We Keep Coding

Can't understand this why we use (S-X)/2

Given the sum and xor of two numbers X and Y, we need to find the numbers.
I came across a solution in GeeksForGeeks (https://www.geeksforgeeks.org/find-two-numbers-sum-xor/).
In this I don't understand why they did A=(S-X)/2 inside compute function. Why not trying to find from S directly? Can any one please explain that?
// Java program to find two numbers with
// given Sum and XOR such that value of
// first number is minimum.
class GFG {
// Function that takes in the sum and XOR
// of two numbers and generates the two
// numbers such that the value of X is
// minimized
static void compute(long S, long X)
{
long A = (S - X)/2;
int a = 0, b = 0;
final int LONG_FIELD_SIZE = 8;
// Traverse through all bits
for (int i=0; i<8*LONG_FIELD_SIZE; i++)
{
long Xi = (X & (1 << i));
long Ai = (A & (1 << i));
if (Xi == 0 && Ai == 0)
{
// Let us leave bits as 0.
}
else if (Xi == 0 && Ai > 0)
{
a = ((1 << i) | a);
b = ((1 << i) | b);
}
else if (Xi > 0 && Ai == 0)
{
a = ((1 << i) | a);
// We leave i-th bit of b as 0.
}
else // (Xi == 1 && Ai == 1)
{
System.out.println("Not Possible");
return;
}
}
System.out.println("a = " + a +"\nb = " + b);
}
// Driver function
public static void main(String[] args) {
long S = 17, X = 13;
compute(S, X);
}
}

Each substring of a certian length of a binary substring should have at least one '1' character

You have been given a binary string containing only the characters '1' and '0'.
Calculate how many characters of the string need to be changed in order to make the binary string such that each of its substrings of at least a certain length contains at least one "1" character.
I came to think of the following idea but it fails for many testcases:
public static int minimumMoves(String s, int d) {
int n = s.length();
int i=0, answer = 0;
while(i<n)
{
boolean hasOne = false;
int j=i;
while(j<n && j<i+d)
{
if(s.charAt(j) == '1')
{
hasOne = true;
break;
}
j++;
}
if(!hasOne) {
answer++;
i += d;
}
else i++;
}
return answer;
}
Also my algorithm runs on O(|s|2) time. Can anyone suggest ideas on O(|s|) time?

Just throwing off an idea:
return s.split("(?<=\\G.{" + String.valueof(d) + "})").stream().filter(str -> str.contains("1")).count()

You just need to break ensure there is no run of d zeros.
public static int minimumMoves(String s, int d) {
int result = 0;
int runLength = 0;
for(char c: s.toCharArray()) {
if (c == '0') {
runLength += 1;
if (runLength == d) { // we need to break this run
result += 1;
runLength = 0;
}
} else {
runLength = 0;
}
}
return result;
}

I used the sliding window technique and Deque to solve this. This is my accepted solution:
public static int minimumMoves(String s, int d) {
int n = s.length();
Deque<Character> dq = new LinkedList<>();
int count = 0, answer = 0;
for(int i=0; i<d; i++)
{
if(s.charAt(i) == '1') count++;
dq.addLast(s.charAt(i));
}
if(count == 0) {
answer++;
count++;
dq.removeLast();
dq.addLast('1');
}
int i=d;
while(i<n)
{
if(dq.getFirst() == '1') count--;
dq.removeFirst();
if(s.charAt(i) == '1') count++;
dq.addLast(s.charAt(i));
if(count == 0)
{
answer++;
dq.removeLast();
dq.addLast('1');
count++;
}
i++;
}
return answer;
}

You just need to use a sliding window and a count of 1s so far at each index. Use a sliding window of d and if you don't see any ones so far, update the last index of that window with 1 and increment the result.
Code below:
public static int minimumMoves(String s, int d) {
int n = s.length();
int[] count = new int[n+1];
int res = 0;
for ( int i = 1; i <= d; i++ ) {
if ( s.charAt(i-1) == '1') count[i] = count[i-1]+1;
else count[i] = count[i-1];
}
if ( count[d] == 0 ) {
res++;
count[d] = 1;
}
for ( int i = d+1; i <= n; i++ ) {
if ( s.charAt(i-1) == '0' ) {
count[i] = count[i-1];
int ones = count[i] - count[i-d];
if ( ones == 0 ) {
count[i] = count[i-1] + 1;
res++;
}
} else {
count[i] = count[i-1] + 1;
}
}
return res;
}

Thought of another implementation you can do for this by working from the maximum possible changes (assumes at start that all values are '0' in String), reduce it when it finds a '1' value, and then jump to the next substring start. This allows it to run in O(n) and Ω(n/m) (n = String length, m = Substring length).
public static int minimumMoves(String s, int d)
{
char[] a = s.toCharArray();
//Total possible changes (not counting tail)
if(a.length < d)
return 0;
int t = (int) a.length / d;
//Total possible changes (counting tail)
//int t = (int) Math.ceil((double) a.length / (double) d);
for(int i = 0; i < a.length; i++)
{
if(a[i] == '1')
{
t--;
//Calculate index for start of next substring
i = (i / d + 1) * d - 1;
}
}
return t;
}

I get java.lang.StringIndexOutOfBoundsException when translating c++ code into java code

I'm trying to solve this problem
https://vjudge.net/problem/UVALive-6805
I found solution but in c++ , Can anybody help me converting it to java code. I'm very newbie to programming
I tried a lot of solutions but non of them work.
Please I need help in this if possible
I don't know for example what is the equivalent for .erase function in c++ in java
Also is is sbstr in c++ provide different result from java ?
#include <iostream>
#include <string>
using namespace std;
int syllable(string word)
{
int L = word.size();
int syllable;
if (L>=7)
{
syllable = 3;
}
else if (L==6)
{
int indicator = 0;
for (int k=0; k<=L-2; k++)
{
string subword = word.substr(k, 2);
if (subword == "ng" || subword == "ny")
{
indicator++;
}
}
if (indicator == 0)
{
syllable = 3;
}
else
{
syllable = 2;
}
}
else if (L == 4 || L == 5)
{
syllable = 2;
}
else if (L == 3)
{
char Char = word[0];
if (Char=='a' || Char=='A' || Char=='e' || Char=='E' || Char=='i' || Char=='I' || Char=='o' || Char=='O' || Char=='u' || Char=='U')
{
syllable = 2;
}
else
{
syllable = 1;
}
}
else
{
syllable = 1;
}
return syllable;
}
int main()
{
string word;
int T;
cin >> T;
for (int i=1; i<=T; i++)
{
int syl[] = {0, -1, -2, -3};
string rhy[] = {"a", "b", "c", "d"};
int verse = 0;
int stop = 0;
while (stop == 0)
{
cin >> word;
int L = word.size();
char end = word[L-1];
if (end == '.')
{
stop = 1;
}
if (word[L-1] == ',' || word[L-1] == '.')
{
word = word.erase(L-1, 1);
L = word.size();
}
if (verse<=3)
{
syl[verse] = syl[verse] + syllable(word);
}
if (end == ',' || end == '.')
{
if (verse<=3)
{
rhy[verse] = word.substr(L-2, 2);
}
verse++;
if (verse<=3)
{
syl[verse] = 0;
}
}
}
int A = 0, B = 0, C = 0, D = 0;
for (int k=0; k<4; k++)
{
if (syl[k] >= 8 && syl[k] <= 12)
{
A = A + 10;
}
}
for (int k=0; k<2; k++)
{
if (rhy[k] == rhy[k+2])
{
B = B + 20;
}
}
for (int k=0; k<2; k++)
{
if (syl[k] == syl[k+2])
{
C = C + 10;
}
}
if (verse > 4)
{
D = (verse - 4) * 10;
}
int E = A + B + C - D;
cout << "Case #" << i << ": " << A << " " << B << " " << C << " " << D << " " << E << endl;
}
}
here is my trying
import java.util.*;
public class First {
public static int syllable(String word) {
int L = word.length();
int syllable;
if (L >= 7) {
syllable = 3;
} else if (L == 6) {
int indicator = 0;
for (int k = 0; k < L - 3; k++) {
String subword = word.substring(k, 2);
if (subword == "ng" || subword == "ny") {
indicator++;
}
}
if (indicator == 0) {
syllable = 3;
} else {
syllable = 2;
}
} else if (L == 4 || L == 5) {
syllable = 2;
} else if (L == 3) {
char Char = word.charAt(0);
if (Char == 'a' || Char == 'A' || Char == 'e' || Char == 'E' || Char == 'i' || Char == 'I' || Char == 'o'
|| Char == 'O' || Char == 'u' || Char == 'U') {
syllable = 2;
} else {
syllable = 1;
}
} else {
syllable = 1;
}
return syllable;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String word;
int T;
T = sc.nextInt();
for (int i = 1; i <= T; i++) {
int syl[] = { 0, -1, -2, -3 };
String rhy[] = { "a", "b", "c", "d" };
int verse = 0;
int stop = 0;
while (stop == 0) {
word = sc.next();
int L = word.length();
char end = word.charAt(L-1);
if (end == '.') {
stop = 1;
}
if (word.charAt(L-1) == ',' || word.charAt(L-1) == '.') {
word.substring(L-1, 1);
L = word.length();
}
if (verse <= 3) {
syl[verse] = syl[verse] + syllable(word);
}
if (end == ',' || end == '.') {
if (verse <= 3) {
rhy[verse] = word.substring(L - 2, 2);
}
verse++;
if (verse <= 3) {
syl[verse] = 0;
}
}
}
int A = 0, B = 0, C = 0, D = 0;
for (int k = 0; k < 4; k++) {
if (syl[k] >= 8 && syl[k] <= 12) {
A = A + 10;
}
}
for (int k = 0; k < 2; k++) {
if (rhy[k] == rhy[k + 2]) {
B = B + 20;
}
}
for (int k = 0; k < 2; k++) {
if (syl[k] == syl[k + 2]) {
C = C + 10;
}
}
if (verse > 4) {
D = (verse - 4) * 10;
}
int E = A + B + C - D;
System.out.println("Case #" + i + ": " + A + " " + B + " " + C + " " + D + " " + E);
}
}
}

The Exception is thrown by your second and your third call of String substring method. Your beginIndex is higher than your endIndex. As you can see in here https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#substring(int,%20int) beginIndex always has to be lower than the endIndex.

Before answering your question, there are some important points to mention in regards to Strings and Java in general.
Strings are immutable (This also applies to C++). This means that no method called on a String will change it, and that all methods simply return new versions of the original String with the operations done on it
The substring method in java has two forms.
One takes in beginIndex and returns everything from beginIndex to str.length() - 1 (where str represents a String)
The other takes in the beginIndex, and the endIndex, and returns everything from beginIndex to endIndex - 1. The beginIndex should never be larger than endIndex otherwise it throws an IndexOutOfBoundsException
C++'s substring method (string::substr()) takes in the beginning "index" and takes in the number of characters after it to include in the substring. So by doing substr(L-2, 2) you get the last two characters of the string.
Java will never allow you to go out of bounds. That means you need to constantly check whether you are within the bounds of anything you are iterating through.
With all this in mind, I would go and verify that all of the substring() method calls are returning the proper range of characters, and that you are properly reassigning the values returned from substring() to the proper variable.
To mimic C++'s string::erase(), depending on what part of the word you want to erase, you want to get the substring of the part before and the substring of the part after it and add them together.
Ex. Lets say I have a String line = "I do not like the movies"; Since it is impossible for anyone to not like movies, we want to cut out the word not
We do this by doing what I said above
String before = line.substring(0, 5); // This gives us "I do " since it goes up to but not including the 5th index.
String after = line.substring(5 + 3); // This gives us the rest of the string starting after the word "not" because not is 3 characters long and this skips to the 3rd index after index 5 (or index 8)
line = before + after; // This'll add those two Strings together and give you "I do like the movies"
Hope this helps!

Can't figure out the error Luhn check

Its supose to tell me if a card is valid or invalid using luhn check
4388576018402626 invalid
4388576018410707 valid
but it keeps telling me that everything is invalid :/
Any tips on what to do, or where to look, would be amazing. I have been stuck for a few hours.
It would also help if people tell me any tips on how to find why a code is not working as intended.
im using eclipse and java
public class Task11 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Enter a credit card number as a long integer: ");
long number = input.nextLong();
if (isValid(number)) {
System.out.println(number + " is valid");
} else {
System.out.println(number + " is invalid");
}
}
public static boolean isValid(long number) {
return (getSize(number) >= 13) && (getSize(number) <= 16)
&& (prefixMatched(number, 4) || prefixMatched(number, 5) || prefixMatched(number, 6) || prefixMatched(number, 37))
&& (sumOfDoubleEvenPlace(number) + sumOfOddPlace(number)) % 10 == 0;
}
public static int sumOfDoubleEvenPlace(long number) {
int result = 0;
long start = 0;
String digits = Long.toString(number);
if ((digits.length() % 2) == 0) {
start = digits.length() - 1;
} else {
start = digits.length() - 2;
}
while (start != 0) {
result += (int) ((((start % 10) * 2) % 10) + (((start % 10) * 2) / 2));
start = start / 100;
}
return result;
}
public static int getDigit(int number) {
return number % 10 + (number / 10);
}
public static int sumOfOddPlace(long number) {
int result = 0;
while (number != 0) {
result += (int) (number % 10);
number = number / 100;
}
return result;
}
public static boolean prefixMatched(long number, int d) {
return getPrefix(number, getSize(d)) == d;
}
public static int getSize(long d) {
int numberOfDigits = 0;
String sizeString = Long.toString(d);
numberOfDigits = sizeString.length();
return numberOfDigits;
}
public static long getPrefix(long number, int k) {
String size = Long.toString(number);
if (size.length() <= k) {
return number;
} else {
return Long.parseLong(size.substring(0, k));
}
}
}

You should modiffy your isValid() method to write down when it doesn't work, like this:
public static boolean isValid(long number) {
System.err.println();
if(getSize(number) < 13){
System.out.println("Err: Number "+number+" is too short");
return false;
} else if (getSize(number) > 16){
public static boolean isValid(long number) {
System.err.println();
if(getSize(number) < 13){
System.out.println("Err: Number "+number+" is too short");
return false;
} else if (getSize(number) > 16){
System.out.println("Err: Number "+number+" is too long");
return false;
} else if (! (prefixMatched(number, 4) || prefixMatched(number, 5) || prefixMatched(number, 6) || prefixMatched(number, 37)) ){
System.out.println("Err: Number "+number+" prefix doesn't match");
return false;
} else if( (sumOfDoubleEvenPlace(number) + sumOfOddPlace(number)) % 10 != 0){
System.out.println("Err: Number "+number+" doesn't have sum of odd and evens % 10. ");
return false;
}
return true;
}
My guess for your problem is on the getPrefix() method, you should add some logs here too.
EDIT: so, got more time to help you (don't know if it's still necessary but anyway). Also, I corrected the method I wrote, there were some errors (like, the opposite of getSize(number) >= 13 is getSize(number) < 13)...
First it will be faster to test with a set of data instead of entering the values each time yourself (add the values you want to check):
public static void main(String[] args) {
long[] luhnCheckSet = {
0, // too short
1111111111111111111L, // too long (19)
222222222222222l // prefix doesn't match
4388576018402626l, // should work ?
};
//System.out.print("Enter a credit card number as a long integer: ");
//long number = input.nextLong();
for(long number : luhnCheckSet){
System.out.println("Checking number: "+number);
if (isValid(number)) {
System.out.println(number + " is valid");
} else {
System.out.println(number + " is invalid");
}
System.out.println("-");
}
}
I don't know the details of this, but I think you should work with String all along, and parse to long only if needed (if number is more than 19 characters, it might not parse it long).
Still, going with longs.
I detailed your getPrefix() with more logs AND put the d in parameter in long (it's good habit to be carefull what primitive types you compare):
public static boolean prefixMatched(long number, long d) {
int prefixSize = getSize(d);
long numberPrefix = getPrefix(number, prefixSize);
System.out.println("Testing prefix of size "+prefixSize+" from number: "+number+". Prefix is: "+numberPrefix+", should be:"+d+", are they equals ? "+(numberPrefix == d));
return numberPrefix == d;
}
Still don't know what's wrong with this code, but it looks like it comes from the last test:
I didn't do it but you should make one method from sumOfDoubleEvenPlace(number) + sumOfOddPlace(number)) % 10 and log both numbers and the sum (like i did in prefixMatched() ). Add logs in both method to be sure it gets the result you want/ works like it should.
Have you used a debugger ? if you can, do it, it can be faster than adding a lot of logs !
Good luck

EDIT:
Here are the working functions and below I provided a shorter, more efficient solution too:
public class CreditCardValidation {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int count = 0;
long array[] = new long [16];
do
{
count = 0;
array = new long [16];
System.out.print("Enter your Credit Card Number : ");
long number = in.nextLong();
for (int i = 0; number != 0; i++) {
array[i] = number % 10;
number = number / 10;
count++;
}
}
while(count < 13);
if ((array[count - 1] == 4) || (array[count - 1] == 5) || (array[count- 1] == 3 && array[count - 2] == 7)){
if (isValid(array) == true) {
System.out.println("\n The Credit Card Number is Valid. ");
} else {
System.out.println("\n The Credit Card Number is Invalid. ");
}
} else{
System.out.println("\n The Credit Card Number is Invalid. ");
}
in.close();
}
public static boolean isValid(long[] array) {
int total = sumOfDoubleEvenPlace(array) + sumOfOddPlace(array);
if ((total % 10 == 0)) {
for (int i=0; i< array.length; i++){
System.out.println(array[i]);}
return true;
} else {
for (int i=0; i< array.length; i++){
System.out.println(array[i]);}
return false;
}
}
public static int getDigit(int number) {
if (number <= 9) {
return number;
} else {
int firstDigit = number % 10;
int secondDigit = (int) (number / 10);
return firstDigit + secondDigit;
}
}
public static int sumOfOddPlace(long[] array) {
int result = 0;
for (int i=0; i< array.length; i++)
{
while (array[i] > 0) {
result += (int) (array[i] % 10);
array[i] = array[i] / 100;
}
}
System.out.println("\n The sum of odd place is " + result);
return result;
}
public static int sumOfDoubleEvenPlace(long[] array) {
int result = 0;
long temp = 0;
for (int i=0; i< array.length; i++){
while (array[i] > 0) {
temp = array[i] % 100;
result += getDigit((int) (temp / 10) * 2);
array[i] = array[i] / 100;
}
}
System.out.println("\n The sum of double even place is " + result);
return result;
}
}
I also found a solution with less lines of logic. I know you're probably searching for an OO approach with functions, building from this could be of some help.
Similar question regarding error in Luhn algorithm logic:
Check Credit Card Validity using Luhn Algorithm
Link to shorter solution:
https://code.google.com/p/gnuc-credit-card-checker/source/browse/trunk/CCCheckerPro/src/com/gnuc/java/ccc/Luhn.java
And here I tested the solution with real CC numbers:
public class CreditCardValidation{
public static boolean Check(String ccNumber)
{
int sum = 0;
boolean alternate = false;
for (int i = ccNumber.length() - 1; i >= 0; i--)
{
int n = Integer.parseInt(ccNumber.substring(i, i + 1));
if (alternate)
{
n *= 2;
if (n > 9)
{
n = (n % 10) + 1;
}
}
sum += n;
alternate = !alternate;
}
return (sum % 10 == 0);
}
public static void main(String[] args){
//String num = "REPLACE WITH VALID NUMBER"; //Valid
String num = REPLACE WITH INVALID NUMBER; //Invalid
num = num.trim();
if(Check(num)){
System.out.println("Valid");
}
else
System.out.println("Invalid");
//Check();
}
}

Does my pseudocode make sense? [duplicate]

This question already has an answer here:
Can you quickly tell me if this pseudocode makes sense or not?
(1 answer)
Closed 9 years ago.
I believe my code is now foolproof. I will write up the pseudocode now. But I do have one question. Why does DRJava ask that I return something outside of my if statements? As you can see I wrote for ex: "return 1;" just because it asked. It will never return that value however. Can someone explain this to me?
public class assignment1question2test {
public static void main(String[] args) {
int[] a = new int[1];
int l = 0;
int r = a.length-1;
for(int i=0; i<=r; i++) {
a[i] = 1;
}
a[0] = 10;
for (int i=0; i<=r; i++) {
System.out.println(a[i]);
}
System.out.print(recursiveSearch(a,l,r));
}
public static int recursiveSearch (int[] a, int l, int r) {
int third1 = (r-l)/3 + l;
int third2 = third1*2 - l + 1;
if (r-l == 0) {
return l;
}
System.out.println("i will be checking compare from " + l + " to " + third1 + " and " + (third1 + 1) + " to " + third2);
int compareResult = compare(a,l,third1,third1 + 1, third2);
if(r-l == 1) {
if (compareResult == 1) {
return l;
}
else {
return r;
}
}
if (compareResult == 0) {
return recursiveSearch(a,third2 + 1, r);
}
if (compareResult == 1) {
return recursiveSearch(a,l,third1);
}
if (compareResult == -1) {
return recursiveSearch(a,third1 + 1, third2);
}
return 1;
}
public static int compare(int[] a, int i, int j, int k, int l) {
int count1 = 0;
int count2 = 0;
for(int g=i; g<=j; g++) {
count1 = count1 + a[g];
}
for(int g=k; g<=l; g++) {
count2 = count2 + a[g];
}
if (count1 == count2) {
return 0;
}
if (count1 > count2) {
return 1;
}
if (count1 < count2) {
return -1;
}
return 0;
}
}
FINAL PSEUDOCODE I THINK
Algorithm: recursiveSearch (a,l,r)
Inputs: An array a, indices l and r which delimit the part of interest.
Output: The index that has the lead coin.
int third1 ← (r - l)/3
int third2 ← third1*2 - l + 1
if (r-l = 0) then
return l
int compareResult ← compare(a,l,third1,third1 + 1,third2)
if (r-l = 1) then
if (compareResult = 1) then
return l
else
return r
if (compareResult = 0) then
return recursiveSearch(a, third2 + 1, r)
if (compareResult = "1") then
return recursiveSearch(a,l,third1)
if (compareResult = "-1") then
return recursiveSearch(a,third1 + 1,third2)

String compareResult ← compare(a,l,mid,mid,r)
Here you check the middle element twice, make it:
String compareResult ← compare(a,l,mid,mid+1,r)
Apart from that your algorithm seems fair enough to me.

You should refine you logic more, it doesn't consider the case where the number of coin is even.
Odd: take a coin out, divide the remaining into 2 equal half and do the comparison.
Even: divide the remaining into 2 equal half and do the comparison.
For recursive function, please also define the base case:
When n=1, return the coin.
When n=2, return the heavier coin.
NumberOfCoin = r-l+1
if (NumberOfCoin = 1)
return l;
if (NumberOfCoin = 2)
compare(a,l,l,r,r)
0: Think it yourself
-1: Think it yourself
1: Think it yourself
if (NumberOfCoin is odd number)
mid = Think it yourself
compare(a, l, mid-1, mid+1, r)
0: Think it yourself
-1: Think it yourself
1: Think it yourself
if (NumberOfCoin is even number)
mid = l+r/2
compare(a, l, mid, mid+1, r)
0: Think it yourself
-1: Think it yourself
1: Think it yourself