class Solution {
//given a location on the matrix, this function recursively find the deepest possible depth, which is the length of a side of a found square
private int isSquare(int row_index, int col_index, int depth, char[][] matrix) {
int last_row = row_index + depth;
int last_col = col_index + depth;
if (row_index >= matrix.length || col_index >= matrix[0].length) {
return 0;
}
if (last_row >= matrix.length || last_col >= matrix[0].length) {
return 0;
}
for (int i = col_index; i < last_col; i++) {
if (matrix[row_index][i] != '1') {
return 0;
}
}
for (int i = row_index; i < last_row; i++ ) {
if (matrix[i][col_index] != '1') {
return 0;
}
}
return Math.max(depth, isSquare(row_index, col_index, depth + 1, matrix));
}
public int maximalSquare(char[][] matrix) {
int max = 0;
for (int row = 0; row < matrix.length; row ++) {
for (int col = 0; col <matrix[0].length; col ++) {
int curr_depth = isSquare(row, col, 1, matrix);
if (curr_depth > max) {
max = curr_depth;
}
}
};
return max * max;
}
}
Hi, I was working on LeetCode 221, and it seems like that my solution is not passing test cases with output 1, where the biggest square on the given matrix is just 1 x 1. To me it looks like those depth 1 cases are not passing the two for loops in function isSquare, which is supposed to catch 0s in the square.
I tried LC debugging tool but it did not help much, and my base cases seem fine to me. Please let me know what is going on here. For the problem, https://leetcode.com/problems/maximal-square/
One of the depth 1 test cases that I am failing is below.
Input:
[["0","1"],["1","0"]]
Output:
0
Expected:
1
This is a recursive check if this is a prime number -- is it correct?
public static boolean isPrimeRecursive (int n,int i){//i eqoual to n
if (n <= 1) {
return false;
}if (i==1){
return false;
}if(n%i==0){
return false;
}
return isPrimeRecursive(n,i--);
}
I wouldn't burden your user with that mysterious second argument but rather present a different method of just one argument, that first deals with numbers less than 2 and even numbers, and then calls into your recursive method with the proper arguments:
private static boolean isPrimeRecursive(int n, int i) {
if (i * i > n) {
return true;
}
if (n % i == 0) {
return false;
}
return isPrimeRecursive(n, i + 2);
}
public static boolean isPrime(int n) {
if (n <= 2 || n % 2 == 0) {
return (n == 2);
}
return isPrimeRecursive(n, 3);
}
public static void main(String[] args) {
System.out.println(isPrime(Integer.parseInt(args[0])));
}
With your code, you should start of i with a value of n-1 since n % n is always true of prime numbers.
Then in your condition (if (i == 1) { ... }, should return true because if the method reaches to 1, then it fulfills all other conditions.
Finally in your return statement return isPrimeRecursive(n, i++);, it is better to use ++i since i++ will increment after the execution of the function with the value of i.
public static boolean isPrimeRecursive (int n,int i){
if (n <= 1) {
return false;
}
if (i == 1) {
return true;
}
if(n % i == 0){
return false;
}
return isPrimeRecursive(n, --i);
}
In your main function, you will then use:
int n = 17;
System.out.println(isPrimeRecursive(n, n-1);
Another way of doing it is to always start i with a value of 2 and increment it's value. From there.
public static boolean isPrimeRecursive (int n, int i) {
if (n <= 2) {
return (n == 2) ? true : false;
}
if (i >= n) {
return true;
}
if (n % i == 0) {
return false;
}
return isPrimeRecursive(n, ++i);
}
Then you simple do:
int n = 17;
System.out.println(isPrimeRecursive(n, 2);
I am having a trouble in completing a homework of mine in C.The task is given an integer n print all the binary numbers with the length n that do not have 2 consecutive zeros int them and that includes leading zeros, note that at least one of the functions has to be recursive. Here is an example if n is 4 then the binary number 10 is treated 0010 and therefore wont be printed because it has 2 leading zeros. My problem is that my code crashes if n = 16 and I do not know why even though I have done a lot of debugging. Here is my code, Thanks for any help here.
void binaries_n_digits_no_00(int n)
{
if(n < 0)
{
return;
}
print_binaries_n_digits_no_00(0,n);
}
void print_binaries_n_digits_no_00(long int current_binary_index,int n)
{
int num_of_leading_zeros;
if((current_binary_index > (power(2,n) - 1)) || n == 0)
{
return;
}
num_of_leading_zeros = n - binary_num_length(current_binary_index);
if((binary_not_contain_00(current_binary_index) == 1) &&
(num_of_leading_zeros == 1 || num_of_leading_zeros == 0)){
if(current_binary_index == 0 && n == 1)
{
printf("\n0");
}
else if(num_of_leading_zeros == 1)
{
if(current_binary_index != 0)
{
printf("\n0");
print_binary(current_binary_index);
}
}
else{
printf("\n");
print_binary(current_binary_index);
}
}
print_binaries_n_digits_no_00(current_binary_index+1,n);
}
int binary_not_contain_00(long int num)
{
if(num/2 == 0)
{
return 1;
}
if(((num%2) == 0) && (((num/2) % 2) == 0))
{
return 0;
}
return binary_not_contain_00(num/2);
}
void print_binary(long int num)
{
if(num > 1)
{
print_binary(num/2);
}
printf("%d",num%2);
}
int binary_num_length(long int num)
{
if(num <= 1)
{
return 1;
}
else{
return (1 + binary_num_length(num/2));
}
}
long int power(int m, int n)
{
if(n == 0)
{
return 1;
}
return m * power(m,n-1);
}
So I am creating a calculator that will compute numbers.I have this code(I will not include very basic codes)
long num1, num2, answer;
boolean mySwitch = false;
boolean do_subtraction_flag = false; // when true we will apply subtraction
boolean multiply = false;
boolean divide = false;
void loop()
{
char keypressed = myKeypad.getKey();
if(keypressed != NO_KEY)
{
Serial.print(keypressed);
if(keypressed > 47 && keypressed < 58) // is between '0' and '9'
{
if(!mySwitch)
{
num1 = (num1 * 10) + (keypressed - 48);
}
else
{
num2 = (num2 * 10) + (keypressed - 48);
}
}
if(keypressed == '=')
{
if(do_subtraction_flag) // we want to subtract the numbers
{
answer = num1 - num2;
}else if(multiply){
answer = num1 * num2;
}else if(divide){
answer = num1 / num2;
}
else // we want to add the numbers
{
answer = num1 + num2;
}
Serial.println(answer);
num1 = 0;
num2 = 0;
mySwitch = false;
do_subtraction_flag = false;
multiply = false;
divide = false;
}
else if(keypressed == '+')
{
mySwitch = true;
}
else if(keypressed == '*'){
mySwitch = true;
multiply = true;
}else if(keypressed == '/'){
mySwitch = true;
divide = true;
}
else if(keypressed == '-')
{
mySwitch = true;
do_subtraction_flag = true;
}else if (keypressed == 'C'){
for (int i=0; i < 80; i++)
{
Serial.write(8); // print 80 times backspace (BS)
}
}
}
}
Im confused here because I want to compute multiple number with multiple operands (eg., 2+1+3 or 2+1-2), but when i add another variable 'num3' what should I suppose to do with it? if i put it into the do_subtraction flag what if the user inputs 2-1+3?Is it possible to compute 3 numbers with this code?I'm getting confused here but let me know if you are also confused what I want to do
Make an array to hold the numbers and operators that you input, then loop through them when the input == '='. Have your num1 always hold the result and num2 hold the next array value.
I'm using this article to write a fluid simulation application.
I can't manage to implement the inside boundaries. As far as i know when I'm setting the boundaries (in the set_bnd function) for each cell that is inside the boundary I should calculate the average value from the adjacent non-boundary cells like this:
for (i = 0 ; i < n ; i++)
{
for (j = 0 ; j < n ; j++)
{
if (isBoundary(i,j)
{
sum = 0;
count = 0;
if (!isBoundary(i+1,j) {
sum += x[i+1][j];
}
if (!isBoundary(i-1,j) {
sum += x[i-1][j];
}
if (!isBoundary(i,j+1) {
sum += x[i][j+1];
}
if (!isBoundary(i,j-1) {
sum += x[i-1][j];
}
x[i][j] = sum / 4;
}
}
}
Unfortunately the smoke is absorbed and disappears in contact with boundary surface.
My math background is not sufficient to understand every part of the calculation, so I'll be very grateful if someone points me the right direction.
Here is some code to explain further.
insideBound is array (1 - boundary, 0 - empty, the fluid can pass trough)
#define FOR_EACH_CELL for ( i=1 ; i<=N ; i++ ) { for ( j=1 ; j<=N ; j++ ) {
void set_bnd ( int N, int b, float * x, int * insideBound )
{
int i, j;
float sum;
int count;
for ( i=1 ; i<=N ; i++ ) {
x[IX(0 ,i)] = b==1 ? -x[IX(1,i)] : x[IX(1,i)];
x[IX(N+1,i)] = b==1 ? -x[IX(N,i)] : x[IX(N,i)];
x[IX(i,0 )] = b==2 ? -x[IX(i,1)] : x[IX(i,1)];
x[IX(i,N+1)] = b==2 ? -x[IX(i,N)] : x[IX(i,N)];
}
x[IX(0 ,0 )] = 0.5f*(x[IX(1,0 )]+x[IX(0 ,1)]);
x[IX(0 ,N+1)] = 0.5f*(x[IX(1,N+1)]+x[IX(0 ,N)]);
x[IX(N+1,0 )] = 0.5f*(x[IX(N,0 )]+x[IX(N+1,1)]);
x[IX(N+1,N+1)] = 0.5f*(x[IX(N,N+1)]+x[IX(N+1,N)]);
if (!b) return;
FOR_EACH_CELL
sum = 0.0f;
count = 0;
if (insideBound[IX(i,j)] == 1)
{
if (insideBound[IX(i-1,j)] != 1)
{
count++;
sum = sum + x[IX(i-1,j)];
}
if (insideBound[IX(i+1,j)] != 1)
{
count++;
sum = sum + x[IX(i+1,j)];
}
if (insideBound[IX(i,j-1)] != 1)
{
count++;
sum = sum + x[IX(i, j-1)];
}
if (insideBound[IX(i,j+1)] != 1)
{
count++;
sum = sum + x[IX(i, j+1)];
}
if (count > 0)
{
x[IX(i,j)] = -sum / count;
} else {
x[IX(i,j)] = 0;
}
}
END_FOR
}
Per book (working):
In the first loop are set top, right, bottom and left boundary cells.
Since for them there is only one adjacent cell that is not bound the cell get its value. (I don't know why its opposite for U and the same value for V)
After the first loop, the corner boundary values are set. Here they get average values from their adjacent cells (i guess since there is no adjacent cell that is not boundary they use boundary cells).
Mine, not working properly:
if (!b) return - ignores calculations for density and updates only velocity.
The loop calculates the values for all boundary cells (again, average values from the adjacent cells that are not boundaries themselves).
I get almost realistic result from this methods, but there is big loses in the density and some bugs with too big boundaries where the fluid disappears completely.
I've managed to find a solution, here it is for potential people with the same problem
void set_bnd ( int N, int b, float * x, int * insideBound )
{
int i, j;
float sum, tmp;
int count;
for ( i=1 ; i<=N ; i++ ) {
x[IX(0 ,i)] = b==1 ? -x[IX(1,i)] : x[IX(1,i)];
x[IX(N+1,i)] = b==1 ? -x[IX(N,i)] : x[IX(N,i)];
x[IX(i,0 )] = b==2 ? -x[IX(i,1)] : x[IX(i,1)];
x[IX(i,N+1)] = b==2 ? -x[IX(i,N)] : x[IX(i,N)];
}
x[IX(0 ,0 )] = 0.5f*(x[IX(1,0 )]+x[IX(0 ,1)]);
x[IX(0 ,N+1)] = 0.5f*(x[IX(1,N+1)]+x[IX(0 ,N)]);
x[IX(N+1,0 )] = 0.5f*(x[IX(N,0 )]+x[IX(N+1,1)]);
x[IX(N+1,N+1)] = 0.5f*(x[IX(N,N+1)]+x[IX(N+1,N)]);
if (!b) return;
for ( i=1 ; i<=N ; i++ ) {
for ( j=1 ; j<=N ; j++ ) {
sum = 0.0f;
count = 0;
if (insideBound[IX(i,j)] == 1)
{
if (insideBound[IX(i-1,j)] != 1)
{
count++;
if (b == 2)
tmp = -x[IX(i-1,j)];
else
tmp = x[IX(i-1,j)];
sum = sum + tmp;
}
if (insideBound[IX(i+1,j)] != 1)
{
count++;
if (b == 2)
tmp = -x[IX(i+1,j)];
else
tmp = x[IX(i+1,j)];
sum = sum + tmp;
}
if (insideBound[IX(i,j-1)] != 1)
{
count++;
if (b == 1)
tmp = - x[IX(i, j-1)];
else
tmp = x[IX(i, j-1)];
sum = sum + tmp;
}
if (insideBound[IX(i,j+1)] != 1)
{
count++;
if (b == 1)
tmp = -x[IX(i, j+1)];
else
tmp = x[IX(i, j+1)];
sum = sum + tmp;
}
if (count > 0)
{
x[IX(i,j)] = -sum / count;
} else {
x[IX(i,j)] = 0;
}
}
}
}
}
insideBound is boolean array (0,1) which indicates the cells that are boundaries. Works with one or more boundary areas, but they should be at least 3 cells wide and high.