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Search algorithm optimization

The code "works", but I'm having problems with optimization.
Algorithms idea is count the number of ways you can navigate n × n grid starting from the top left corner, always moving one step to the left, right, up or down, visiting each square. For example, when n = 3, there are 8 possible paths.
For example the algorithm gets unbearably slow with p.count(7), any hints/tips?
public class Paths {
static int[][] grid;
static int n,counter;
public int count(int n) {
grid = new int[n][n];
counter = 0;
search(0,0,1,n);
return counter;
}
void search(int y, int x, int k, int n) {
if (y < 0 || x < 0 || y >= n || x >= n) return;
if (grid[y][x] != 0) return;
if (x-1 > 0 && x+1 < n && y == n && grid[y][x-1] == 0 && grid[y][x+1] == 0) return;
else if (x-1 > 0 && x+1 < n && y == 0 && grid[y][x-1] == 0 && grid[y][x+1] == 0) return;
else if (y-1 > 0 && y+1 < n && x == n && grid[y-1][x] == 0 && grid[y+1][x] == 0) return;
else if (y-1 > 0 && y+1 < n && x == 0 && grid[y-1][x] == 0 && grid[y+1][x] == 0) return;
else if (y==n && x==n && grid[y][x-1] == 0 && grid[y-1][x] == 0) return;
else if (y==n && x==0 && grid[y-1][x] == 0 && grid[y][x+1] == 0) return;
else if (y==0 && x==n && grid[y+1][x] == 0 && grid[y][x-1] == 0) return;
if (k == n*n) {
counter++;
return;
}
grid[y][x] = k;
search(y,x-1,k+1,n);
search(y,x+1,k+1,n);
search(y+1,x,k+1,n);
search(y-1,x,k+1,n);
grid[y][x] = 0;
}
}
Output:
public static void main(String[] args) {
Paths p = new Paths();
System.out.println(p.count(1)); // 1
System.out.println(p.count(2)); // 2
System.out.println(p.count(3)); // 8
System.out.println(p.count(4)); // 52
}
}

I'm facing an issue with control structures

My problem is with output of my code. When I enter 20, the output must be weird, but I am getting not weird. Same with the value 18.
import java.util.Scanner;
public class conditional {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int n = sc.nextInt();
String ans = "";
if(n%2 == 1){
ans = "Weird";
} else {
if(n <= 2 && n >= 5){
ans="Not weird";
} else if(n <= 6 && n >= 20){
ans = "Weird";
} else{
ans = "Not Weird";
}
}
System.out.println(ans);
}
}

the output must be weird,but i am getting not weird
Because, if(n%2 == 1) return false and fall to else block where
if(n <= 2 && n >= 5) is `false`
and
else if(n <= 6 && n >= 20) is also `false`
So, again falls to else block. You probably change
if(n <= 2 && n >= 5)
to
if(n >= 2 && n <= 5)
and
else if(n <= 6 && n >= 20)
to
else if(n >= 6 && n <= 20)
Otherwise, they will never be true and always falls to else.

In your program last else is being executed. Change && (logical AND) to || (logical OR) which will check if number is less than something OR higher than something, instead of checking if something is less or equal 5 AND higher or equal to 20 in the same time as it doesn't have a possibility to evaluate in any case.

I have come up with two solutions and also i see a flaw:
1. if(n%2 == 1) this code can be altered to if(n%2 == 0)
2. The flaw is **(n <= 2 && n >= 5)** . No number can be <2 and >5 at the same time. Try changing that to (n <= 2 || n >= 5) and same goes for (n <= 6 && n >= 20)
import java.util.Scanner;
public class conditional {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int n = sc.nextInt();
String ans = "";
if(n%2 == 1){
ans = "Weird";
} else {
if(n <= 6 || n >= 20){
ans="Not weird";
} else if(n <= 2 || n >= 5){
ans = "Weird";
} else{
ans = "Not Weird";
}
}
System.out.println(ans);
}
}

check if array contains certain int BEFORE another int

I need to check if an array contains a 1 and then later in the array contains a 2.
What I have coded only checks if both are in there, not if one is before the other. How could I do this?
if(array[i] == 1)
count++;
else if(array[i] == 2)
count++;
}
if(count > 1)
System.out.print("true");
else
System.out.print("false");
Comparing the index of the values works!
if (nums[i] == 1)
value1 = i;
else if(nums[i] == 2)
value2 = i;
}
if (value2 > value1)
System.out.print("true");
else
System.out.print("false");

This oughta do it!
public void hasOneThenTwo(int[] a) {
bool hasOne = false;
for (int i = 0; i < a.length; ++i) {
if (!hasOne && a[i] == 1) {
hasOne = true;
} else if (hasOne && a[i] == 2) {
return true;
}
}
return false;
}

How can I get this program to print out ten items per line seperated by 1 space?

Here's my code so far
public class DivisibleBy5and6
{
public static void main (String []args)
{
for (int i = 100; i <= 200; i++)
{
boolean num = (i % 5 == 0 || i % 6 == 0) && !(i % 5 == 0 && i % 6 == 0);
if (num == true)
System.out.println(i + " is divisible");
}
}
}
Like stated previously how can I get the output to print out 10 items per line separated by a space?

How about:
int count = 0;
for (int i = 100; i <= 200; i++) {
boolean num = (i % 5 == 0 || i % 6 == 0) && !(i % 5 == 0 && i % 6 == 0);
if (num == true) {
count++;
System.out.print(i + " is divisible ");
if(count >= 10) {
System.out.println();
count -= 10;
}
}
}

Why isn't my FizzBuzz code processing both if statements when they both match? [duplicate]

This question already has answers here:
Conditional statement true in both parts of if-else-if ladder
(4 answers)
Closed 2 years ago.
For those who don't know, FizzBuzz is the following problem:
Write a program that prints the numbers from 1 to 100. But for
multiples of three print "Fizz" instead of the number and for the
multiples of five print "Buzz". For numbers which are multiples of
both three and five print "FizzBuzz".
Every FizzBuzz solution I find is either some crazy esoteric solution made for the sake of being original, or your basic if-else chain:
for(int i = 1; i <= 100; i++) {
if(i % 3 == 0 && i % 5 == 0) {
System.out.println("FizzBuzz");
} else if (i % 3 == 0) {
System.out.println("Fizz");
} else if (i % 5 == 0) {
System.out.println("Buzz");
} else {
System.out.println(i);
}
}
I am looking for a simple solution that aims to take out the "FizzBuzz" if statement. I have this in mind:
for(int i = 1; i <= 100; i++) {
if (i % 3 == 0)
System.out.print("Fizz");
if (i % 5 == 0)
System.out.println("Buzz")
else
System.out.println(i);
}
But this doesn't work. I assume it would be able to print FizzBuzz by entering both ifs, for Fizz and for Buzz, but if the number is, for example, 3, it would print Fizz3. How do I avoid this?

What you're trying to do is
if (a)
...
if (b)
...
else // if neigther a nor b
...
This is simply not possible. An else can only belong to a single if. You have to go with the slightly longer variant.
To avoid doing redundant evaluations of the modulo operator, you could formulate the loop body as
boolean fizz = i % 3 == 0;
boolean buzz = i % 5 == 0;
if (fizz)
System.out.print("Fizz");
if (buzz)
System.out.print("Buzz");
if (!(fizz || buzz))
System.out.print(i);
System.out.println();
Another one would be
String result = "";
if (i % 3 == 0) result = "Fizz";
if (i % 5 == 0) result += "Buzz";
if (result == "") result += i;
System.out.println(result);

Your first if statement is all alone.
So, your code hits the first statement, which is ONLY an if statement, and then goes on to the next, which is an if/else statement.
RosettaCode has a good example without using AND operators.
int i;
for (i = 0; i <= 100; i++) {
if ((i % 15) == 0)
cout << "FizzBuzz" << endl;
else if ((i % 3) == 0)
cout << "Fizz" << endl;
else if ((i % 5) == 0)
cout << "Buzz" << endl;
else
cout << i << endl;
}

If your only goal is to avoid using &&, you could use a double negation and DeMorgan's laws:
for(int i = 1; i <= 100; i++) {
if(!(i % 3 != 0 || i % 5 != 0)) {
System.out.println("FizzBuzz");
} else if (i % 3 == 0) {
System.out.println("Fizz");
} else if (i % 5 == 0) {
System.out.println("Buzz");
} else {
System.out.println(i);
}
}
You can avoid && using the fact that i % 3 == 0 and i % 5 == 0 implies i % 15 == 0, as per RFC1337's answer.
Another solution is to use a switch on the remainder (mod 15, which is 5 times 3):
for(int i = 1; i <= 100; i++) {
final int mod = i % 15;
switch (mod) {
case 0:
case 3:
case 6:
case 9:
case 12:
System.out.print("Fizz");
if (mod != 0) break;
case 5:
case 10:
System.out.print("Buzz");
break;
default:
System.out.print(i);
}
System.out.println();
}

This is my solution. Granted, it's a bit convoluted (as in roundabout), but I believe it suits your requirement.
int main()
{
char fizzpass=0;
unsigned short index=0;
for(index=1;index<=100;index++)
{
if(0 == (index%3))
{
printf("Fizz");
fizzpass = 1;
}
if(0 == (index%5))
{
if(1 == fizzpass)
{
fizzpass = 0;
}
printf("Buzz\n");
continue;
}
if(1 == fizzpass)
{
fizzpass = 0;
printf("\n");
continue;
}
printf("%d\n",index);
}
return 0;
}
Regards.

Just add a flag variable and use System.out.print:
package com.stackoverflow;
public class FizzBuzz {
public static void main(String[] args) {
for (int i = 1; i <= 100; i++) {
boolean printed = false;
if (i % 3 == 0) {
printed = true;
System.out.print("Fizz");
}
if (i % 5 == 0) {
printed = true;
System.out.print("Buzz");
}
if (printed) {
System.out.println();
} else {
System.out.println(i);
}
}
}
}

This doesn't take out the if statements but does not use the && (and) operator, you could flip the binary operators.
//FizzBuzz Case
if(!(a % 3 != 0 || a % 5 != 0)){ //flips
result[index] = "FizzBuzz";
index++;
}

Don't use an if statement at all.
import java.util.*;
import java.lang.*;
import java.io.*;
class FizzBuzz
{
public static void main (String[] args) throws java.lang.Exception
{
String[] words = {"", "Fizz", "Buzz"};
String[] nwords = {"", ""};
for(int i = 1; i < 101; ++i)
{
int fp = (i % 3 == 0) ? 1 : 0;
int bp = ((i % 5 == 0) ? 1 : 0) * 2;
int np = ((fp > 0 || bp > 0) ? 1: 0);
nwords[0] = Integer.toString(i);
System.out.print(words[fp]);
System.out.print(words[bp]);
System.out.println(nwords[np]);
}
}
}
See it on ideone.

public class fizzbuzz
{
public static void main(String[] args)
{
String result;
for(int i=1; i<=100;i++)
{
result=" ";
if(i%3==0)
{
result=result+"Fizz";
}
if(i%5==0)
{
result=result+"Buzz";
}
if (result==" ")
{
result=result+i;
}
System.out.println(result);
}
}
}
This is the most efficient way I could come up with. Hope it helps! :)

Crazy albeit unrelated solution done in Python3
#!/usr/bin/python3
for i in range(1,100):
msg = "Fizz" * bool(i%3==0)
msg += "Buzz" * bool(i%5==0)
if not msg:
msg = i
print(msg)