here is my function:
int repeatedNTimes(int* A, int ASize)
{
int i, count, j, temp;
for(i = 0; i < ASize; ++i)
{
count = 0;
temp = A[i];
for(j = i; j < ASize; ++j)
{
if(A[i] == A[j])
count++;
}
if(count == ASize / 2)
return A[i];
else
continue;
}
return 0;
}
Can I use return 1, or return (any integer) instead of return 0?
And secondly, what if I don't return an integer?
If you do not return an integer, then the behavior is not well defined (probably undefined, but I don't have the standard memorized). Your compiler will likely emit a warning if you have warnings on.
As for returning an integer other than 0, yes, you can do that. What matters is the return type of the function when it comes to what you can and cannot return. That said, returning a different result may not have the effect you want depending on what your function does. Sometimes values like zero are reserved for special conditions like not found.
I'm trying to learn pointers in C++, but seems that it get more complicated...
In the main loop
int i;
for (i = 0; i < 5; ++i){
if (fun == arrfun[i]) break;
}
How is that fun==arrfun[i] at fun2 if both fun and arrfun start looping form 0? Hence they should equal at log(x) instead. How could I loop to sin or cos, etc?
#include <iostream>
#include <cmath>
using namespace std;
typedef double(*FUNDtoD)(double);
typedef FUNDtoD ARRFUN[];
FUNDtoD funmax(ARRFUN, double);
double fun0(double x) { return log(x); }
double fun1(double x) { return x*x; }
double fun2(double x) { return exp(x); }
double fun3(double x) { return sin(x); }
double fun4(double x) { return cos(x); }
int main() {
ARRFUN arrfun = { fun0, fun1, fun2, fun3, fun4 };
FUNDtoD fun = funmax(arrfun, 1);
int i;
for (i = 0; i < 5; ++i){
if (fun == arrfun[i]) break;
}
cout.precision(14);
cout << "Largest value at x=1 assumed by function # "
<< i << ".\nThe value is " << fun(2) << endl;
return 0;
}
FUNDtoD funmax(ARRFUN f, double x){
double m = f[0](x), z;
int k = 0;
for (int i = 1; i < 5; i++){
if ((z = f[i](x)) > m) {
m = z;
k = i;
}
}
return f[k];
}
I don't understand how function FUNDtoD funmax is working at the bottom, could somebody clarify it please, many thanks.
How is that fun==arrfun[i] at fun2 if both fun and arrfun start looping form 0? > Hence they should equal at log(x) instead. How could I loop to sin or cos, etc?
fun is not being looped, the same pointer is checked against each function pointer stored in the array (arrfun). It is simply trying to find the index of the returned function pointer. If sin(x) gave the highest value then that loop would finish with a 4 in i.
don't understand how function FUNDtoD funmax is working at the bottom, could
somebody clarify it please, many thanks.
It breaks down as follows:
Firstly it performs m = f0;
f[0] -> fun0 so the result is m = log( x );
Next it steps through the other 4 functions and tests whether the operation on x results in a higher valuer than the previous. It then stores the index of the function that returned the highest value.
Finally it returns that function pointer.
Suppose we have an array of integers. We've written a function to fetch the index of the first specified value in the array, or -1 if the array does not contain the value..
So for example, if the array = { 4, 5, 4, 4, 7 }, then getFirstIndexOf(4) would return 0, getFirstIndexOf(7) would return 4, and getFirstIndexOf(8) would return -1.
Below, I have presented three different ways to write this function. It is a widely accepted coding standard that returns in the middle of functions, and breaks in the middle of loops are poor practice. It seems to me that this might be an acceptable use for them.
public int getFirstIndexOf(int specifiedNumber) {
for (int i = 0; i < array.length; i++) {
if (array[i] == specifiedNumber) {
return i;
}
}
return -1;
}
VS.
public int getFirstIndexOf(int specifiedNumber) {
int result = -1;
for (int i = 0; i < array.length; i++) {
if (array[i] == specifiedNumber) {
result = i;
break;
}
}
return result;
}
VS.
public int getFirstIndexOf(int specifiedNumber) {
int result = -1;
for (int i = 0; i < array.length; i++) {
if (array[i] == specifiedNumber && result == -1) {
result = i;
}
}
return result;
}
What do you think? Which is best? Why? Is there perhaps another way to do this?
I think it's poor practice to run a full loop when you have already found your result...
If you really want to avoid using return from the middle of the loop, I would sugest to use a "sentinel" to stop your loop.
public int getFirstIndexOf(int specifiedNumber, int[] array) {
boolean found = false;
boolean exit = false;
int i = 0;
int arraySize = array.length();
while(!found && !exit) {
if(array[i] == specifiedNumber) {
found = true;
} else {
if(i++ > arraySize) {
exit = true;
}
}
if(found ==true) {
return i;
} else {
return 99999;
}
}
edit: I hate to indent code using spaces in StackOverflow...
That's why do...while & while loop was invented.
As requested:
public int getFirstIndexOf(int specifiedNumber) {
int i = array.Length;
while(--i > -1 && array[i] != specifiedNumber);
return i;
}
I am looking for an optimal algorithm to find out remaining all possible permutation
of a give binary number.
For ex:
Binary number is : ........1. algorithm should return the remaining 2^7 remaining binary numbers, like 00000001,00000011, etc.
Thanks,
sathish
The example given is not a permutation!
A permutation is a reordering of the input.
So if the input is 00000001, 00100000 and 00000010 are permutations, but 00000011 is not.
If this is only for small numbers (probably up to 16 bits), then just iterate over all of them and ignore the mismatches:
int fixed = 0x01; // this is the fixed part
int mask = 0x01; // these are the bits of the fixed part which matter
for (int i=0; i<256; i++) {
if (i & mask == fixed) {
print i;
}
}
to find all you aren't going to do better than looping over all numbers e.g. if you want to loop over all 8 bit numbers
for (int i =0; i < (1<<8) ; ++i)
{
//do stuff with i
}
if you need to output in binary then look at the string formatting options you have in what ever language you are using.
e.g.
printf("%b",i); //not standard in C/C++
for calculation the base should be irrelavent in most languages.
I read your question as: "given some binary number with some bits always set, create the remaining possible binary numbers".
For example, given 1xx1: you want: 1001, 1011, 1101, 1111.
An O(N) algorithm is as follows.
Suppose the bits are defined in mask m. You also have a hash h.
To generate the numbers < n-1, in pseudocode:
counter = 0
for x in 0..n-1:
x' = x | ~m
if h[x'] is not set:
h[x'] = counter
counter += 1
The idea in the code is to walk through all numbers from 0 to n-1, and set the pre-defined bits to 1. Then memoize the resulting number (iff not already memoized) by mapping the resulting number to the value of a running counter.
The keys of h will be the permutations. As a bonus the h[p] will contain a unique index number for the permutation p, although you did not need it in your original question, it can be useful.
Why are you making it complicated !
It is as simple as the following:
// permutation of i on a length K
// Example : decimal i=10 is permuted over length k= 7
// [10]0001010-> [5] 0000101-> [66] 1000010 and 33, 80, 40, 20 etc.
main(){
int i=10,j,k=7; j=i;
do printf("%d \n", i= ( (i&1)<< k + i >>1); while (i!=j);
}
There are many permutation generating algorithms you can use, such as this one:
#include <stdio.h>
void print(const int *v, const int size)
{
if (v != 0) {
for (int i = 0; i < size; i++) {
printf("%4d", v[i] );
}
printf("\n");
}
} // print
void visit(int *Value, int N, int k)
{
static level = -1;
level = level+1; Value[k] = level;
if (level == N)
print(Value, N);
else
for (int i = 0; i < N; i++)
if (Value[i] == 0)
visit(Value, N, i);
level = level-1; Value[k] = 0;
}
main()
{
const int N = 4;
int Value[N];
for (int i = 0; i < N; i++) {
Value[i] = 0;
}
visit(Value, N, 0);
}
source: http://www.bearcave.com/random_hacks/permute.html
Make sure you adapt the relevant constants to your needs (binary number, 7 bits, etc...)
If you are really looking for permutations then the following code should do.
To find all possible permutations of a given binary string(pattern) for example.
The permutations of 1000 are 1000, 0100, 0010, 0001:
void permutation(int no_ones, int no_zeroes, string accum){
if(no_ones == 0){
for(int i=0;i<no_zeroes;i++){
accum += "0";
}
cout << accum << endl;
return;
}
else if(no_zeroes == 0){
for(int j=0;j<no_ones;j++){
accum += "1";
}
cout << accum << endl;
return;
}
permutation (no_ones - 1, no_zeroes, accum + "1");
permutation (no_ones , no_zeroes - 1, accum + "0");
}
int main(){
string append = "";
//finding permutation of 11000
permutation(2, 6, append); //the permutations are
//11000
//10100
//10010
//10001
//01100
//01010
cin.get();
}
If you intend to generate all the string combinations for n bits , then the problem can be solved using backtracking.
Here you go :
//Generating all string of n bits assuming A[0..n-1] is array of size n
public class Backtracking {
int[] A;
void Binary(int n){
if(n<1){
for(int i : A)
System.out.print(i);
System.out.println();
}else{
A[n-1] = 0;
Binary(n-1);
A[n-1] = 1;
Binary(n-1);
}
}
public static void main(String[] args) {
// n is number of bits
int n = 8;
Backtracking backtracking = new Backtracking();
backtracking.A = new int[n];
backtracking.Binary(n);
}
}
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Closed 11 years ago.
Not so long ago I was in an interview, that required solving two very interesting problems. I'm curious how would you approach the solutions.
Problem 1 :
Product of everything except current
Write a function that takes as input two integer arrays of length len, input and index, and generates a third array, result, such that:
result[i] = product of everything in input except input[index[i]]
For instance, if the function is called with len=4, input={2,3,4,5}, and index={1,3,2,0}, then result will be set to {40,24,30,60}.
IMPORTANT: Your algorithm must run in linear time.
Problem 2 : ( the topic was in one of Jeff posts )
Shuffle card deck evenly
Design (either in C++ or in C#) a class Deck to represent an ordered deck of cards, where a deck contains 52 cards, divided in 13 ranks (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K) of the four suits: spades (?), hearts (?), diamonds (?) and clubs (?).
Based on this class, devise and implement an efficient algorithm to shuffle a deck of cards. The cards must be evenly shuffled, that is, every card in the original deck must have the same probability to end up in any possible position in the shuffled deck.
The algorithm should be implemented in a method shuffle() of the class Deck:
void shuffle()
What is the complexity of your algorithm (as a function of the number n of cards in the deck)?
Explain how you would test that the cards are evenly shuffled by your method (black box testing).
P.S. I had two hours to code the solutions
First question:
int countZeroes (int[] vec) {
int ret = 0;
foreach(int i in vec) if (i == 0) ret++;
return ret;
}
int[] mysticCalc(int[] values, int[] indexes) {
int zeroes = countZeroes(values);
int[] retval = new int[values.length];
int product = 1;
if (zeroes >= 2) { // 2 or more zeroes, all results will be 0
for (int i = 0; i > values.length; i++) {
retval[i] = 0;
}
return retval;
}
foreach (int i in values) {
if (i != 0) product *= i; // we have at most 1 zero, dont include in product;
}
int indexcounter = 0;
foreach(int idx in indexes) {
if (zeroes == 1 && values[idx] != 0) { // One zero on other index. Our value will be 0
retval[indexcounter] = 0;
}
else if (zeroes == 1) { // One zero on this index. result is product
retval[indexcounter] = product;
}
else { // No zeros. Return product/value at index
retval[indexcounter] = product / values[idx];
}
indexcouter++;
}
return retval;
}
Worst case this program will step through 3 vectors once.
For the first one, first calculate the product of entire contents of input, and then for every element of index, divide the calculated product by input[index[i]], to fill in your result array.
Of course I have to assume that the input has no zeros.
Tnilsson, great solution ( because I've done it the exact same way :P ).
I don't see any other way to do it in linear time. Does anybody ? Because the recruiting manager told me, that this solution was not strong enough.
Are we missing some super complex, do everything in one return line, solution ?
A linear-time solution in C#3 for the first problem is:-
IEnumerable<int> ProductExcept(List<int> l, List<int> indexes) {
if (l.Count(i => i == 0) == 1) {
int singleZeroProd = l.Aggregate(1, (x, y) => y != 0 ? x * y : x);
return from i in indexes select l[i] == 0 ? singleZeroProd : 0;
} else {
int prod = l.Aggregate(1, (x, y) => x * y);
return from i in indexes select prod == 0 ? 0 : prod / l[i];
}
}
Edit: Took into account a single zero!! My last solution took me 2 minutes while I was at work so I don't feel so bad :-)
Product of everything except current in C
void product_except_current(int input[], int index[], int out[],
int len) {
int prod = 1, nzeros = 0, izero = -1;
for (int i = 0; i < len; ++i)
if ((out[i] = input[index[i]]) != 0)
// compute product of non-zero elements
prod *= out[i]; // ignore possible overflow problem
else {
if (++nzeros == 2)
// if number of zeros greater than 1 then out[i] = 0 for all i
break;
izero = i; // save index of zero-valued element
}
//
for (int i = 0; i < len; ++i)
out[i] = nzeros ? 0 : prod / out[i];
if (nzeros == 1)
out[izero] = prod; // the only non-zero-valued element
}
Here's the answer to the second one in C# with a test method. Shuffle looks O(n) to me.
Edit: Having looked at the Fisher-Yates shuffle, I discovered that I'd re-invented that algorithm without knowing about it :-) it is obvious, however. I implemented the Durstenfeld approach which takes us from O(n^2) -> O(n), really clever!
public enum CardValue { A, Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten, J, Q, K }
public enum Suit { Spades, Hearts, Diamonds, Clubs }
public class Card {
public Card(CardValue value, Suit suit) {
Value = value;
Suit = suit;
}
public CardValue Value { get; private set; }
public Suit Suit { get; private set; }
}
public class Deck : IEnumerable<Card> {
public Deck() {
initialiseDeck();
Shuffle();
}
private Card[] cards = new Card[52];
private void initialiseDeck() {
for (int i = 0; i < 4; ++i) {
for (int j = 0; j < 13; ++j) {
cards[i * 13 + j] = new Card((CardValue)j, (Suit)i);
}
}
}
public void Shuffle() {
Random random = new Random();
for (int i = 0; i < 52; ++i) {
int j = random.Next(51 - i);
// Swap the cards.
Card temp = cards[51 - i];
cards[51 - i] = cards[j];
cards[j] = temp;
}
}
public IEnumerator<Card> GetEnumerator() {
foreach (Card c in cards) yield return c;
}
System.Collections.IEnumerator System.Collections.IEnumerable.GetEnumerator() {
foreach (Card c in cards) yield return c;
}
}
class Program {
static void Main(string[] args) {
foreach (Card c in new Deck()) {
Console.WriteLine("{0} of {1}", c.Value, c.Suit);
}
Console.ReadKey(true);
}
}
In Haskell:
import Array
problem1 input index = [(left!i) * (right!(i+1)) | i <- index]
where left = scanWith scanl
right = scanWith scanr
scanWith scan = listArray (0, length input) (scan (*) 1 input)
Vaibhav, unfortunately we have to assume, that there could be a 0 in the input table.
Second problem.
public static void shuffle (int[] array)
{
Random rng = new Random(); // i.e., java.util.Random.
int n = array.length; // The number of items left to shuffle (loop invariant).
while (n > 1)
{
int k = rng.nextInt(n); // 0 <= k < n.
n--; // n is now the last pertinent index;
int temp = array[n]; // swap array[n] with array[k] (does nothing if k == n).
array[n] = array[k];
array[k] = temp;
}
}
This is a copy/paste from the wikipedia article about the Fisher-Yates shuffle. O(n) complexity
Tnilsson, I agree that YXJuLnphcnQ solution is arguably faster, but the idee is the same. I forgot to add, that the language is optional in the first problem, as well as int the second.
You're right, that calculationg zeroes, and the product int the same loop is better. Maybe that was the thing.
Tnilsson, I've also uset the Fisher-Yates shuffle :). I'm very interested dough, about the testing part :)
Trilsson made a separate topic about the testing part of the question
How to test randomness (case in point - Shuffling)
very good idea Trilsson:)
YXJuLnphcnQ, that's the way I did it too. It's the most obvious.
But the fact is, that if you write an algorithm, that just shuffles all the cards in the collection one position to the right every time you call sort() it would pass the test, even though the output is not random.
Shuffle card deck evenly in C++
#include <algorithm>
class Deck {
// each card is 8-bit: 4-bit for suit, 4-bit for value
// suits and values are extracted using bit-magic
char cards[52];
public:
// ...
void shuffle() {
std::random_shuffle(cards, cards + 52);
}
// ...
};
Complexity: Linear in N. Exactly 51 swaps are performed. See http://www.sgi.com/tech/stl/random_shuffle.html
Testing:
// ...
int main() {
typedef std::map<std::pair<size_t, Deck::value_type>, size_t> Map;
Map freqs;
Deck d;
const size_t ntests = 100000;
// compute frequencies of events: card at position
for (size_t i = 0; i < ntests; ++i) {
d.shuffle();
size_t pos = 0;
for(Deck::const_iterator j = d.begin(); j != d.end(); ++j, ++pos)
++freqs[std::make_pair(pos, *j)];
}
// if Deck.shuffle() is correct then all frequencies must be similar
for (Map::const_iterator j = freqs.begin(); j != freqs.end(); ++j)
std::cout << "pos=" << j->first.first << " card=" << j->first.second
<< " freq=" << j->second << std::endl;
}
As usual, one test is not sufficient.