[i] = (int)(Math.random() * 100);
for (int i = 0; i < total; i++)
{
System.out.print(arr[i] + " ");
// Check if 10th value on the line, if so, display line break
// ** UNLESS it's also the last value - in that case, don't bother, special handling for that
if (i % 10 == 9 && i != total - 1)
printLine("Random Numbers");
}
// Display number of displayed elements above the last line
if (total < 10 || total % 10 != 0)
printLine(total % 10);
else
printLine(10);
}
To print 10 indexes on a line then those elements of an array, use two String variables to build the lines, then print them in two nested loops:
for (int i = 0; i < array.length; i += 10) {
String indexes = "", elements = "";
for (int j = 0; j < 10 && i * 10 + j < array.length; j++) {
int index = i * 10 + j;
indexes += (index + 1) + " "; // one-based as per example in question
elements += array[index] + " ";
}
System.out.println(indexes);
System.out.println(elements);
}
I want to get the fibonacci sequence entered by the user in array. The task given to me was "Ask the user for 2 integer input which will be taken for first and second array elements of size 10 array."
Here is my code.
int limit = 10;
int[] fib = new int[limit];
fib[0] = 0;
fib[1] = 1;
for (int j = 1; j < 2; j++)
{
System.out.print("Enter number " + "[" + j + "]: ");
num[j] = reader.nextInt();
num[j] = fib[j+1] + fib[j+2];
System.out.println("");
}
System.out.print("Result: ");
for(int j = 0; j < limit; j++ )
{
System.out.print(fib[j] + " ");
System.out.print("");
}
I badly need help for this one, been searching for solution for hours and still don't get it.
I'll just make some corrections to your code and explain them:
int limit = 10;
int[] fib = new int[limit];
// fib[0] = 0;
// fib[1] = 1;
// The two lines above are wrong. Even though the real fibonacci sequence starts
// with 0 and 1, the question asks for the first two terms to come from user
// inputs. Instead, you can initialize them below:
// In your old code, you had "j = 1; j < 2; j++". However, that only loops once.
// So, have your condition to be j <= 2 instead: (I'm assuming that you want 1
// and 2 and not zero-based because it should print out "Enter number [1]:" and
// "Enter number [2]:"
for (int j = 0; j < 2; j++) // Not "j < 2"
{
System.out.print("Enter number " + "[" + j + "]: ");
fib[j] = reader.nextInt(); // not num[j] = ..., it's fib[j] = ...
// num[j] = fib[j+1] + fib[j+2];
// You don't need this ^^^
System.out.println("");
}
// Now you need to fill in the array:
for (int j = 2; j < limit; j++)
{
fib[j] = fib[j - 1] + fib[j - 2];
}
System.out.print("Result: ");
for(int j = 0; j < limit; j++)
{
System.out.print(fib[j] + " ");
System.out.print("");
}
You have a number of mistakes here. Try to answer the following things:
What is the purpose of num variable here? You are using it to take input (num[j] = reader.nextInt();), then you are using it to store the Fibonacci sequence as well (num[j] = fib[j+1] + fib[j+2];)!!!
The first loop is only running one time. Assuming that your calculations are correct (which is not in this case!), it will have values for the first fibonacci number only.
Finally when you are printing the Fibonacci numbers, you are not using the num variable! Why?
Anyway, here is a solution to your problem. But, it will not help you unless you understand the logic behind it and you know which line is doing what!!!
int limit = 10;
int[] fib = new int[limit];
for (int j = 0; j < 2; j++)
{
System.out.print("Enter number " + "[" + j + 1 + "]: ");
fib[j] = reader.nextInt(); // These are the first two fibonacci numbers provided by the user
System.out.println("");
}
System.out.print("Result: ");
for(int j = 0; j < limit; j++ )
{
if( j > 1 ) // You only calculate from the third Fibonacci, as the first two were given by user
{
fib[j] = fib[j-1] + fib[j-2];
}
System.out.print(fib[j] + " ");
System.out.print("");
}
In this case, I want to add two numbers in this array in to obtain a specific sum when added, let’s say, 4. I also want to output what indices are being added in order to obtain that specific sum, just to see the inner workings of my code. What am I doing wrong?
public static int addingNumbers(int[] a) {
int i1 = 0, i2 = 0;
for(int i = 0, j = i + 1; i < a.length && j < a.length; i++, j++) {
if(a[i] + a[j] == 4) { // index 0 and index 2 when added gives you a sum 4
i1 = i;
i2 = j;
}
}
System.out.println("The indices are " + i1 + " and " + i2);
return i1;
}
public static void main(String args[]) {
int[] a = {1, 2, 3, 4, 5, 6};
System.out.println(addingNumbers(a));
}
The error you are making is using only one loop that iterates over the array once:
for(int i = 0, j = i + 1; i < a.length && j < a.length; i++, j++) {
In your loop you are setting i to 0 and j to 1, then you increment them with every step. So you are only comparing adjacent places in your array:
iteration: a[0] + a[1]
iteration: a[1] + a[2]
iteration: a[2] + a[3]
etc. pp
Since your array doesn't have two adjacent elements that sum up to 4 your if(a[i] + a[j] == 4) will never be entered and i1, i2 will still be 0 when the loop is finished.
To compare every array element with each other you should use 2 nested loops:
public static int addingNumbers(int[] a) {
int i1 = -1, i2 = -1;
for(int i = 0; i < a.length ; i++) {
for(int j = i+1; j < a.length ; j++) {
if(a[i] + a[j] == 4) { // index 0 and index 2 when added gives you a sum 4
i1 = i;
i2 = j;
}
}
}
if(i1>=0 && i2 >=0) {
System.out.println("The indices are " + i1 + " and " + i2);
}
return i1;
}
Note that this will only print out the last detected 2 indices that add up to 4. If you want to be able to detect multiple possible solutions and print them out could for example move the System.out.println into the if block.
It can never be == 4 because 1+2=3 then 2+3=5. So it does nothing.
There is a logic error in your code. The sum you are checking in your code is never for.
I added some debug output for easy checking:
public static int addingNumbers(int[] a) {
int i1 = 0, i2 = 0;
for(int i = 0, j = i + 1; i < a.length && j < a.length; i++, j++) {
int sum = a[i] + a[j];
System.out.println(sum);
if(sum == 4) { // index 0 and index 2 when added gives you a sum 4
i1 = i;
i2 = j;
}
}
System.out.println("The indices are " + i1 + " and " + i2);
return i1;
}
Output is: 3
5
7
9
11
The indices are 0 and 0
0
this algorithm will never be able to add a[0] to a[2], be cause when you put j=i+1 it will always be 0+1 then 1+2 ... The sum of tow adjacent numbers is never pair.
An other matter is the condition to stop your loop must be j < a.length-1
try to explain more of what you want from your algorithm.
Are you over complicating this on purpose?
Trying to figure out your intention for this task.
Why don't you just do (this is pseudo):
for length of i {
if (a[i] + a[i+1] == 4) {
System.out.println("The indices are " + a[i] + " and " + a[i+1]);
}
}
I'm trying to print the following:
2
2 4
2 4 6 ..etc
The code I have written (below) prints the following:
2
4 6
8 10 12 ...etc
Can anyone spot where I'm going wrong? The n variable comes from the main method which I am not including.
public static void printEvenTable(int n) {
int i;
int j;
int k = 0;
for (i = 1; i <= n; i++) {
for (j = 0; j < i; j++)
System.out.print(" " + (k += 2));
System.out.println(" ");
}
}
You need to prevent the variable k from using its old value by reassigning 0 to it right after the second for loop. Placing k = 0; before the second for loop makes redundant reassigning because it has already been assigned right before the loop. Ensure program optimization. If you are using a good editor, it show you some warning if placed before the second for loop.
for (i = 1; i <= n; i++) {
for (j = 0; j < i; j++) {
System.out.print(" " + (k += 2));
}
k=0;
System.out.println(" ");
}
Here, the inner loop is increasing the variable k by 2 in each steps after the first execution of the inner loop k becomes 2 from the initial value 0. In the second iteration of the outer loop k starts as 2. After k+=2, k becomes 4 so second line of output starts from 4. That's why we need to re-initialize k to 0 before each inner loop.
public static void printEvenTable(int n) {
int i;
int j;
int k = 0;
for (i = 1; i <= n; i++) {
k = 0;
for (j = 0; j < i; j++)
System.out.print(" " + (k += 2));
System.out.println(" ");
}
}