I'm trying to solve an algorithmic problem where the premises revolves around an input sequence that consists of 0s and 1s along with a minimum length l for the output subsequence. The problem asks for the subsequence which has the highest rate of 1s (number of ones in the subsequence divided by the length of the subsequence). Further background and sample input/output to the problem can be found here.
I have come up with a solution that passes all tests except for the last one and I am trying to figure out what my current implementation is lacking. My approach is to use a dynamically resizeable sliding window while storing the maximum rate of the sliding window along with the length of that maximum rated window. I think the way that I'm moving (growing and shrinking) my window is the problem, but I'm having trouble figuring out what to change.
This is how I move my window:
static void max_rate(long min_len, string sequence) {
long left_window = 0, right_window = min_len - 1;
long best_left = 0, best_len = 0, most_ones = 0;
long double best_success_rate = -1;
for (;;) {
auto tmp = sequence.substr(left, right - left + 1);
long n_ones = count_ones(tmp);
long double success_rate = (long double)n_ones / (long double)tmp.length();
if (success_rate >= best_success_rate) {
best_success_rate = success_rate;
best_left = left;
best_len = right - left + 1;
most_ones = n_ones;
}
// Window sliding starts here
bool can_move_right = (right + 1) < (long)sequence.length();
bool can_move_left = (right - left + 1 - 1) >= min_len;
if (can_move_right && sequence.at(right + 1) == '1') {
++(right);
} else if (can_move_right && sequence.at(right + 1) == '1') {
++(right);
} else if (can_move_left && (sequence.at(left + 1) == '0')) {
++left;
} else if (can_move_right) {
++(right);
} else {
break;
}
cout << best_left + 1 << " ";
cout << best_len << endl;
I'm basically checking:
Grow the window if you can increase the rate
Otherwise, if possible (considering our minimum size requirement), shrink the window if you can increase the rate
Otherwise, if possible (-), shrink the window
Otherwise, if possible (we are not at the end of the sequence) grow the window
I must be missing something here I think
According to the code you posted, for input
8, "0011101011"
the sequence would seem to be
0 0 1 1 1 0 1 0 1 1
l r
l r
l r
l r
which gives us:
zero-based index 1, interval length 9
and ratio 6 / 9 = 0.6666666666666666
but the correct answer is
ratio 0.75
zero-based index 2, interval length 8
Related
These are my current errors, I think I did something wrong with the maths but everything I tried didn't work.
Ps: Sorry if my question's formatting is bad, first time using stackflow.
:) credit.c exists
:) credit.c compiles
:) identifies 378282246310005 as AMEX
:) identifies 371449635398431 as AMEX
:) identifies 5555555555554444 as MASTERCARD
:) identifies 5105105105105100 as MASTERCARD
:) identifies 4111111111111111 as VISA
:) identifies 4012888888881881 as VISA
:) identifies 4222222222222 as VISA
:) identifies 1234567890 as INVALID
:) identifies 369421438430814 as INVALID
:) identifies 4062901840 as INVALID
:) identifies 5673598276138003 as INVALID
:( identifies 4111111111111113 as INVALID
expected "INVALID\n", not "VISA\n"
:( identifies 4222222222223 as INVALID
expected "INVALID\n", not "VISA\n"
#include <cs50.h>
#include <math.h>
// Prompt user for credit card number
int main(void)
{
long credit_card, credit_number;
do
{
credit_card = get_long("Enter credit card number: ");
}
while (credit_card < 0);
credit_number = credit_card;
// Calculate total number of digits
int count = (credit_number == 0) ? 1 : (log10(credit_number) + 1);
int summation = 0;
while (credit_number == 0)
{
int x = credit_number % 10; summation += x;
int y = 2 * ((credit_number / 10) % 10);
int r = (y % 10) + floor((y / 10) % 10); summation += r; credit_number /= 100;
}
string card;
// Identify which card type you get after inputing your credit card number
int test = cc / pow(10, count - 2);
if ((count == 13 || count == 16) && test / 10 == 4)
{
card = "VISA";
}
else if (count == 16 && test >= 51 && test <= 55)
{
card = "MASTERCARD";
}
else if (count == 15 && (test == 34 || test == 37))
{
card = "AMEX";
}
else
{
card = "INVALID";
}
// Final verification
if (sum % 10 == 0)
{
printf("%s\n", card);
}
else
{
printf("INVALID\n");
}
}```
Your algorithm is maybe not fully correct. I would therefore propose a different approach. You can look at each single digit in a loop. And, you can also do the whole checksum calculation in one step.
I will show you how to do and explain the algorithm behind it.
BTW. Chosing the right algorithm is always the key for success.
So, first we need to think on how we can extract digits from a number. This can be done in a loop by repeating the follwoing steps:
Perform a modulo 10 division to get a digit
Do a integer division by 10
Repeat
Let us look at the example 1234.
Step 1 will get the 4 -- (1234 % 10 = 4)
Step 2 will convert original number into 123 -- (1234 / 10 = 123)
Step 1 will get the 3 -- (123 % 10 = 3)
Step 2 will convert the previous number into 12 -- (123 / 10 = 12)
Step 1 will get the 2 -- (12 % 10 = 2)
Step 2 will convert the previous number into 1 -- (12 / 10 = 1)
Step 1 will get the 1 -- (1 % 10 = 1)
Step 2 will convert the previous number into 0 -- (1 / 10 = 0)
Then the loop stops. Additionally we can observe that the loop stops, when the resulting divided becomes 0. And, we see addtionally that the number of loop executions is equal to the number of digits in the number. But this is somehow obvious.
OK, then let us look, what we learned so far
while (creditCardNumber > 0) {
unsigned int digit = creditCardNumber % 10;
creditCardNumber /= 10;
++countOfDigits;
}
This will get all digits and count them.
Good. Lets go to next step.
For later validation and comparison purpose we need to get the most significant digit (the first digit) and the second most significant digit (the second digit) of the number.
For this, we define 2 variables which will hold the number. We simply assign the current evaluated digit (and override it in each loop execution) to the "mostSignificantDigit". At the end of the loop, we will have it in our desired variable.
For the "secondMostSignificantDigit" we will simple copy the "old" or "previous" value of the "mostSignificantDigit", before assigning a new value to "mostSignificantDigit". With that, we will always have both values available.
The loop looks now like this:
while (creditCardNumber > 0) {
const unsigned int digit = creditCardNumber % 10;
secondMostSignificantDigit = mostSignificantDigit;
mostSignificantDigit = digit;
creditCardNumber /= 10;
++countOfDigits;
}
OK, now we come to the maybe more complex part. The cheksum. The calculation method is.
Start with the least significant (the last) digit
Do not multiply the digit, which is equivalent with multiplying it with 1, and add it to the checksum
Goto the next digit. Multiply it by 2. If the result is greater than 10, then get again the single digits and add both digits to the checksum
Repeat
So, the secret is, to analyze the somehow cryptic specification, given here. If we start with the last digit, we do not multiply it, the next digit will be multiplied, the next not and so on and so on.
To "not multiply" is the same as multiplying by 1. This means: In the loop we need to multiply alternating with 1 or with 2.
How to get alternating numbers in a loop? The algorithm for that is fairly simple. If you need alternating numbers, lets say, x,y,x,y,x,y,x..., Then, build the sum of x and y and perform the subtratcion "value = sum - value". Example:
We need alternating values 1 and 2. The sum is 3. To get the next value, we subtract the current value from the sum.
initial value = 1
sum = 3
current value = initial value = 1
next value = 3 - 1 = 2. Current value = 2
next value = 3 - 2 = 1. Current value = 1
next value = 3 - 1 = 2. Current value = 2
next value = 3 - 2 = 1. Current value = 1
next value = 3 - 1 = 2. Current value = 2
next value = 3 - 2 = 1. Current value = 1
. . .
Good, now we understand, how to make alternating values.
Next, If we multiply a digit with 2, then the maximum result maybe a 2 digit value. We get the single digits with a modulo and an integer division by 10.
And, now important, it does not matter, if we multiply or not, because, if we do not multiply, then the upper digit will always be 0. And this will not contribute to the sum.
With all that, we can always do a multiplication and always split the result into 2 digits (many of them having the upper digit 0).
The result will be:
checkSum += (digit * multiplier) % 10 + (digit * multiplier) / 10;
multiplier = 3 - multiplier;
An astonishingly simple formula.
Next, if we know C or C++ we also know that a multiplication with 2 can be done very efficiently with a bit shift left. And, additionally, a "no-multiplication" can be done with a bit shift 0. That is extremely efficient and faster than multiplication.
x * 1 is identical with x << 0
x * 2 is identical with x << 1
For the final result we will use this mechanism, alternate the multiplier between 0 and 1 and do shifts.
This will give us a very effective checksum calculation.
At the end of the program, we will use all gathered values and compare them to the specification.
Thsi will lead to:
int main() {
// Get the credit card number. Unfortunately I do not know CS50. I use the C++ standard iostream lib.
// Please replace the following 4 lines with your CS50 equivalent
unsigned long long creditCardNumber;
std::cout << "Enter credit card number: ";
std::cin >> creditCardNumber;
std::cout << "\n\n";
// We need to count the number of digits for validation
unsigned int countOfDigits = 0;
// Here we will calculate the checksum
unsigned int checkSum = 0;
// We need to multiply digits with 1 or with 2
unsigned int multiplier = 0;
// For validation purposes we need the most significant 2 digits
unsigned int mostSignificantDigit = 0;
unsigned int secondMostSignificantDigit = 0;
// Now we get all digits from the credit card number in a loop
while (creditCardNumber > 0) {
// Get the least significant digits (for 1234 it will be 4)
const unsigned int digit = creditCardNumber % 10;
// Now we have one digit more. In the end we will have the number of all digits
++countOfDigits;
// Simply remember the most significant digits
secondMostSignificantDigit = mostSignificantDigit;
mostSignificantDigit = digit;
// Calculate the checksum
checkSum += (digit << multiplier) % 10 + (digit << multiplier) / 10;
// Multiplier for next loop
multiplier = 1 - multiplier;
creditCardNumber /= 10;
}
// Get the least significant digit of the checksum
checkSum %= 10;
// Validate all calculated values and show the result
if ((0 == checkSum) && // Checksum must be correct AND
(15 == countOfDigits) && // Count of digits must be correct AND
((3 == mostSignificantDigit) && // Most significant digits must be correct
((4 == secondMostSignificantDigit) || (7 == secondMostSignificantDigit)))) {
std::cout << "AMEX\n";
}
else if ((0 == checkSum) && // Checksum must be correct AND
(16 == countOfDigits) && // Count of digits must be correct AND
((5 == mostSignificantDigit) && // Most significant digits must be correct
((secondMostSignificantDigit > 0) && (secondMostSignificantDigit < 6)))) {
std::cout << "MASTERCARD\n";
}
else if ((0 == checkSum) && // Checksum must be correct AND
((16 == countOfDigits) || (13 == countOfDigits)) && // Count of digits must be correct AND
((4 == mostSignificantDigit))) { // Most significant digit must be correct
std::cout << "VISA\n";
}
else {
std::cout << "INVALID\n";
}
return 0;
}
What we learn with this example, is integer division and modulo division and the smart usage of the identity element for binary operations.
In case of questions, please ask
Just to be complete, I will show you a C++ solution, based on a std::string and using modern C++ elements and algorithms.
For example, the whole checksum calculation will be done with one statement. The whole program does not contain any loop.
#include <iostream>
#include <string>
#include <regex>
#include <numeric>
int main() {
// ---------------------------------------------------------------------------------------------------
// Get user input
// Inform user, what to do. Enter a credit card number. We are a little tolerant with the input format
std::cout << "\nPlease enter a credit card number:\t";
// Get the number, in any format from the user
std::string creditCardNumber{};
std::getline(std::cin, creditCardNumber);
// Remove the noise, meaning, all non digits from the credit card number
creditCardNumber = std::regex_replace(creditCardNumber, std::regex(R"(\D)"), "");
// ---------------------------------------------------------------------------------------------------
// Calculate checksum
unsigned int checksum = std::accumulate(creditCardNumber.rbegin(), creditCardNumber.rend(), 0U,
[multiplier = 1U](const unsigned int sum, const char digit) mutable -> unsigned int {
multiplier = 1 - multiplier; unsigned int value = digit - '0';
return sum + ((value << multiplier) % 10) + ((value << multiplier) / 10); });
// We are only interested in the lowest digit
checksum %= 10;
// ---------------------------------------------------------------------------------------------------
// Validation and output
if ((0 == checksum) && // Checksum must be correct AND
(15 == creditCardNumber.length()) && // Count of digits must be correct AND
(('3' == creditCardNumber[0]) && // Most significant digits must be correct
(('4' == creditCardNumber[1]) || ('7' == creditCardNumber[1])))) {
std::cout << "AMEX\n";
}
else if ((0 == checksum) && // Checksum must be correct AND
(16 == creditCardNumber.length()) && // Count of digits must be correct AND
(('5' == creditCardNumber[0]) && // Most significant digits must be correct
((creditCardNumber[1] > '0') && (creditCardNumber[1] < '6')))) {
std::cout << "MASTERCARD\n";
}
else if ((0 == checksum) && // Checksum must be correct AND
((16 == creditCardNumber.length()) || (13 == creditCardNumber.length())) && // Count of digits must be correct AND
(('4' == creditCardNumber[0]))) { // Most significant digit must be correct
std::cout << "VISA\n";
}
else {
std::cout << "INVALID\n";
}
return 0;
In a binary representation of a number is there a simpler way than this
long half(long patten, bool exclusive) {
if (patten == 0) {
return 1;
}
long newPatten = 0;
for (int p = 0; p < MAX_STEPS; p++) {
long check = 1 << p;
if ((check & patten) > 0) {
int end = (p + MAX_STEPS) - 1;
for (int to = p+1; to <= end; to++) {
long checkTo = 1 << (to % MAX_STEPS);
if ( (checkTo & patten) > 0 || to == end ) {
int distance = to - p;
long fullShift = (int)round( ((float)p) + ((float)distance)/2 );
long shift = fullShift % MAX_STEPS;
long toAdd = 1 << shift;
newPatten = newPatten | toAdd;
break;
}
}
}
}
return exclusive ? patten ^ newPatten : patten | newPatten;
}
To add a bit in the middle of any two other bits with wrapping around and rounding to one side when there is an even number of positions between?
E.g.
010001 to
110101
Or
1001 to
1101
Update: These bits aren't coming from anywhere else, such as another number, just want a new True bit to be in the middle of any other two True bits, so if there was a 101, then the middle would be the 0 and the result would be 111. Or if we started with 10001, then the middle would be the center 0 and the result would be 10101.
When i say wrapping i also mean adding bits as if the bit array was a circle so if we had 00100010 representing the positions with letters:
00100010
hgfedcba
So we have True bits at b and f we would put a middle bit between b and f going left to right at d but also going right to left, wrapping around and putting it at h resulting in:
10101010
hgfedcba
I understand this isn't a usual problem.
Are there known tricks to do things like this without loops?
I am writing code in Hackerrank. And recently the problem said, convert decimal to base 2 and then count the max consecutive 1's in the binary number. And first I come with following solution. It works fine. But I do not understand the counting part of it, even though I wrote it.
The code is
int main(){
int n,ind=0, count=0, mmax=0;
char bin[100];
cin >> n;
while(n){
if(n%2==0) {
bin[ind]='0';
n = n / 2;
ind = ind + 1;
}
else if(n%2==1) {
bin[ind]='1';
n = n / 2;
ind = ind + 1;
}
}
for(int i=0; i<=(ind-1); i++){
if(bin[i] == '1' && bin[i+1] == '1'){
count++;
if(mmax < count)
mmax = count;
}
else
count=0;
}
cout << mmax + 1 << endl;
return 0;
}
In the above code, I guess that variable mmax will give me the max consecutive number of 1's but it gives me value that has (max consecutive - 1), So I just wrote like that and submitted the code. But I am curious about. why it is working that way. I am little bit of confused the way that code works like this.
Thanks
Lets say you have this binary sequence:
11110
Your code will compare starting from the first and second:
|11|110 1 && 1 -> max = 1
1|11|10 1 && 1 -> max = 2
11|11|0 1 && 1 -> max = 3
111|10| 1 && 0 -> max = 3
you can see, that although there are 4 1's you only do 3 comparisons, so your max will always be -1 of the actual max. You can fix this by adding mmax += 1 after your for loop.
Just a little bit of trace using small example will show why.
First, lets say there is only 1 '1' in your array.
Since you require both the current position and your next position to be '1', you will always get 0 for this case.
Let's say I have "11111". At the first '1', since next position is also '1', you increment count once. This repeats until 4th '1' and you increment your count 4 times in total so far. When you reach 5th '1', your next position is not '1', thus your count stops at 4.
In general, your method is like counting gaps between fingers, given 5 fingers, you get 4 gaps.
Side note: your code will fail for the case when there is no '1' in your array.
Ok some background
I have been working on this project, which I had started back in college, (no longer in school but want to expand on it to help me improve my understanding of C++). I digress... The problem is to find the Best path through a matrix. I generate a matrix filled with a set integer value lets say 9. I then create a path along the outer edge (Row 0, Col length-1) so that all values along it are 1.
The goal is that my program will run through all the possible paths and determine the best path. To simplify the problem I decide to just calculate the path SUM and then compare that to what the SUM computed by the application.
(The title is miss leading S=single-thread P=multi-threads)
OK so to my question.
In one section the algorithm does some simple bit-wise shifts to come up with the bounds for iteration. My question is how exactly do these shifts work so that the entire matrix (or MxN array) is completely traversed?
void AltitudeMapPath::bestPath(unsigned int threadCount, unsigned int threadIndex) {
unsigned int tempPathCode;
unsigned int toPathSum, toRow, toCol;
unsigned int fromPathSum, fromRow, fromCol;
Coordinates startCoord, endCoord, toCoord, fromCoord;
// To and From split matrix in half along the diagonal
unsigned int currentPathCode = threadIndex;
unsigned int maxPathCode = ((unsigned int)1 << (numRows - 1));
while (currentPathCode < maxPathCode) {
tempPathCode = currentPathCode;
// Setup to path iteration
startCoord = pathedMap(0, 0);
toPathSum = startCoord.z;
toRow = 0;
toCol = 0;
// Setup from path iteration
endCoord = pathedMap(numRows - 1, numCols - 1);
fromPathSum = endCoord.z;
fromRow = numRows - 1;
fromCol = numCols - 1;
for (unsigned int index = 0; index < numRows - 1; index++) {
if (tempPathCode % 2 == 0) {
toCol++;
fromCol--;
}
else {
toRow++;
fromRow--;
}
toCoord = pathedMap(toRow, toCol);
toPathSum += toCoord.z;
fromCoord = pathedMap(fromRow, fromCol);
fromPathSum += fromCoord.z;
tempPathCode = tempPathCode >> 1;
}
if (toPathSum < bestToPathSum[threadIndex][toRow]) {
bestToPathSum[threadIndex][toRow] = toPathSum;
bestToPathCode[threadIndex][toRow] = currentPathCode;
}
if (fromPathSum < bestFromPathSum[threadIndex][fromRow]) {
bestFromPathSum[threadIndex][fromRow] = fromPathSum;
bestFromPathCode[threadIndex][fromRow] = currentPathCode;
}
currentPathCode += threadCount;
}
}
I simplified the code since all the extra stuff just detracts from the question. Also if people are wondering I wrote most of the application but this idea of using the bit-wise operators was given to me by my past instructor.
Edit:
I added the entire algorithm for which each thread executes on. The entire project is still a work a progress but here is the source code for the whole thing if any one is interested [GITHUB]
A right bit shift is equivalent to dividing by 2 to the power of the number of bits shifted. IE 1 >> 2 = 1 / (2 ^ 2) = 1 / 4
A left bit shift is equivalent to multiplying by 2 to the power of the number of bits shifted. IE 1 << 2 = 1 * 2 ^ 2 = 1 * 4
I'm not entirely sure what that algorithm does and why it needs to multiply by 2^ (num rows - 1) and then progressively divide by 2.
I am trying to make a simple Arduino code for when a photocell reading is less than 900, it will add 1 to the CurrentNumber and display it on a 4 digit 7 segment display. The problem is that it wont stop adding one even though it's reading more than 1000.
void loop() {
photocellReading = analogRead(photocellPin);
Serial.print("Analog reading = ");
Serial.println(photocellReading); // the raw analog reading
photocellReading = 1023 - photocellReading;
if(photocellReading << 10){
CurrentNumber = CurrentNumber + 1;
}
displayNumber(CurrentNumber);
}
Your problem is in your if condition:
if(photocellReading << 10){
CurrentNumber = CurrentNumber + 1;
}
What you're essentially doing he is: shifting the bits of photocellReading to the left by 10 (which is equivalent to multiplying by 2^10 aka 1024). Most likely this means the only time this is ever going to be false is if the value of photocellReading was 0 to start with. (I say most likely because it depends on if the bits loop back around or not, but this is not entirely relevant).
tl;dr your code is conceptually equivalent to:
if((photocellReading * 1024) != 0){
CurrentNumber = CurrentNumber + 1;
}
What I'm guessing you wanted to do (considering you subtracted 1023, which coincidentally is 1024 - 1) is:
if(photocellReading < 1024){ // again 1024 == 2^10
CurrentNumber = CurrentNumber + 1;
}