We Keep Coding

Down to Zero II

This is the question:
You are given Q queries. Each query consists of a single number N . You can perform any of the operations on in each move:
If we take 2 integers a and b where N=a*b (a ,b cannot be equal to 1), then we can change N=max(a,b)
Decrease the value of N by 1 .
Determine the minimum number of moves required to reduce the value of to .
Input Format
The first line contains the integer Q.
The next Q lines each contain an integer,N .
Output Format
Output Q lines. Each line containing the minimum number of moves required > to reduce the value of N to 0.
I have written the following code. This code is giving some wrong answers and also giving time limit exceed error . Can you tell what are the the mistakes present in my code ? where or what I am doing wrong here?
My code:
public static int downToZero(int n) {
// Write your code here
int count1=0;
int prev_i=0;
int prev_j=0;
int next1=0;
int next2=Integer.MAX_VALUE;
if (n==0){
return 0;
}
while(n!=0){
if(n==1){
count1++;
break;
}
next1=n-1;
outerloop:
for (int i=1;i<=n;i++){
for (int j=1;j<=n;j++){
if (i*j==n){
if (prev_i ==j && prev_j==i){
break outerloop;
}
if (i !=j){
prev_i=i;
prev_j=j;
}
int max=Math.max(i,j);
if (max<next2){
next2=max;
}
}
}
}
n=Math.min(next1,next2);
count1++;
}
return count1;
}
This is part is coded for us:
public class Solution {
public static void main(String[] args) throws IOException {
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));
int q = Integer.parseInt(bufferedReader.readLine().trim());
for (int qItr = 0; qItr < q; qItr++) {
int n = Integer.parseInt(bufferedReader.readLine().trim());
int result = Result.downToZero(n);
bufferedWriter.write(String.valueOf(result));
bufferedWriter.newLine();
}
bufferedReader.close();
bufferedWriter.close();
}
}
Ex: it is not working for number 7176 ....

To explore all solution tree and find globally optimal solution, we must choose the best result both from all possible divisor pairs and from solution(n-1)
My weird translation to Java (ideone) uses bottom-up dynamic programming to make execution faster.
We calculate solutions for values i from 1 to n, they are written into table[i].
At first we set result into 1 + best result for previous value (table[i-1]).
Then we factor N into all pairs of divisors and check whether using already calculated result for larger divisor table[d] gives better result.
Finally we write result into the table.
Note that we can calculate table once and use it for all Q queries.
class Ideone
{
public static int makezeroDP(int n){
int[] table = new int[n+1];
table[1] = 1; table[2] = 2; table[3] = 3;
int res;
for (int i = 4; i <= n; i++) {
res = 1 + table[i-1];
int a = 2;
while (a * a <= i) {
if (i % a == 0)
res = Math.min(res, 1 + table[i / a]);
a += 1;
}
table[i] = res;
}
return table[n];
}
public static void main (String[] args) throws java.lang.Exception
{
int n = 145;//999999;
System.out.println(makezeroDP(n));
}
}
Old part
Simple implementation (sorry, in Python) gives answer 7 for 7176
def makezero(n):
if n <= 3:
return n
result = 1 + makezero(n - 1)
t = 2
while t * t <= n:
if n % t == 0:
result = min(result, 1 + makezero(n // t))
t += 1
return result
In Python it's needed to set recursion limit or change algorithm. Now use memoization, as I wrote in comments).
t = [-i for i in range(1000001)]
def makezeroMemo(n):
if t[n] > 0:
return t[n]
if t[n-1] < 0:
res = 1 + makezeroMemo(n-1)
else:
res = 1 + t[n-1]
a = 2
while a * a <= n:
if n % a == 0:
res = min(res, 1 + makezeroMemo(n // a))
a += 1
t[n] = res
return res
Bottom-up table dynamic programming. No recursion.
def makezeroDP(n):
table = [0,1,2,3] + [0]*(n-3)
for i in range(4, n+1):
res = 1 + table[i-1]
a = 2
while a * a <= i:
if i % a == 0:
res = min(res, 1 + table[i // a])
a += 1
table[i] = res
return table[n]

We can construct the directed acyclic graph quickly with a sieve and
then compute shortest paths. No trial division needed.
Time and space usage is Θ(N log N).
n_max = 1000000
successors = [[n - 1] for n in range(n_max + 1)]
for a in range(2, n_max + 1):
for b in range(a, n_max // a + 1):
successors[a * b].append(b)
table = [0]
for n in range(1, n_max + 1):
table.append(min(table[s] for s in successors[n]) + 1)
print(table[7176])
Results:
7

EDIT:
The algorithm uses Greedy approach and doesn't return optimal results, it just simplifies OP's approach. For 7176 given as example, below algorithm returns 10, I can see a shorter chain of 7176 -> 104 -> 52 -> 13 -> 12 -> 4 -> 2 -> 1 -> 0 with 8 steps, and expected answer is 7.
Let's review your problem in simple terms.
If we take 2 integers a and b where N=a*b (a ,b cannot be equal to 1), then we can change N=max(a,b)
and
Determine the minimum number of moves required to reduce the value of to .
You're looking for 2 factors of N, a and b and, if you want the minimum number of moves, this means that your maximum at each step should be minimum. We know for a fact that this minimum is reached when factors are closest to N. Let me give you an example:
36 = 1 * 36 = 2 * 18 = 3 * 12 = 4 * 9 = 6 * 6
We know that sqrt(36) = 6 and you can see that the minimum of 2 factors you can get at this step is max(6, 6) = 6. Sure, 36 is 6 squared, let me take a number without special properties, 96, with its square root rounded down to nearest integer 9.
96 = 2 * 48 = 3 * 32 = 4 * 24 = 6 * 16 = 8 * 12
You can see that your minimum value for max(a, b) is max(8, 12) = 12, which is, again, attained when factors are closest to square root.
Now let's look at the code:
for (int i=1;i<=n;i++){
for (int j=1;j<=n;j++){
if (i*j==n){
You can do this in one loop, knowing that n / i returns an integer, therefore you need to check if i * (n / i) == n. With the previous observation, we need to start at the square root, and go down, until we get to 1. If we got i and n / i as factors, we know that this pair is also the minimum you can get at this step. If no factors are found and you reach 1, which obviously is a factor of n, you have a prime number and you need to use the second instruction:
Decrease the value of N by 1 .
Note that if you go from sqrt(n) down to 1, looking for factors, if you find one, max(i, n / i) will be n / i.
Additionally, if n = 1, you take 1 step. If n = 2, you take 2 steps (2 -> 1). If n = 3, you take 3 steps (3 -> 2 -> 1). Therefore if n is 1, 2 or 3, you take n steps to go to 0. OK, less talking, more coding:
static int downToZero(int n) {
if (n == 1 || n == 2 || n == 3) return n;
int sqrt = (int) Math.sqrt(n);
for (int i = sqrt; i > 1; i--) {
if (n / i * i == n) {
return 1 + downToZero(n / i);
}
}
return 1 + downToZero(n - 1);
}
Notice that I'm stopping when i equals 2, I know that if I reach 1, it's a prime number and I need to go a step forward and look at n - 1.
However, I have tried to see the steps your algorithm and mine takes, so I've added a print statement each time n changes, and we both have the same succession: 7176, 92, 23, 22, 11, 10, 5, 4, 2, 1, which returns 10. Isn't that correct?

So, I found a solution which is working for all the test cases -
static final int LIMIT = 1_000_000;
static int[] solutions = buildSolutions();
public static int downToZero(int n) {
// Write your code here
return solutions[n];
}
static int[] buildSolutions() {
int[] solutions = new int[LIMIT + 1];
for (int i = 1; i < solutions.length; i++) {
solutions[i] = solutions[i - 1] + 1;
for (int j = 2; j * j <= i; j++) {
if (i % j == 0) {
solutions[i] = Math.min(solutions[i], solutions[i / j] + 1);
}
}
}
return solutions;
}
}

Java:Three digit Sum - Find out all the numbers between 1 and 999 where the sum of 1st digit and 2nd digit is equal to 3rd digit

Problem statement: Three digit sum - Find all the numbers between 1 and 999 where the sum of the 1st digit and the 2nd digit is equal to the 3rd digit.
Examples:
123 : 1+2 = 3
246 : 2+4 = 6
Java:
public class AssignmentFive {
public static void main(String[] args) {
int i=1;
int valuetwo;
int n=1;
int sum = 0;
int valuethree;
int valueone = 0;
String Numbers = "";
for (i = 1; i <= 999; i++) {
n = i;
while (n > 1) {
valueone = n % 10;/*To get the ones place digit*/
n = n / 10;
valuetwo = n % 10;/*To get the tens place digit*/
n = n / 10;
valuethree = n;/*To get the hundreds place digit*/
sum = valuethree + valuetwo;/*adding the hundreds place and
tens place*/
}
/*Checking if the ones place digit is equal to the sum and then print
the values in a string format*/
if (sum == valueone) {
Numbers = Numbers + n + " ";
System.out.println(Numbers);
}
}
}
}
I got my result :
1
10
100
1000
10000
100000
1000000
10000000
100000000
1000000000
10000000001
100000000011
1000000000111
10000000001111
100000000011111
1000000000111111
10000000001111111
100000000011111111
1000000000111111111
Process finished with exit code 0
The result is not showing the actual result like it should be which should show values like: 123, 246 (Please refer to the problem statement above.)
Please let me know what seems to be the issue with the code and how to tweak it.

Don't know what you're trying to do with that while loop, or why you are building up a space-separated string of numbers.
Your code should be something like:
for (int n = 1; n <= 999; n++) {
int digit1 = // for you to write code here
int digit2 = // for you to write code here
int digit3 = // for you to write code here
if (digit1 + digit2 == digit3) {
// print n here
}
}

So basically your question is how to calculate the numbers, right?
My first hint for you would be how to get the first, second and third value from a 2 or 3 digit number.
For example for 3 digits you can do int hundretDigit = (n - (n % 100)) % 100. Of course this is really inefficient. But just get code working before optimizing it ;)
Just think about a way to get the "ten-digit" (2nd number). Then you add them and if they equal the third one you write System.out.println(<number>);
EDIT:
For 2 digit numbers I will give you the code:
if(i >= 10 && i <= 99) {
int leftDigit = (i - (i % 10)) / 10;
if(leftDigit == (i % 10)) {
//Left digit equals right digit (for example 33 => 3 = 3
System.out.println(i);
}
}
Try again and edit your source code. If you have more questions I will edit my (this) answer to give you a little bit more help if you need!

Terminated due to timeout error [duplicate]

I am working on a program that takes an integer and finds the number of combinations of consecutive sums that the integer has:
The number 13 can be expressed as a sum of consecutive positive
integers 6 + 7. Fourteen can be expressed as 2 + 3 + 4 + 5, also a sum
of consecutive positive integers. Some numbers can be expressed as a
sum of consecutive positive integers in more than one way. For
example, 25 is 12 + 13 and is also 3 + 4 + 5 + 6 + 7.
I researched and read that it's the number of odd factors minus one. So I wrote a program that finds the number of odd factors and my answer is still wrong in certain cases. Any insight?
Code seems to work fine but there is a crash due to Timeout which is probably due to optimization error.
The constraints for possible input size is
1 to 10^(12)
The code below is copied from alfasin's answer below:
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
static long consecutive(long num) {
while (num % 2 == 0) num /= 2;
return consecutiveHelper(num);
}
public static long consecutiveHelper(long num) {
return LongStream.rangeClosed(3, (num / 2)).parallel().filter(x -> x % 2 != 0).map(fn -> (num % fn == 0) ? 1 : 0).sum();
}
public static void main(String[] args) throws IOException {
Scanner in = new Scanner(System.in);
final String fileName = System.getenv("OUTPUT_PATH");
BufferedWriter bw = null;
if (fileName != null) {
bw = new BufferedWriter(new FileWriter(fileName));
}
else {
bw = new BufferedWriter(new OutputStreamWriter(System.out));
}
int res;
long num;
num = Long.parseLong(in.nextLine().trim());
res = consecutive(num);
bw.write(String.valueOf(res));
bw.newLine();
bw.close();
}
}
This is what i currently have

As the post i answered to was duplicate, I copied my answer here as well.Let's try to find a pseudo-optimized method to resolve your problem :
What you need to do is to decompose your number in prime factors.
For example, if you take 1200 :
1200 = 2*2*2*2*3*5*5 = 1 * 2^4 * 3^1 * 5^2
You can then analyze how you could get odd factors with those prime factors. A quick analyze will tell you that :
odd * odd = odd
odd * even = even
even * even = even
With that in mind, let's find all the factors we get with odd * odd :
1 * 1 = 1
3 * 1 = 3
5 * 1 = 5
5 * 3 = 15
5 * 5 = 25
5 * 5 * 3 = 75
A quick way to find these combinations without writing them all is the "plus 1 method" : add 1 to the number of occurences of each prime odd factor, and multiply them together :
We found that 1200 = 1 * 2^4 * 3^1 * 5^2, so we can do :
("number of 3" + 1) ("number of 5" + 1) = (1 + 1) ( 2 + 1) = 6
There are 6 odd factors for the number 1200, and as you stated, remove 1 from that number to get the number of combinations of consecutive sums that 1200 has :
6 - 1 = 5 <-- woohoo ! finally got the result !
Now, let's look at the code. What we want to have is a Map, the keys being the prime factors and the values being the number of their occurences :
/*
If number is odd,
find the number in the keys and add 1 to its value.
If the number is not in the keys, add it with value = 1.
*/
public static void addValue(Map<Integer, Integer> factors, int i) {
if(i % 2 != 0) {
int count = factors.containsKey(i) ? factors.get(i) : 0;
factors.put(i, ++count);
}
}
/*
Classic algorithm to find prime numbers
*/
public static Map<Integer, Integer> oddPrimeFactors(int number) {
int n = number;
Map<Integer, Integer> factors = new HashMap<>();
for (int i = 2; i <= n / i; i++) {
while (n % i == 0) {
addValue(factors, i);
n /= i;
}
}
if(n > 1) addValue(factors, n);
return factors;
}
With that, let's try to print what the map contains for number 1200 :
public static void main(String[] args) {
int n = 1200;
System.out.println(oddPrimeFactors(n));
}
$n : {3=1, 5=2}
Good ! Now let's finish the program with the method we developed before :
public static int combinations = 1;
public static void main(String[] args) {
int n = 1200;
oddPrimeFactors(n).forEach((key, value) -> combinations *= (value + 1));
combinations--;
System.out.println(combinations);
}
$combinations = 5
Finished ! feel free to ask if you did not understand something !
Note : I tried my program with the max value Integer can handle and it took less than one second for my program to proceed, which seems pretty fast to me. It could probably be faster though, it's up to you to find the most optimized version of this code !

Here are the optimizations that we discussed in the comments section, see comments as markers:
static int consecutive(long num) {
while (num % 2 == 0) num /= 2; // 1st opt.
return consecutiveHelper(num)-1;
}
public static int consecutiveHelper(long num) {
long factorNumber = 1;
int count = 0;
while(factorNumber <= num / 2) { // 2nd opt.
if(num % factorNumber == 0) {
count++;
}
factorNumber += 2; // 3rd opt.
}
if (num % 2 != 0) {
count++;
}
return count;
}
UPDATE
I managed to reduce ~50% runtime for big-numbers (10^12) by using Java 8 Stream interface and running in parallel:
static long consecutive(long num) {
while (num % 2 == 0) num /= 2;
return consecutiveHelper(num);
}
public static long consecutiveHelper(long num) {
return LongStream
.rangeClosed(3, (num / 2))
.parallel()
.filter(x -> x % 2 != 0)
.map(fn -> (num % fn == 0) ? 1 : 0)
.sum();
}
That said, parallel will be more expensive when you're dealing with smaller numbers. If you want your answer to be optimal you should use both methods: for smaller numbers use the first and for large numbers use the latter.

14 - Project Euler - Java - Incorrect Answer - Bug?

I cannot find the bug in this code. If it were not for Project Euler ruling my answer as incorrect, I would swear to the heavens that my code is right.
I could use another approach, but this problem is not all that complex, and yet I have been utterly defeated trying to find the bug.
The question is :
The following iterative sequence is defined for the set of positive integers:
n → n/2 (n is even)
n → 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.
Which starting number, under one million, produces the longest chain?
My code is :
public class CollatzSequence014 {
public static void main(String [] args){
long start = System.currentTimeMillis();
long maxTotal = 0;
long inputNum = 0;
for (long i = 3; i < 1000000 ; i++){
long total = generateSequence(i);
if (total > maxTotal){
inputNum = i;
maxTotal = total;
}
}
long end = System.currentTimeMillis();
System.out.println("The input number was : " + inputNum + " and the total was " + maxTotal);
System.out.println("Time taken : " + (end - start) + "ms");
}
public static long generateSequence(long n){
long counter = 0;
long currentDigit = n;
while (currentDigit > 1){
counter++;
if (n % 2 == 0){
currentDigit = currentDigit / 2;
}
else {
currentDigit = (3 * currentDigit) + 1;
}
}
return counter;
}
}

You have to check if the currentDigit is even, not n.
if(currentDigit % 2 == 0)

First, how you should look for the bug: try outputting the sequence (not just the length) for small n (e.g. 13, since the problem already gives you the correct sequence there). You'll see that you get 13, 40, 121... and that should already tell you where the bug is.

error is here :
if (n % 2 == 0){
should be :
if (currentDigit % 2 == 0){

using mod 3 in Java

So, I want to find what numbers between 1 and 100 are divisible by 3 and 7. I got it to work, except for one of the numbers. For some reason, 3 % 3 is giving me 3 as a remainder, but 6 % 3 is giving me 0. This is my code:
public class factors
{
public static void main(System args[])
{
//Variables
int integer, remainder;
//Displays header
System.out.print("Integers less than 100 that are \nevenly divisible by 3 or 7");
//Loops through each integer
for (integer = 1; integer <= 100; integer++)
{
remainder = integer % 3; //determines if 3 is a factor
if (remainder == 0) //displays integer
{
System.out.println(integer + " is divisible by 3");
}
remainder = integer % 7; //determines if 7 is a factor
if (remainder == 0) //displays integer
{
System.out.println(integer + " is divisible by 7");
}
}
}
}Does anyone know why this isn't working for the number 3?

You code is actually doing
remainder = 3 % 7; // equals 3.
The best way to determine why your code is not doing what you think is to step through your code using a debugger.
All the multiples of 3 & 7 will be multiples of 21, i.e. 21, 42, 63, 84.

Your 3 is getting tacked onto the end of the line of text above. You'll be seeing
Integers less than 100 that are
evenly divisible by 3 or 73
because you wrote print instead of println for this line of text. The % operator is working just fine, and 3 % 3 is indeed 0, not 3.

You are not outputting a remainder - you are displaying integer. So for 3 it should print 3.
Make you print statements more definite:
System.out.println(integer + " is divisible by 3"); // for the first `if`
and
System.out.println(integer + " is divisible by 7"); // for the second `if`
This should clear your confusion.
Your logic prints number divisible by 3 or 7.

Firstly, your code can be shortened to:
//and
for (int i = 1; i <= 100; i++){
if(i % 3 == 0 && i % 7 == 0) {
System.out.println(i);
}
}
//or
for (int i = 1; i <= 100; i++){
if(i % 3 == 0 || i % 7 == 0) {
System.out.println(i);
}
}
Also I note you're not declaring a type for your integer, remainder variables. I didn't attempt to recreate with those issues; start by solving that.