I tried using 6%2, but its always giving the value as 2 and not 0. Why and how can I get a solution to this?
if(!(y%x))
{
...
}
In your case !(6%2) would return true.
(Answer very similar to the original in the question)
I'm asuming that you want to find out if Y=kX has integer values of k for a given X so that Y=5, X=3 fails (k is 5/3), but Y=6, X=2 passes (k is exactly 3). You are happy that k is either positive or negative.
That way, using Y remainder X == 0 is a good test. As an aside, be careful of negative remainders (e.g. Y % 2 == 1 as a test for oddness fails for negative numbers, use Y % 2 != 0 to be sure)
Code example in Java
public class Example {
public static void main(String[] args) {
System.out.println(isIntegerFactor(5,3)); // k is not an integer
System.out.println(isIntegerFactor(6,3)); // k is 2
System.out.println(isIntegerFactor(-6,-3)); // k is 2
System.out.println(isIntegerFactor(-6,3)); // k is -2
System.out.println(isIntegerFactor(6,-3)); // k is -2
}
public static boolean isIntegerFactor(int y, int x) {
return (y % x) == 0;
}
}
bool prime = PrimeTool.IsPrime(input_Number);
if (!prime)
{
Console.Write("multiple of other number");
Console.WriteLine(i);
}
Related
Problem Statement
Given an integer array A of size N. Find the sum of GCD (Greatest Common Divisor) of all elements with their frequency.
Input
First line contains an integers N.
Next line contains N space separated integers denoting elements of array.
Constraints
1 <= N <= 1000
0 <= Ai<= 10^5
Output
Print the sum of GCD of all elements with their frequency.
Example
Sample Input 1:
3
1 2 3
Output
3
Explanation:
gcd(1, 1) + gcd(2, 1) + gcd(3, 1) = 1+1+1 = 3
Sample Input 2:
6
3 6 6 9 3 3
Output
14
Explanation
gcd(3, 3) + gcd(6, 2) + gcd(6, 2) + gcd(9, 1) + gcd(3, 3) + gcd(3, 3)= 3+2+2+1+3+3= 14
import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
class Main {
public static void main (String[] args) {
Scanner s = new Scanner(System.in);
int N = s.nextInt();
int [] a= new int[N];
HashMap<Integer,Integer> map = new HashMap<>();
for (int i=0;i<N;i++) {
a[i]=s.nextInt();
map.put(a[i],map.getOrDefault(a[i],0)+1);
}
long sum=0;
for (int i=0;i<N;i++) {
sum += gcd(a[i],map.get(a[i]));
}
System.out.println(sum);
}
public static int gcd(int a, int b) {
if (b == 0)
return a;
return gcd(b,a%b);
}
}
the origin code is :
public static int numberOfLeadingZeros(int i) {
// HD, Figure 5-6
if (i == 0)
return 32;
int n = 1;
if (i >>> 16 == 0) { n += 16; i <<= 16; }
if (i >>> 24 == 0) { n += 8; i <<= 8; }
if (i >>> 28 == 0) { n += 4; i <<= 4; }
if (i >>> 30 == 0) { n += 2; i <<= 2; }
n -= i >>> 31;
return n;
}
I think it can be optimized ,should add following condition:
if (i < 0)
return 0;
the fully optimized code is :
public static int numberOfLeadingZeros(int i) {
if(i<=0) {
return i < 0 ? 0 : 32;
}
int n = 1;
if (i >>> 16 == 0) { n += 16; i <<= 16; }
if (i >>> 24 == 0) { n += 8; i <<= 8; }
if (i >>> 28 == 0) { n += 4; i <<= 4; }
if (i >>> 30 == 0) { n += 2; i <<= 2; }
n -= i >>> 31;
return n;
}
In theory yes, your suggestion makes sense.
In practice, unless you use an exotic JVM, it will not make any difference because the method is intrinsic, so the code that is executed is not the code you can find in the Java class.
For example on x86/64 cpus, the code is here and uses the bsrl CPU instruction, which is as fast as you can hope for.
Besides the fact that this method will likely get replaced by an intrinsic operation for hot spots, this check for negative numbers is only an improvement, if the number is negative. For positive numbers, it is just an additional condition to be evaluated.
So the worth of this optimization depends on the likelihood of negative arguments at this function. When I consider typical use cases of this function, I’d consider negative values a corner case rather than typical argument.
Note that the special handling of zero at the beginning is not an optimization, but a requirement as the algorithm wouldn’t return the correct result for zero without that special handling.
Since your bug report yield to finding an alternative (also shown in your updated question) which improves the negative number case without affecting the performance of the positive number case, as it fuses the required zero test and the test for negative numbers into a single pre-test, there is nothing preventing the suggested optimization.
Bug has been created on oracle bug database: http://bugs.java.com/bugdatabase/view_bug.do?bug_id=JDK-8189230
So i've tried interpreting this pseudocode a friend made and i wasn't exactly sure that my method returns the right result. Anyone who's able to help me out?
I've done some test cases where e.g. an array of [2,0,7] or [0,1,4] or [0, 8, 0] would return true, but not cases like: [1,7,7] or [2,6,0].
Array(list, d)
for j = 0 to d−1 do
for i = 0 to d−1 do
for k = 0 to d−1 do
if list[j] + list[ i] + list[k] = 0 then
return true
end if
end for
end for
end for
return false
And i've made this in java:
public class One{
public static boolean method1(ArrayList<String> A, int a){
for(int i = 0; i < a-1; i++){
for(int j = 0; j < a-1; j++){
for(int k = 0; k < a-1; k++){
if(Integer.parseInt(A.get(i)+A.get(j)+A.get(k)) == 0){
return true;
}
}
}
}
return false;
}
}
Thanks in advance
For a fix to your concrete problem, see my comment. A nicer way to write that code would be to actually use a list of Integer instead of String, because you will then want to convert the strings back to integers. So, your method looks better like this:
public static boolean method(List<Integer> A) {
for (Integer i : A)
for (Integer j : A)
for (Integer k : A)
if (i + j + k == 0)
return true;
return false;
}
See that you don't even need the size as parameter, since any List in Java embeds its own size.
Somehow offtopic
You're probably trying to solve the following problem: "Find if a list of integers contains 3 different ones that sum up to 0". The solution to this problem doesn't have to be O(n^3), like yours, it can be solved in O(n^2). See this post.
Ok, so here is what I believe the pseudo code is trying to do. It returns true if there is a zero in your list or if there are three numbers that add up to zero in your list. So it should return true for following test cases. (0,1,2,3,4,5), (1,2,3,4,-3). It will return false for (1,2,3,4,5). I just used d=5 as a random example. Your code is good for the most part - you just need to add the ith, jth and kth elements in the list to check if their sum equals zero for the true condition.
Can someone give me example of an algorithm that has square root(n) time complexity. What does square root time complexity even mean?
Square root time complexity means that the algorithm requires O(N^(1/2)) evaluations where the size of input is N.
As an example for an algorithm which takes O(sqrt(n)) time, Grover's algorithm is one which takes that much time. Grover's algorithm is a quantum algorithm for searching an unsorted database of n entries in O(sqrt(n)) time.
Let us take an example to understand how can we arrive at O(sqrt(N)) runtime complexity, given a problem. This is going to be elaborate, but is interesting to understand. (The following example, in the context for answering this question, is taken from Coding Contest Byte: The Square Root Trick , very interesting problem and interesting trick to arrive at O(sqrt(n)) complexity)
Given A, containing an n elements array, implement a data structure for point updates and range sum queries.
update(i, x)-> A[i] := x (Point Updates Query)
query(lo, hi)-> returns A[lo] + A[lo+1] + .. + A[hi]. (Range Sum Query)
The naive solution uses an array. It takes O(1) time for an update (array-index access) and O(hi - lo) = O(n) for the range sum (iterating from start index to end index and adding up).
A more efficient solution splits the array into length k slices and stores the slice sums in an array S.
The update takes constant time, because we have to update the value for A and the value for the corresponding S. In update(6, 5) we have to change A[6] to 5 which results in changing the value of S1 to keep S up to date.
The range-sum query is interesting. The elements of the first and last slice (partially contained in the queried range) have to be traversed one by one, but for slices completely contained in our range we can use the values in S directly and get a performance boost.
In query(2, 14) we get,
query(2, 14) = A[2] + A[3]+ (A[4] + A[5] + A[6] + A[7]) + (A[8] + A[9] + A[10] + A[11]) + A[12] + A[13] + A[14] ;
query(2, 14) = A[2] + A[3] + S[1] + S[2] + A[12] + A[13] + A[14] ;
query(2, 14) = 0 + 7 + 11 + 9 + 5 + 2 + 0;
query(2, 14) = 34;
The code for update and query is:
def update(S, A, i, k, x):
S[i/k] = S[i/k] - A[i] + x
A[i] = x
def query(S, A, lo, hi, k):
s = 0
i = lo
//Section 1 (Getting sum from Array A itself, starting part)
while (i + 1) % k != 0 and i <= hi:
s += A[i]
i += 1
//Section 2 (Getting sum from Slices directly, intermediary part)
while i + k <= hi:
s += S[i/k]
i += k
//Section 3 (Getting sum from Array A itself, ending part)
while i <= hi:
s += A[i]
i += 1
return s
Let us now determine the complexity.
Each query takes on average
Section 1 takes k/2 time on average. (you might iterate atmost k/2)
Section 2 takes n/k time on average, basically number of slices
Section 3 takes k/2 time on average. (you might iterate atmost k/2)
So, totally, we get k/2 + n/k + k/2 = k + n/k time.
And, this is minimized for k = sqrt(n). sqrt(n) + n/sqrt(n) = 2*sqrt(n)
So we get a O(sqrt(n)) time complexity query.
Prime numbers
As mentioned in some other answers, some basic things related to prime numbers take O(sqrt(n)) time:
Find number of divisors
Find sum of divisors
Find Euler's totient
Below I mention two advanced algorithms which also bear sqrt(n) term in their complexity.
MO's Algorithm
try this problem: Powerful array
My solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1E6 + 10, k = 500;
struct node {
int l, r, id;
bool operator<(const node &a) {
if(l / k == a.l / k) return r < a.r;
else return l < a.l;
}
} q[N];
long long a[N], cnt[N], ans[N], cur_count;
void add(int pos) {
cur_count += a[pos] * cnt[a[pos]];
++cnt[a[pos]];
cur_count += a[pos] * cnt[a[pos]];
}
void rm(int pos) {
cur_count -= a[pos] * cnt[a[pos]];
--cnt[a[pos]];
cur_count -= a[pos] * cnt[a[pos]];
}
int main() {
int n, t;
cin >> n >> t;
for(int i = 1; i <= n; i++) {
cin >> a[i];
}
for(int i = 0; i < t; i++) {
cin >> q[i].l >> q[i].r;
q[i].id = i;
}
sort(q, q + t);
memset(cnt, 0, sizeof(cnt));
memset(ans, 0, sizeof(ans));
int curl(0), curr(0), l, r;
for(int i = 0; i < t; i++) {
l = q[i].l;
r = q[i].r;
/* This part takes O(n * sqrt(n)) time */
while(curl < l)
rm(curl++);
while(curl > l)
add(--curl);
while(curr > r)
rm(curr--);
while(curr < r)
add(++curr);
ans[q[i].id] = cur_count;
}
for(int i = 0; i < t; i++) {
cout << ans[i] << '\n';
}
return 0;
}
Query Buffering
try this problem: Queries on a Tree
My solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10, k = 333;
vector<int> t[N], ht;
int tm_, h[N], st[N], nd[N];
inline int hei(int v, int p) {
for(int ch: t[v]) {
if(ch != p) {
h[ch] = h[v] + 1;
hei(ch, v);
}
}
}
inline void tour(int v, int p) {
st[v] = tm_++;
ht.push_back(h[v]);
for(int ch: t[v]) {
if(ch != p) {
tour(ch, v);
}
}
ht.push_back(h[v]);
nd[v] = tm_++;
}
int n, tc[N];
vector<int> loc[N];
long long balance[N];
vector<pair<long long,long long>> buf;
inline long long cbal(int v, int p) {
long long ans = balance[h[v]];
for(int ch: t[v]) {
if(ch != p) {
ans += cbal(ch, v);
}
}
tc[v] += ans;
return ans;
}
inline void bal() {
memset(balance, 0, sizeof(balance));
for(auto arg: buf) {
balance[arg.first] += arg.second;
}
buf.clear();
cbal(1,1);
}
int main() {
int q;
cin >> n >> q;
for(int i = 1; i < n; i++) {
int x, y; cin >> x >> y;
t[x].push_back(y); t[y].push_back(x);
}
hei(1,1);
tour(1,1);
for(int i = 0; i < ht.size(); i++) {
loc[ht[i]].push_back(i);
}
vector<int>::iterator lo, hi;
int x, y, type;
for(int i = 0; i < q; i++) {
cin >> type;
if(type == 1) {
cin >> x >> y;
buf.push_back(make_pair(x,y));
}
else if(type == 2) {
cin >> x;
long long ans(0);
for(auto arg: buf) {
hi = upper_bound(loc[arg.first].begin(), loc[arg.first].end(), nd[x]);
lo = lower_bound(loc[arg.first].begin(), loc[arg.first].end(), st[x]);
ans += arg.second * (hi - lo);
}
cout << tc[x] + ans/2 << '\n';
}
else assert(0);
if(i % k == 0) bal();
}
}
There are many cases.
These are the few problems which can be solved in root(n) complexity [better may be possible also].
Find if a number is prime or not.
Grover's Algorithm: allows search (in quantum context) on unsorted input in time proportional to the square root of the size of the input.link
Factorization of the number.
There are many problems that you will face which will demand use of sqrt(n) complexity algorithm.
As an answer to second part:
sqrt(n) complexity means if the input size to your algorithm is n then there approximately sqrt(n) basic operations ( like **comparison** in case of sorting). Then we can say that the algorithm has sqrt(n) time complexity.
Let's analyze the 3rd problem and it will be clear.
let's n= positive integer. Now there exists 2 positive integer x and y such that
x*y=n;
Now we know that whatever be the value of x and y one of them will be less than sqrt(n). As if both are greater than sqrt(n)
x>sqrt(n) y>sqrt(n) then x*y>sqrt(n)*sqrt(n) => n>n--->contradiction.
So if we check 2 to sqrt(n) then we will have all the factors considered ( 1 and n are trivial factors).
Code snippet:
int n;
cin>>n;
print 1,n;
for(int i=2;i<=sqrt(n);i++) // or for(int i=2;i*i<=n;i++)
if((n%i)==0)
cout<<i<<" ";
Note: You might think that not considering the duplicate we can also achieve the above behaviour by looping from 1 to n. Yes that's possible but who wants to run a program which can run in O(sqrt(n)) in O(n).. We always look for the best one.
Go through the book of Cormen Introduction to Algorithms.
I will also request you to read following stackoverflow question and answers they will clear all the doubts for sure :)
Are there any O(1/n) algorithms?
Plain english explanation Big-O
Which one is better?
How do you calculte big-O complexity?
This link provides a very basic beginner understanding of O() i.e., O(sqrt n) time complexity. It is the last example in the video, but I would suggest that you watch the whole video.
https://www.youtube.com/watch?v=9TlHvipP5yA&list=PLDN4rrl48XKpZkf03iYFl-O29szjTrs_O&index=6
The simplest example of an O() i.e., O(sqrt n) time complexity algorithm in the video is:
p = 0;
for(i = 1; p <= n; i++) {
p = p + i;
}
Mr. Abdul Bari is reknowned for his simple explanations of data structures and algorithms.
Primality test
Solution in JavaScript
const isPrime = n => {
for(let i = 2; i <= Math.sqrt(n); i++) {
if(n % i === 0) return false;
}
return true;
};
Complexity
O(N^1/2) Because, for a given value of n, you only need to find if its divisible by numbers from 2 to its root.
JS Primality Test
O(sqrt(n))
A slightly more performant version, thanks to Samme Bae, for enlightening me with this. 😉
function isPrime(n) {
if (n <= 1)
return false;
if (n <= 3)
return true;
// Skip 4, 6, 8, 9, and 10
if (n % 2 === 0 || n % 3 === 0)
return false;
for (let i = 5; i * i <= n; i += 6) {
if (n % i === 0 || n % (i + 2) === 0)
return false;
}
return true;
}
isPrime(677);
So I tried to write a code that finds the largest palindromic number from two (3 spaces long) multiplied numbers. Does my code work fine or are there no palindromes for this?
function checkPalindrom(str) {
return str === str.split('').reverse().join('');
}; //Declares the funciton to check if a string is a palindrome
var x = 999;
var y = 999;
var z = 0;
var n = z.toString(); //Declares that n is the string of z
for (i=0; i<899; i++) { //For loop: counts from 0 to 899
x*y===z; //Is this correct? z is set equal to x*y
if(checkPalindrom(n) === true) { //If n is a palindrome,
console.log(n); //Write out the palindrome
} else {
x-=1; //subtract 1 from x and run again
}
};
Also, what is the best way to check for all combinations of 3 digit numbers? Because right now I am just checking for any number from 100 to 999, but I actually need to check for all combinations...
Your post has a few problems, as well as multiple questions in it. I'll try to hone in on the major stuff but, as this is a fairly standard type of Programming 101 homework question, I'm not going to give you an exact answer right out.
First off, there are three different 'equals' in javascript, =, ==, and ===. A single = is an assignment operator and it always works from right to left. Thus,
var x = 2;
assigns the value of 2 to the variable x. In your code,
x*y === z;
has a couple of problems. First off, it is backwards. Secondly, it uses === but should be using =.
z = x*y;
That is what you were trying to put here.
In javascript, == and === are both comparitives. The triple === adds type comparison and is stronger but generally unnecessary. In almost all cases, == is sufficient. But, what it does is compare the values like inside an if statement:
if(x == 2)
This just checks if the value of x is equal to the value of 2, but the values themselves do not change.
Ok, for your other question: "number from 100 to 999, but I actually need to check for all combinations..."
The best way to handle this is a double loop:
var z;
for(var x = 100; x < 1000; x++)
for(var y = x; y < 1000; y++)
z = x*y;
This will first let x = 100, then check 100 * every number from 100 to 999. Then you let x = 101 and check 101* every number from 101 to 999.
function checkPalindrom(str) {
return str === str.split('').reverse().join('');
}; //Declares the funciton to check if a string is a palindrome
var x;
var y;
var z;
var n;
var max = 0;
for (x=999; x >= 100; x--) {
for (y=999; y >= 100; y--) {
z = x*y;
n = z.toString();
if(checkPalindrom(n) === true && max < z) {
console.log(n);
max = z;
}
}
}