This question already has answers here:
Explain this dynamic programming climbing n-stair code
(3 answers)
Closed 6 years ago.
Question is: You are climbing a stair case. It takes n steps to reach to the top.
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
And I saw a java code which is correct, but I don't understand the logic. Can anyone explain it to me? What's a,b,c stand of?
public int climbStairs(int n) {
if (n<2) return 1;
int a = 1;
int b = 1;
int c = 1;
for (int i=2; i<=n; i++){
c = b;
b = a + b;
a = c;
}
return b;
}
The code itself is basically a fibonacci-number generator.
int a = 1;
int b = 1;
int c = 1;
for (int i=2; i<=n; i++){
c = b;
b = a + b;
a = c;
}
creates the nth fibonacci-number, starting with 1 for n = 0.
So the more important question:
How does the number of possible ways correspond to the fibonacci-row?
For n = 0 and n = 1, the answer is pretty simple: there's exactly one way: don't move (0), take a single step (1). For any other n, we can use a recursive approach: there are two ways to reach step n: take a short step from n - 1 or a long step from n - 2. Which is the same as the fibonacci-sequence: fib(n + 2) = fib(n + 1) + fib(n).
The recursive formula is
f(n) = f(n-1) + f(n-2)
f(0) = f(1) = 1
Translated to a, b, c in the code
if (n<2) return 1;
int a = 1;
int b = 1;
int c = 1;
The above defines f(0) = f(1) = 1 and below calculates f(n) = f(n-1) + f(n-2) for n >= 2
for (int i=2; i<=n; i++){
c = b; // f(i-1) is temporary saved in c
b = a + b; // f(i-2) + f(i-1) is saved in b
a = c; // f(i-1) is saved in a for the next iteration
}
Related
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;
}
}
This question already has answers here:
Is there a difference between x++ and ++x in java?
(18 answers)
Closed 2 years ago.
I want to know the difference between post-increment and pre-increment, basically, I wrote two same codes together by just changing the variables to check whether there is any difference in post and pre-increment. but it's the same??
int tableOf = 4;
for (int i = 1; i <= 10; ++i) {
System.out.println(tableOf * i);
System.out.println(i);
}
int tableof = 4;
for (int y = 1; y <= 10; y++) {
System.out.println(tableof * y);
System.out.println(y);
}
post-increment uses the value and increments it after usage. pre-increment first increments and then uses the value.
in the example of a for-loop the procedure is
i = i++;
or
i = ++i;
in this case it doesnt matter if you are using pre or post in the end i is incremented by 1.
An relevant example would be an addition:
int a = 1;
int b = 2;
int c = a + b++; -> c = 3
or
int c = a + ++b; -> c = 4
I just finished Project Euler problem 9 (warning spoilers):
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a^2 + b^2 = c^2
For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
Here's my solution:
public static int specPyth(int num)
{
for (int a = 1; a < num; a++)
for (int b = 2; b < a; b++)
{
if (a*a +b*b == (num-a-b)*(num-a-b))
return a*b*(num-a-b); //ans = 31875000
}
return -1;
}
I can't help but think that there's a solution that involves only one loop. Anyone have ideas? I'd prefer answers using only one loop, but anything that's more efficient than what I currently have would be nice.
if a + b +c = 1000
then
a + b + sqroot(a² + b²) = 1000
-> (a² + b²) = (1000 - a - b)²
-> a² + b² = 1000000 - 2000*(a+b) + a² + 2*a*b + b²
-> 0 = 1000000 - 2000*(a+b) + 2*a*b
-> ... (easy basic maths)
-> a = (500000 - 1000*b) / (1000 - b)
Then you try every b until you find one that makes a natural number out of a.
public static int specPyth(int num)
{
double a;
for (int b = 1; b < num/2; b++)
{
a=(num*num/2 - num*b)/(num - b);
if (a%1 == 0)
return (int) (a*b*(num-a-b));
}
return -1;
}
EDIT: b can't be higher than 499, because c>b and (b+c) would then be higher than 1000.
I highly recommend reading http://en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple and writing a function that will generate the Pythagorean triples one by one.
Not to give too much of a spoiler, but there are a number of other PE problems that this function will come in handy for.
(I don't consider this giving away too much, because part of the purpose of PE is to encourage people to learn about things like this.)
First, since a is the smallest, you need not to count it up to num, num/3 is sufficient, and even num/(2+sqrt(2)).
Second, having a and constraints
a+b+c=num
a^2+b^2=c^2
we can solve this equations and find b and c for given a, which already satisfy this equations and there is no need to check if a^2+b^2=c^2 as you do now. All you need is to check if b and c are integer. And this is done in one loop
for (int a = 1; a < num/3; a++)
Runs in 62 milli seconds
import time
s = time.time()
tag,n=True,1000
for a in xrange (1,n/2):
if tag==False:
break
for b in xrange (1,n/2):
if a*a + b*b - (n-a-b)*(n-a-b) ==0:
print a,b,n-a-b
print a*b*(n-a-b)
tag=False
print time.time() - s
C solution
Warning : solution assumes that GCD(a, b) = 1. It works here but may not always work. I'll fix the solution in some time.
#include <stdio.h>
#include <math.h>
int main(void)
{
int n = 1000; // a + b + c = n
n /= 2;
for(int r = (int) sqrt(n / 2); r <= (int) sqrt(n); r++)
{
if(n % r == 0)
{
int s = (n / r) - r;
printf("%d %d %d\n", r*r - s*s, 2*r*s, r*r + s*s);
printf("Product is %d\n", (2*r*s) * (r*r - s*s) * (r*r + s*s));
}
}
return 0;
}
Solution uses Euclid's formula for triplets which states that any primitive triple is of form a = r^2 - s^2, b = 2rs, c = r^2 + s^2.
Certain restrictions like sqrt(n / 2) <= r <= sqrt(n) can be added based on the fact that s is positive and r > s.
Warning: you may need long long if the product is large
Definitely not the most optimal solution, but my first instinct was to use a modified 3SUM. In Python,
def problem_9(max_value = 1000):
i = 0
range_of_values = [n for n in range(1, max_value + 1)]
while i < max_value - 3:
j = i + 1
k = max_value - 1
while j < k:
a = range_of_values[i]
b = range_of_values[j]
c = range_of_values[k]
if ((a + b + c) == 1000) and (a*a + b*b == c*c):
return a*b*c
elif (a + b + c) < 1000:
j += 1
else:
k -= 1
i += 1
return -1
You say a < b < c, then b must always be bigger than a, so your starting point in the second loop could be b = a + 1; that would lead certainly to fewer iterations.
int specPyth(int num)
{
for (int a = 1; a < num/3; a++)
for (int b = a + 1; b < num/2; b++)
{
int c = num - a - b;
if (a * a + b * b == c * c)
return a * b * c; //ans = 31875000
}
return -1;
}
In the first given equation, you have three variables a, b, c. If you want to find-out matching values for this equation, you have to run 3 dimension loop. Fortunately there is another equation a+b+c=N where N is known number.
Using this, you can reduce down the dimension to two because if you know two among the three, you can calculate the rest. For instance, if you know a and b, c equals N - a - b.
What if you can reduce one more dimension of the loop? It is possible if you fiddle with the two given equations. Get a pen and paper. Once you get the additional equation with two variables and one constant (N), you will be able to acquire the result in O(n). Solve the two equations a+b+c=n; a^2+b^2=c^2 taking n and a to be constant and solve for b and c:
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
for(int a0 = 0; a0 < t; a0++){
int n = in.nextInt();
int max=-1;
int multi=0;
int b=1,c=1;
for(int i=1;i<=n;i++)
{
if(2*(i-n)!=0)
b=(2*i*n-(n*n))/(2*(i-n));
c=n-b-i;
if( (i*i+b*b==c*c)&& i+b+c==n && b>0 && c>=0 && i+b>c && c+i>b && b+c>i)
{
multi=i*b*c;
if(max<multi)
max=multi;
}
}
if(max==-1)
System.out.println(-1);
else
System.out.println(max);
}
}
Python:
Before jump into the code, do a small exercise in algebra.
a^2 + b^2 = c ^2
a + b + c = 1000 --> c = 1000 - (a+b)
a^2 + b^2 = (1000 - (a+b))^2
a = 1000*(500-b)/(1000-b)
Now the time to write the code.
for b in range(2, 500):
a = 1000*(500 - b)/(1000 - b)
if a.is_integer():
c = 1000 - (a+b)
print(a, b, c, a*b*c)
break
I have this Java problem, which I suspect it relates to a higher-level algorithm, but my searches haven't been able to come up with anything practical.
You construct an array as follows:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
Basically, Ai,j = Ai-1,j-1+Ai-1,j. It's supposed to return the element at index (l, c): for (4, 1) it should return 4, (5, 2) returns 10, etc. My solution is straightforward, but it is not enough:
static long get(int l, int c){
long[][] matrix = new long[l+1][l+1];
matrix[0][0]=1;
matrix[1][0]=1;
matrix[1][1]=1;
for(int i=2;i<=l;i++){
matrix[i][0]=1;
for(int j=1;j<=i;j++){
matrix[i][j] = matrix[i-1][j-1]+matrix[i-1][j];
}
}
return matrix[l][c];
}
It doesn't work for large values of l and c. Using BigInteger doesn't work. My searches have led me to loop skewing and scalarization, but I don't know where to start. Any steer in the right direction is really appreciated.
PS: Sorry for the newbie-vibe, this is my first question!
You are describing Pascal's triangle, for which a closed formula exists:
matrix[n][k] = n!/(k! * (n-k)!)
P.S. If these numbers seem familiar, it is because they are also from the binomial theorem, where the common examples are:
(x+y)^2 = 1* x^2 + 2xy + 1*y^2
(x+y)^3 = 1*x^3 + 3*xy^2 + 3yx^2 + 1*y^3
You don't need to use a loop, this is simply pascal's triangle, the formula:
(n, k) = n! / ( k! * (n-k)!)
Will generate your answer for the position (n, k).
Try this:
static long get(int l, int c) throws Exception {
if (l != c)
throw new Exception("l != c");
long[][] matrix = new long[l+1][l+1];
matrix[0][0]=1;
for (int i = 1; i <= l; ++i) {
for (int j = 0; j <= i; ++j) {
if (j - 1 >= 0) {
matrix[i][j] = matrix[i - 1][j] + matrix[i - 1][j - 1];
} else {
matrix[i][j] = matrix[i - 1][j];
}
}
}
return matrix[l][c];
}
I was just going through the iterative version of fibonacci series algorithm. I found this following code
int Fibonacci(int n)
{
int f1 = 0;
int f2 = 1;
int fn;
for ( int i = 2; i < n; i++ )
{
fn = f1 + f2;
f1 = f2;
f2 = fn;
}
}
A silly question just raised in my mind. The function above adds two previous numbers and returns the third one and then get variables ready for the next iteration. What if it would be something like this. "Return a number of series which is the sum of previous three numbers" how we can change the above code to find such a number.u
As a hint, notice that the above algorithm works by "cycling" the numbers through some variables. In the above code, at each point you are storing
F_0 F_1
a b
You then "shift" them over by one step in the loop:
F_1 F_2
a b
You then "shift" them again in the next loop iteration:
F_2 F_3
a b
If you want to update the algorithm sum the last three values, think about storing them like this:
T_0 T_1 T_2
a b c
Then shift them again:
T_1 T_2 T_3
a b c
Then shift them again:
T_2 T_3 T_4
a b c
Converting this intuition into code is a good exercise, so I'll leave those details to you.
That said - there is a much, much faster way to compute the nth term of the Fibonacci and "Tribonacci" sequences. This article describes a very clever trick using matrix multiplication to compute terms more quickly than the above loop, and there is code available here that implements this algorithm.
Hope this helps!
I like recursion. Call me a sadist.
static int rTribonacci (int n, int a, int b, int c) {
if (n == 0) return a;
return rTribonacci (n-1, b, c, a + b + c);
}
int Tribonacci (int n) { return rTribonacci(n, 0, 0, 1); }
I don't normally answer questions that "smell" like homework, but since someone else already replied this is what I would do:
int Tribonacci(int n)
{
int last[3] = { 0, 0, 1 }; // the start of our sequence
for(int i = 3; i <= n; i++)
last[i % 3] = last[i % 3] + last[(i + 1) % 3] + last[(i + 2) % 3];
return last[n % 3];
}
It can be improved a bit to avoid all the ugly modular arithmetic (which I left in to make the circular nature of the last[] array clear) by changing the loop to this:
for(int i = 3; i <= n; i++)
last[i % 3] = last[0] + last[1] + last[2];
It can be optimized a bit more and frankly, there are much better ways to calculate such sequences, as templatetypedef said.
If you want to use recursion, you don't need any other parameters:
int FibonacciN(int position)
{ if(position<0) throw new ArgumentException("invalid position");
if(position==0 || position ==1) return position;
return FibonacciN(position-1) + FibonacciN(position-2);
}