I have a problem with my Objective-C code. I would like to have an if statement in a formula saying that the condition is checked just as if a float is equal to one of the other 5 I have defined. But shorter and simpler than:
if (float1 == float5 || float1 == float2 || float1 == float3 || float1 == float11)
{
//something to do
}
Thank you very much for your answers.
Assuming that you can guarantee that the float is an integer, either by definition or applying roundf(), floorf(), ceilf(), this is a possible solution:
const NSUInteger LENGTH = 7;
float myNumbersToCheck[LENGTH] = {1,13,6,8,99,126,459674598};
for (NSUInteger index = 0; index < LENGTH; index++) {
if (float1 == myNumbersToCheck[index]) {
// do something
break;
}
}
Related
I got a question to answer with the best complexity we can think about.
We got one sorted array (int) and X value. All we need to do is to find how many places in the array equals the X value.
This is my solution to this situation, as i don't know much about complexity.
All i know is that better methods are without for loops :X
class Question
{
public static int mount (int [] a, int x)
{
int first=0, last=a.length-1, count=0, pointer=0;
boolean found=false, finish=false;
if (x < a[0] || x > a[a.length-1])
return 0;
while (! found) **//Searching any place in the array that equals to x value;**
{
if ( a[(first+last)/2] > x)
last = (first+last)/2;
else if ( a[(first+last)/2] < x)
first = (first+last)/2;
else
{
pointer = (first+last)/2;
count = 1;
found = true; break;
}
if (Math.abs(last-first) == 1)
{
if (a[first] == x)
{
pointer = first;
count = 1;
found = true;
}
else if (a[last] == x)
{
pointer = last;
count = 1;
found = true;
}
else
return 0;
}
if (first == last)
{
if (a[first] == x)
{
pointer = first;
count = 1;
found = true;
}
else
return 0;
}
}
int backPointer=pointer, forwardPointer=pointer;
boolean stop1=false, stop2= false;
while (!finish) **//Counting the number of places the X value is near our pointer.**
{
if (backPointer-1 >= 0)
if (a[backPointer-1] == x)
{
count++;
backPointer--;
}
else
stop1 = true;
if (forwardPointer+1 <= a.length-1)
if (a[forwardPointer+1] == x)
{
count++;
forwardPointer++;
}
else
stop2 = true;
if (stop1 && stop2)
finish=true;
}
return count;
}
public static void main (String [] args)
{
int [] a = {-25,0,5,11,11,99};
System.out.println(mount(a, 11));
}
}
The print command count it right and prints "2".
I just want to know if anyone can think about better complexity for this method.
Moreover, how can i know what is the time/space-complexity of the method?
All i know about time/space-complexity is that for loop is O(n). I don't know how to calculate my method complexity.
Thank a lot!
Editing:
This is the second while loop after changing:
while (!stop1 || !stop2) //Counting the number of places the X value is near our pointer.
{
if (!stop1)
{
if ( a[last] == x )
{
stop1 = true;
count += (last-pointer);
}
else if ( a[(last+forwardPointer)/2] == x )
{
if (last-forwardPointer == 1)
{
stop1 = true;
count += (forwardPointer-pointer);
}
else
forwardPointer = (last + forwardPointer) / 2;
}
else
last = ((last + forwardPointer) / 2) - 1;
}
if (!stop2)
{
if (a[first] == x)
{
stop2 = true;
count += (pointer - first);
}
else if ( a[(first+backPointer)/2] == x )
{
if (backPointer - first == 1)
{
stop2 = true;
count += (pointer-backPointer);
}
else
backPointer = (first + backPointer) / 2;
}
else
first = ((first + backPointer) / 2) + 1;
}
}
What do you think about the changing? I think it would change the time complexity to O(long(n)).
First let's examine your code:
The code could be heavily refactored and cleaned (which would also result in more efficient implementation, yet without improving time or space complexity), but the algorithm itself is pretty good.
What it does is use standard binary search to find an item with the required value, then scans to the back and to the front to find all other occurrences of the value.
In terms of time complexity, the algorithm is O(N). The worst case is when the entire array is the same value and you end up iterating all of it in the 2nd phase (the binary search will only take 1 iteration). Space complexity is O(1). The memory usage (space) is unaffected by growth in input size.
You could improve the worst case time complexity if you keep using binary search on the 2 sub-arrays (back & front) and increase the "match range" logarithmically this way. The time complexity will become O(log(N)). Space complexity will remain O(1) for the same reason as before.
However, the average complexity for a real-world scenario (where the array contains various values) would be very close and might even lean towards your own version.
I have a question considering Big O runtime and the indexOf method within LinkedList, and ArrayList. How can I come up with a Big O runtime assumption and how would it be different in a Linked List as opposed to an Array List?
LinkedList indexOf()
public int indexOf(Object value)
{
int results = -1;
boolean done = false;
Node<E> ref = head.next;
for(int i = 0; i < size && !done; i++)
{
if(value.equals(ref.value))
{
results = i;
done = true;
}
ref = ref.next;
}
return results;
}
ArrayList indexOf()
if (o == null) {
for (int i = 0; i < size; i++)
if (Values[i]==null)
return i;
} else {
for (int i = 0; i < size; i++)
if (o.equals(Values[i]))
return i;
}
return -1;
I apologize if this is a trivial question to some but I am going to need to understand how to come up with a Big O runtime of a method.
In both these implementations you have nothing better to do than go over the list one element at a time and compare it to the value you're looking for. At the worst case, you'd be going through the entire list, giving a complexity of O(n).
Examples:
"Something %d" and "Something else %d" // Compatible
"Something %d" and "Something else %f" // Not Compatible
"Something %d" and "Something %d else %d" // Not Compatible
"Something %d and %f" and "Something %2$f and %1$d" // Compatible
I figured there should be some C function for this, but I'm not getting any relevant search results. I mean the compiler is checking that the format string and the arguments match, so the code for checking this is already written. The only question is how I can call it.
I'm using Objective-C, so if there is an Objective-C specific solution that's fine too.
Checking if 2 printf() format strings are compatible is an exercise in format parsing.
C, at least, has no standard run-time compare function such as:
int format_cmp(const char *f1, const char *f2); // Does not exist
Formats like "%d %f" and "%i %e" are obviously compatible in that both expect an int and float/double. Note: float are promoted to double as short and signed char are promoted to int.
Formats "%*.*f" and "%i %d %e" are compatible, but not obvious: both expect an int,int and float/double.
Formats "%hhd" and "%d" both expect an int, even though the first will have it values cast to signed char before printing.
Formats "%d" and "%u" are not compatible. Even though many systems will behaved as hoped. Note: Typically char will promote to int.
Formats "%d" and "%ld" are not strictly compatible. On a 32-bit system there are equivalent, but not in general. Of course code can be altered to accommodate this. OTOH "%lf" and "%f" are compatible due to the usual argument promotions of float to double.
Formats "%lu" and "%zu" may be compatible, but that depends on the implementation of unsigned long and size_t. Additions to code could allow this or related equivalences.
Some combinations of modifiers and specifiers are not defined like "%zp". The following does not dis-allow such esoteric combinations - but does compare them.
Modifiers like "$" are extensions to standard C and are not implemented in the following.
The compatibility test for printf() differs from scanf().
#include <ctype.h>
#include <limits.h>
#include <stdio.h>
#include <string.h>
typedef enum {
type_none,
type_int,
type_unsigned,
type_float,
type_charpointer,
type_voidpointer,
type_intpointer,
type_unknown,
type_type_N = 0xFFFFFF
} type_type;
typedef struct {
const char *format;
int int_queue;
type_type type;
} format_T;
static void format_init(format_T *state, const char *format);
static type_type format_get(format_T *state);
static void format_next(format_T *state);
void format_init(format_T *state, const char *format) {
state->format = format;
state->int_queue = 0;
state->type = type_none;
format_next(state);
}
type_type format_get(format_T *state) {
if (state->int_queue > 0) {
return type_int;
}
return state->type;
}
const char *seek_flag(const char *format) {
while (strchr("-+ #0", *format) != NULL)
format++;
return format;
}
const char *seek_width(const char *format, int *int_queue) {
*int_queue = 0;
if (*format == '*') {
format++;
(*int_queue)++;
} else {
while (isdigit((unsigned char ) *format))
format++;
}
if (*format == '.') {
if (*format == '*') {
format++;
(*int_queue)++;
} else {
while (isdigit((unsigned char ) *format))
format++;
}
}
return format;
}
const char *seek_mod(const char *format, int *mod) {
*mod = 0;
if (format[0] == 'h' && format[1] == 'h') {
format += 2;
} else if (format[0] == 'l' && format[1] == 'l') {
*mod = ('l' << CHAR_BIT) + 'l';
format += 2;
} else if (strchr("ljztL", *format)) {
*mod = *format;
format++;
} else if (strchr("h", *format)) {
format++;
}
return format;
}
const char *seek_specifier(const char *format, int mod, type_type *type) {
if (strchr("di", *format)) {
*type = type_int;
format++;
} else if (strchr("ouxX", *format)) {
*type = type_unsigned;
format++;
} else if (strchr("fFeEgGaA", *format)) {
if (mod == 'l') mod = 0;
*type = type_float;
format++;
} else if (strchr("c", *format)) {
*type = type_int;
format++;
} else if (strchr("s", *format)) {
*type = type_charpointer;
format++;
} else if (strchr("p", *format)) {
*type = type_voidpointer;
format++;
} else if (strchr("n", *format)) {
*type = type_intpointer;
format++;
} else {
*type = type_unknown;
exit(1);
}
*type |= mod << CHAR_BIT; // Bring in modifier
return format;
}
void format_next(format_T *state) {
if (state->int_queue > 0) {
state->int_queue--;
return;
}
while (*state->format) {
if (state->format[0] == '%') {
state->format++;
if (state->format[0] == '%') {
state->format++;
continue;
}
state->format = seek_flag(state->format);
state->format = seek_width(state->format, &state->int_queue);
int mod;
state->format = seek_mod(state->format, &mod);
state->format = seek_specifier(state->format, mod, &state->type);
return;
} else {
state->format++;
}
}
state->type = type_none;
}
// 0 Compatible
// 1 Not Compatible
// 2 Not Comparable
int format_cmp(const char *f1, const char *f2) {
format_T state1;
format_init(&state1, f1);
format_T state2;
format_init(&state2, f2);
while (format_get(&state1) == format_get(&state2)) {
if (format_get(&state1) == type_none)
return 0;
if (format_get(&state1) == type_unknown)
return 2;
format_next(&state1);
format_next(&state2);
}
if (format_get(&state1) == type_unknown)
return 2;
if (format_get(&state2) == type_unknown)
return 2;
return 1;
}
Note: only minimal testing done. Lots of additional considerations could be added.
Known shortcomings: hh,h,l,ll,j,z,t modifiers with n. l with s,c.
[Edit]
OP comments about security concerns. This changes the nature of the post and the compare from an equality one to a security one. I'd imagine that one of the patterns (A) would be a reference pattern and the next (B) would be the test. The test would be "is B at least as secure as A?". Example A = "%.20s" and B1 = "%.19s", B2 = "%.20s", B3 = "%.21s". B1 and B2 both pass the security test as they do not extract more the 20 char. B3 is a problem as it goes pass the reference limit of 20 char. Further any non-width qualified with %s %[ %c is a security problem - in the reference or test pattern. This answer's code does not address this issue.
As mentioned, code does not yet handle modifiers with "%n".
[2018 edit]
Concerning "Formats "%d" and "%u" are not compatible.": This is for values to be printed in general. For values in the [0..INT_MAX] range, either format may work per C11dr ยง6.5.2.2 6.
My understanding of what you want, is that, you basically want a method which can look at two strings and detect if they both have the same types of values in them. Or something a long those lines.... If so, then try this (or something along the lines of this):
-(int)checkCompatible:(NSString *)string_1 :(NSString *)string_2 {
// Separate the string into single elements.
NSArray *stringArray_1 = [string_1 componentsSeparatedByString:#" "];
NSArray *stringArray_2 = [string_2 componentsSeparatedByString:#" "];
// Store only the numbers for comparison in a new array.
NSMutableArray *numbers_1 = [[NSMutableArray alloc] init];
NSMutableArray *numbers_2 = [[NSMutableArray alloc] init];
// Make sure the for loop below, runs for the appropriate
// number of cycles depending on which array is bigger.
int loopMax = 0;
if ([stringArray_1 count] > [stringArray_2 count]) {
loopMax = (int)[stringArray_1 count];
}
else {
loopMax = (int)[stringArray_2 count];
}
// Now go through the stringArray's and store only the
// numbers in the mutable array's. This will be used
// during the comparison stage.
for (int loop = 0; loop < loopMax; loop++) {
NSCharacterSet *notDigits = [[NSCharacterSet decimalDigitCharacterSet] invertedSet];
if (loop < [stringArray_1 count]) {
if ([[stringArray_1 objectAtindex:loop] rangeOfCharacterFromSet:notDigits].location == NSNotFound) {
// String consists only of the digits 0 through 9.
[numbers_1 addObject:[stringArray_1 objectAtindex:loop]];
}
}
if (loop < [stringArray_2 count]) {
if ([[stringArray_2 objectAtindex:loop] rangeOfCharacterFromSet:notDigits].location == NSNotFound) {
// String consists only of the digits 0 through 9.
[numbers_2 addObject:[stringArray_2 objectAtindex:loop]];
}
}
}
// Now look through the mutable array's
// and perform the type comparison,.
if ([numbers_1 count] != [numbers_2 count]) {
// One of the two strings has more numbers
// than the other, so they are NOT compatible.
return 1;
}
else {
// Both string have the same number of numbers
// numbers so lets go through them to make
// sure the numbers are of the same type.
for (int loop = 0; loop < [numbers_1 count]; loop++) {
// Check to see if the number in the current array index
// is a float or an integer. All the numbers in the array have
// to be the SAME type, in order for the strings to be compatible.
BOOL check_float_1 = [[NSScanner scannerWithString:[numbers_1 objectAtindex:loop]] scanFloat:nil];
BOOL check_int_1 = [[NSScanner scannerWithString:[numbers_1 objectAtindex:loop]] scanInt:nil];
BOOL check_float_2 = [[NSScanner scannerWithString:[numbers_2 objectAtindex:loop]] scanFloat:nil];
BOOL check_int_2 = [[NSScanner scannerWithString:[numbers_2 objectAtindex:loop]] scanInt:nil];
if (check_float_1 == YES) {
if (check_float_2 == NO) {
return 1;
}
}
else if (check_int_1 == YES) {
if (check_int_2 == NO) {
return 1;
}
}
else {
// Error of some sort......
return 1;
}
}
// All the numbers in the strings are of the same
// type (otherwise we would NOT have reached
// this point). Therefore the strings are compatible.
return 0;
}
}
I am trying to write a program that will take an integer input, and then convert it to words. for example: 123, one two three. Also -3908, negative three nine zero eight.
My code works 90% of the time, the only issue coming along when i am putting one or more zeros on the end of the integer. eg. 70800 will come up as seven zero eight. It completely misses the end zeros. I understand why that is happening but does anybody know if there is a way around it.
PS(i am not allowed as a part of this task to accept the input as a string and split it into an array, so it would be best if the answer is based off this code).
int main(int argc, const char * argv[])
{
#autoreleasepool {
float abNumber;
int i = 0;
float number;
float result;
float firstNumber;
printf("type a number: ");
scanf("%f", &firstNumber);
abNumber = abs(firstNumber);
if (firstNumber < 0) {
printf("negative ");
}
number = abNumber;
while (number >= 10) {
number = number / 10;
i++;
}
do {
float countNumber = abNumber;
float power = powf(10, -i);
float powerNo2 = powf(10, i);
countNumber = countNumber * power;
result = floorf(countNumber);
if (result == 9){
printf("nine ");
}
if (result == 8){
printf("eight ");
}
if (result == 7){
printf("seven ");
}
if (result == 6){
printf("six ");
}
if (result == 5){
printf("five ");
}
if (result == 4){
printf("four ");
}
if (result == 3){
printf("three ");
}
if (result == 2){
printf("two ");
}
if (result == 1){
printf("one ");
}
if (result == 0){
printf("zero ");
}
while (abNumber > powerNo2) {
abNumber = abNumber - powerNo2;
}
i--;
} while (i >= 0);
}
return 0;
}
The main error seems to be that
while (abNumber > powerNo2) {
should be
while (abNumber >= powerNo2) {
But I would recommend not to use floating point arithmetic at all, to avoid
possible rounding errors. The same can be achieved with simple integer arithmetic
(I have omitted the "negative case" for simplicity):
int number;
printf("type a number: ");
scanf("%d", &number);
// Determine highest power of 10 that is <= the given number:
int power = 1;
while (10 * power <= number) {
power *= 10;
}
// Extract each digit:
while (power > 0) {
int digit = (number / power) % 10;
/*
* Use switch/case to print 'digit' as a string ...
*/
power /= 10;
}
I would go for recursive solution, like that
int print(int num)
{
if( num )
{
int mod = num%10;
print(num/10);
switch(mod)
{
case 0:printf(" zero");break;
case 1:printf(" one");break;
case 2:printf(" two");break;
case 3:printf(" three");break;
}
}
return 0;
}
Recursivy divide the number untill nothing left of it, on the way back print the mod.
Why don't you just input the number as a string then loop through each character:
Exemple: http://ideone.com/E8QspN
Input:
-12003200
Output:
negative one two zero zero tree two zero zero
Code:
#include <stdio.h>
int main(int argc, char *argv[])
{
char input[25];
scanf("%s", input);
int i = 0;
while (input[i] != '\0') {
switch(input[i]) {
case '-' :
printf("negative");
break;
case '0' :
printf("zero");
break;
case '1' :
printf("one");
break;
case '2' :
printf("two");
break;
case '3' :
printf("tree");
break;
case '4' :
printf("four");
break;
case '5' :
printf("five");
break;
case '6' :
printf("six");
break;
case '7' :
printf("seven");
break;
case '8' :
printf("eight");
break;
case '9' :
printf("nine");
break;
default :
break;
}
printf(" ");
i++;
}
return 0;
}
Consider the following:
Code
#include <stdio.h>
#include <stdlib.h>
const char *numbers[10] = {"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
void printNum(int num);
int main(void)
{
int num;
printf("Enter a number: ");
scanf("%u", &num);
printNum(num);
printf("\n");
return 0;
}
void printNum(int num)
{
int absNum = abs(num);
if(absNum > 9)
printNum(num / 10);
if((absNum < 10) && (num < 0))
printf("negative");
printf(" %s", numbers[absNum % 10]);
}
Example Output
Enter a number: 2582
two five eight two
Enter a number: -943
negative nine four three
Enter a number: 1000
one zero zero zero
Enter a number: -1000
negative one zero zero zero
Logic
Get an integer from user.
Send to recursive function.
Keep recursing until the least significant digit is left. This is to print in the correct order.
Print digit as a string by using a lookup table.
Keep someone a high five.
To Do
Error checking
const char *sentence = "He was not in the cab at the time.";
printf("\"%s\" has %d spaces\n", sentence, (int) ^ {
int i = 0;
int countSpaces = 0;
while (sentence[i] != '\0') {
if (sentence[i] == 0x20) {
countSpaces++;
}
i++;
}
return countSpaces;
});
This code simply counts the white space in a string, but for some reason it says 1606416608 spaces rather than 8. I'm not exactly sure what is going wrong, so thanks for any help!
You're passing the actual block to printf, not the result of the block. Instead, try
const char *sentence = "He was not in the cab at the time.";
printf("\"%s\" has %d spaces\n", sentence, (int) ^ {
int i = 0;
int countSpaces = 0;
while (sentence[i] != '\0') {
if (sentence[i] == 0x20) {
countSpaces++;
}
i++;
}
return countSpaces;
}()); // <-- note the extra parentheses here, indicating that you're calling the block