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How can I find the time complexity of an algorithm?
(10 answers)
Closed 5 years ago.
how do you get the time complexity of the functions count and printall in ThreeSum.java?
public static void printAll(int[] a) {
int n = a.length;
for (int i = 0; i < n; i++) {
for (int j = i+1; j < n; j++) {
for (int k = j+1; k < n; k++) {
if (a[i] + a[j] + a[k] == 0) {
System.out.println(a[i] + " " + a[j] + " " + a[k]);
}
}
}
}
}
public static int count(int[] a) {
int n = a.length;
int count = 0;
for (int i = 0; i < n; i++) {
for (int j = i+1; j < n; j++) {
for (int k = j+1; k < n; k++) {
if (a[i] + a[j] + a[k] == 0) {
count++;
}
}
}
}
Although time complexity questions are sometimes very tough but this one is an easy cake. You can simply check it out as follows,
for (int i = 0; i < n; i++) // it runs for n times
for (int j = i+1; j < n; j++) // it runs for n-i-1 time
for (int k = j+1; k < n; k++) // it runs for n-j-1 times
Now since they are nested loops that means each of inner loop runs as many number of times as outer loop.
total = n * ( n-i-1 ) * ( n-j-1 )
= n^3 ... // ignoring all other lower power terms
So time complexity of this code will be O(n^3).
Related
Would it be possible to produce this layout with nested loops? I'm still new to nested to Java/loops and cannot solve this issue.
*****====+
*****===++
*****==+++
*****=++++
*****+++++
====++++++
===+++++++
==++++++++
=+++++++++
I'm having trouble looping through five times with the "*" character without allowing the "+" to increment.
Here is my code:
class Main {
public static void main(String[] args) {
for (int k = 4; k > 0; k--) {
System.out.print("*****");
for (int l = 0; l < k; l++) {
System.out.print("=");
}
for (int m = 0; m < 1; m++) {
for (int n = 0; n < m; n++) {
System.out.print("+");
}
}
System.out.println();
}
System.out.print("*****+++++");
}
}
there are could be multiple approaches to this problem, here is how I would think about it:
you need to display 9 lines of "something", so lets have top level loop:
for (int i=0; i<9; i++) {...}
now, each iteration of this loop you need to display X stars, Y equal signs, Z plus signs:
for (int i=0; i<9; i++) {
for (int j=0; j< #X# ; j++) System.out.print("*");
for (int j=0; j< #Y# ; j++) System.out.print("=");
for (int j=0; j< #Z# ; j++) System.out.print("+");
System.out.println();
}
now you need to determine rules how X,Y,Z are changed, here is the logic I came up with:
if (stars > 0 && equals == 0) {
stars = 0;
equals = 5;
}
equals--;
pluses++;
so, final code will look like:
public static void main(String[] args) throws InterruptedException {
int stars = 5; // initial state
int equals = 4;
int pluses = 1;
for (int i=0; i<9; i++) {
for (int j=0; j<stars; j++) System.out.print("*");
for (int j=0; j<equals; j++) System.out.print("=");
for (int j=0; j<pluses; j++) System.out.print("+");
System.out.println();
if (stars > 0 && equals == 0) {
stars = 0;
equals = 5;
}
equals--;
pluses++;
}
}
If you notice there is a pattern.
It prints a * starting from left to right.
It prints a + for each increasing number starting from right to left.
It prints a = instead of * if you are past mid point (first the mid point of left to right, then the mid point of top to bottom).
The logic can be applied as following,
class Main {
public static void main(String[] args) {
int max = 10;
int switchPoint = max - 1;
for(int i = 1; i <= max -1; i++) {
for(int j = 1; j <= max; j++) {
if(j > switchPoint)
System.out.print("+");
else if( i > max/2 || j > max / 2)
System.out.print("=");
else
System.out.print("*");
}
switchPoint--;
System.out.println();
}
}
}
/* Output:
*****====+
*****===++
*****==+++
*****=++++
*****+++++
====++++++
===+++++++
==++++++++
=+++++++++
*/
As you may notice the behavior of the "*" and "=" is different in the first five lines than in the next five lines, so you may either divide the loop in two loops or make a single one and check wether you are printing the first five lines or the last ones, that check may be done either by an if statement or in the loops conditions.
So the code will look like:
class Main {
public static void main(String[] args) {
for (int i = 1; i < 10; i++){
for (int j = 1; i <= 5 && j <= 5; j++){
System.out.print("*");
}
for (int k = 5; (i <= 5 && k > i) || (i > 5 && k > i-5); k--){
System.out.print("=");
}
for (int l = 1; l <= i; l++){
System.out.print("+");
}
System.out.println();
}
}
}
public class A {
public static void main(String[] args) {
for (int k = 4; k > 0; k--) {
System.out.print("*****");
for (int l = 0; l < k; l++) {
System.out.print("=");
}
for (int m = 0; m < 5 - k; m++) {
System.out.print("+");
}
System.out.println();
}
System.out.print("*****+++++");
System.out.println();
for (int s = 4; s > 0; s--) {
for (int k = 0; k < s; k++) {
System.out.print("=");
}
for (int l = 0; l < 5; l++) {
System.out.print("+");
}
for (int m = 0; m < 5 - s; m++) {
System.out.print("+");
}
System.out.println();
}
}
}
I am trying to figure out how to increment a pyramid based on input from the user.\
If a user enters the number 3, my program will print a pyramid of height 3, three times.\
What I would like it to do instead, is to print 3 pyramids, but the first pyramid should have a height of 1, and the second a height of 2, and the third a height of 3.
Here is my code:
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
for (int l = 0; l < n; l++) {
System.out.println("Pyramid " + n);
for (int i = 1; i <= n; i++) {
for (int j = 0; j < n - i; j++) {
System.out.print(".");
}
for (int j = 0; j < i; j++) {
System.out.print("#");
}
for (int j = 1; j < i; j++) {
System.out.print("#");
}
for (int k = 0; k < n - i; k++) {
System.out.print(".");
}
System.out.println();
}
}
You use l to keep track of the number of pyramid you're working on, so you could just use it in the loop instead of n. Note that l starts with 0, not 1, so you may want to amend the loop accordingly and run from 1 to n, not from 0 to n-1
for (int l = 1; l <= n; l++) { // Note the loop starts at 1
System.out.println("Pyramid " + l);
for (int i = 1; i <= l; i++) { // Note the usage of l instead on n
for (int j = 0; j < l - i; j++) {
System.out.print(".");
}
for (int j = 0; j < i; j++) {
System.out.print("#");
}
for (int j = 1; j < i; j++) {
System.out.print("#");
}
for (int k = 0; k < l - i; k++) { // Note the usage of l instead on n
System.out.print(".");
}
System.out.println();
}
}
All you have to change is this part of code
System.out.println("Pyramid " + (l+1));
for (int i = 1; i <= l+1; i++) {
I was trying to find time complexity of these 2 codes but i am not sure about my answers .
code1
int i = 1;
int count = 0;
while (i < n) {
for (int j = 0; j < i; j++) {
count++;
}
i *= 2;
}
I calculated the number of steps in loops and I reached to (log n)^2 but i am not sure about it .
Code2
int k=0;
for (int i = 2; i <= n; i++) {
for (int j = 2; j * j <= i; j++) {
if (i % j == 0) {
k++;
break;
}
}
}
and for this one I got ( n * log n)
actually I am new to calculating time complexity and I am not sure about them , could you help me find the correct answer .
I need to print a table that looks like this if the user entered a 5 using nested for loops:
****5
***45
**345
*2345
12345
I've been working on this for hours and the closest I got was:
int size = scan.nextInt();
for (int i = 1; i <= size; i++)
{
for (int star = size-1; star >= i; star--)
System.out.print("*");
for (int k = 1; k <= i; k++)
System.out.print(i);
System.out.println();
}
Which outputs this:
****1
***12
**123
*1234
12345
You have too many loops; I find it easier to reason about zero based looping so I'm going to use that. Iterate i and j from 0 to size. If j + 1 is greater than size - i - 1 then we want to print j + 1. Otherwise, we want a star. Like,
for (int i = 0; i < size; i++) {
for (int j = 0; j < size; j++) {
if (j + 1 > size - i - 1) {
System.out.print(j + 1);
} else {
System.out.print('*');
}
}
System.out.println();
}
For size = 5 that outputs (as requested)
****5
***45
**345
*2345
12345
If you simply must have one based indices, that would be
for (int i = 1; i <= size; i++) {
for (int j = 1; j <= size; j++) {
if (j > size - i) {
System.out.print(j);
} else {
System.out.print('*');
}
}
System.out.println();
}
If you want to keep your loops and avoid if statements, you can tweak last loop by changing
for (int k = 1; k <= i; k++)
into
for (int k = 1+size-i; k <= size; k++)
Btw I also find it way easier to start loops from 0, so updated code would look like this:
int size = scan.nextInt();
for (int i = 0; i < size; i++)
{
for (int star = size-1; star > i; star--)
System.out.print("*");
for (int k = size-i; k <= size; k++)
System.out.print(k);
System.out.println();
}
I hope it helps
My output is coming wrong and I realised the reason that I am running "for loop" for "k" a value of "for loop" of j..
Running for loops for "j and k" simultaneously would solve my problem. How to do that?
public void yestimatedvalue() {
for (int i = 1; i < bmarray.length; i++) {
double g = 0;
s = i;
int p = missingrowcolindex(s);
System.out.println("p::\t" + p);
for (int j = 0; j < bmarray[i].length; j++) {
if (j != p) {
g = arraybeta[0][p];
for (int k = 1; k < arraybeta.length; k++) {
g = g + ((bmarray[i][j]) * (arraybeta[k][p]));
}
}
}
System.out.println("g::\t" + g);
}
}
output::
g:: 23.99999999999998
g:: 4.9999999999999964
g:: 5.999999999999995
g:: 1.0
Answering this question Running for loops for "j and k" simultaneously is:
for(int j=0, k=1; (j < bmarray[i].length && k < arraybeta.length); j++, k++){
//....
}
Yes, you can. Java (and all c-like languages) support syntax like this:
for(int i=0, j=5, k=2; i < 10 && j < 50; i++, j+=2, k+=3) {
}