This is a function if the endValueFixed is equal for example 12.0 I want to print the number without zero so I want it to be 12.
void calculateIncrease() {
setState(() {
primerResult = (startingValue * percentage) / 100;
endValue = startingValue + primerResult;
endValueFixe`enter code here`d = roundDouble(endValue, 2);
});
}
This may be an overkill but it works exactly as you wish:
void main() {
// This is your double value
final end = 98.04;
String intPart = "";
String doublePart = "";
int j = 0;
for (int i = 0; i < end.toString().length; i++) {
if (end.toString()[i] != '.') {
intPart += end.toString()[i];
} else {
j = i + 1;
break;
}
}
for (int l = j; l < end.toString().length; l++) {
doublePart += end.toString()[l];
}
if (doublePart[0] == "0" && doublePart[1] != "0") {
print(end);
} else {
print(end.toString());
}
}
You may use this code as a function and send whatever value to end.
if (endValueFixed==12) {
print('${endValueFixed.toInt()}');
}
conditionally cast it to an int and print it then :)
I implemented select sort in Flutter and in SwiftUI. I made the implementations as similar as possible.
Swift:
func selectSort(list: inout [Double]) -> [Double] {
for i in 0..<list.count {
var minElPos = i;
for j in (minElPos + 1)..<list.count {
if list[j] < list[minElPos] {
minElPos = j;
}
}
// swap
let temp = list[i];
list[i] = list[minElPos];
list[minElPos] = temp;
}
return list;
}
// Measuring time
func generateRandomList(size: Int) -> [Double] {
var res = Array<Double>(repeating: 0.0, count: size)
for i in 0..<size {
res[i] = Double.random(in: 0...1)
}
return res;
}
var arrayToTest: [Double] = generateRandomList(size: 8000);
let startingPoint = Date()
selectSort(list: &arrayToTest);
let time = startingPoint.timeIntervalSinceNow * -1;
Flutter:
class SelectSort {
static List<double> call(List<double> list) {
for(int i = 0; i < list.length - 1; i++) {
int minElPos = i;
for(int j = minElPos + 1; j < list.length; j++) {
if(list[j] < list[minElPos]) {
minElPos = j;
}
}
// swap
double temp = list[i];
list[i] = list[minElPos];
list[minElPos] = temp;
}
return list;
}
}
// Measuring time
class Utils {
static List<double> generateRandomList(int nbOfElements) {
var random = new Random();
List<double> res = List(nbOfElements);
for (var i = 0; i < nbOfElements; i++) {
res[i] = random.nextDouble();
}
return res;
}
}
List<double> arrayToTest = Utils.generateRandomList(8000);
final stopwatch = Stopwatch()..start();
SelectSort.call(arrayToTest);
stopwatch.stop();
int time = stopwatch.elapsedMilliseconds;
I measured the execution time for an array of random numbers. The array size is 8000. Flutter needs 0.053s and SwiftUI needs 0.141s. Does anyone have a clue why flutter as a hybrid framework has better performance than a native solution?
Both apps were run in release mode on a physical device.
I have to make a connect-four game using scala. I have attached the code but everytime the game runs and gets to row 3 it just continues to change the second rows entry instead of going to the next row. Any help would be appreciated. I found this code on another thread on here and couldn't figure out how to get it to work:
// makes the board
val table = Array.fill(9,8)('-')
var i = 0;
while(i < 8){
table(8)(i) = (i+'0').toChar
i = i+1;
}
// prints starting board
def printBoard(table: Array[Array[Char]]) {
table.foreach( x => println(x.mkString(" ")))
}
//player 1 moves
def playerMove1(){
val move = readInt
var currentRow1 = 7
while (currentRow1 >= 0)
if (table(currentRow1)(move) != ('-')) {
currentRow1 = (currentRow1-1)
table(currentRow1)(move) = ('X')
return (player2)}
} else {
table(currentRow1)(move) = ('X')
return (player2)
}
}
//player 2 moves
def playerMove2(){
val move = readInt
var currentRow2 = 7
while (currentRow2 >= 0)
if (table(currentRow2)(move) != ('-')) {
currentRow2 = (currentRow2-1)
table(currentRow2)(move) = ('O')
return (player1)}
} else {
table(currentRow2)(move) = ('O')
return (player1)
}
}
//player 1
def player1(){
printBoard(table)
println("Player 1 it is your turn. Choose a column 0-7")
playerMove1()
}
//player 2
def player2(){
printBoard(table)
println("Player 2 it is your turn. Choose a column 0-7")
playerMove2()
}
for (turn <- 1 to 32){
player1
player2
}
I've tried to make your code readable and compiling and also tried to fix some logic.
However, I've never worked with Scala so this is just a first sketch where you might want to continue ...
Some functions can be merged and the currentRow needed a fix. See here:
object ConnectFour{
val table = Array.fill(9,8)('-')
val currentRow = Array.fill(8)(8)
def main(args: Array[String]) {
var i = 0;
while(i < 8) {
table(8)(i) = (i+'0').toChar
i = i+1;
}
player(1)
}
def printBoard(table: Array[Array[Char]]) {
table.foreach( x => println(x.mkString(" ")))
}
def player(playerNr : Int){
printBoard(table)
println("Player " + playerNr + " it is your turn. Choose a column 0-7")
var column = readAndVerifyInt
var nextUser = 1 : Int
var symbol = 'O' : Char
if(playerNr == 1) {
symbol = 'X'
nextUser = 2
}
var curR = currentRow(column)
while (curR >= 0) {
if (table(curR)(column) != ('-')) {
curR = curR-1
currentRow(column) = curR
}
table(curR)(column) = symbol
player(nextUser)
}
}
def readAndVerifyInt() : Int = {
var column = readInt
if (column >= 0 && column <= 7) {
return column
} else {
println(" > Please try again")
return readAndVerifyInt
}
}
}
I believe this can be achieved by counting the instances for each character in that string. Even if a single character in that string is repeated at least twice, we can declare that string as a palindrome.
For example: bbcccc can be rewritten as bccccb or ccbbcc.
edified can be rewritten as deified.
Some book mentioned we should be using hash table. I think we can just use a list and check for the character count.
Do you think the logic is correct?
Yes, the main idea is to count the times of each char existing in the string. And it will be true if the string has at most one char occurs odd times and all others even times.
For example:
aabbcc => acbbca
aabcc => acbca
aabbb => abbba
No. You don't have to use a hash map (as some of the other answers suggest). But the efficiency of the solution will be determined by the algorithm you use.
Here is a solution that only tracks odd characters. If we get 2 odds, we know it can't be a scrambled palindrome. I use an array to track the odd count. I reuse the array index 0 over and over until I find an odd. Then I use array index 1. If I find 2 odds, return false!
Solution without a hash map in javascript:
function isScrambledPalindrome(input) {
// TODO: Add error handling code.
var a = input.split("").sort();
var char, nextChar = "";
var charCount = [ 0 ];
var charIdx = 0;
for ( var i = 0; i < a.length; ++i) {
char = a[i];
nextChar = a[i + 1] || "";
charCount[charIdx]++;
if (char !== nextChar) {
if (charCount[charIdx] % 2 === 1) {
if (charCount.length > 1) {
// A scrambled palindrome can only have 1 odd char count.
return false;
}
charIdx = 1;
charCount.push(0);
} else if (charCount[charIdx] % 2 === 0) {
charCount[charIdx] = 0;
}
}
}
return true;
}
console.log("abc: " + isScrambledPalindrome("abc")); // false
console.log("aabbcd: " + isScrambledPalindrome("aabbcd")); // false
console.log("aabbb: " + isScrambledPalindrome("aabbb")); // true
console.log("a: " + isScrambledPalindrome("a")); // true
Using a hash map, I found a cool way to only track the odd character counts and still determine the answer.
Fun javascript hash map solution:
function isScrambledPalindrome( input ) {
var chars = {};
input.split("").forEach(function(char) {
if (chars[char]) {
delete chars[char]
} else {
chars[char] = "odd" }
});
return (Object.keys(chars).length <= 1);
}
isScrambledPalindrome("aba"); // true
isScrambledPalindrome("abba"); // true
isScrambledPalindrome("abca"); // false
Any string can be palindrome only if at most one character occur odd no. of times and all other characters must occur even number of times.
The following program can be used to check whether a palindrome can be string or not.
vector<int> vec(256,0); //Vector for all ASCII characters present.
for(int i=0;i<s.length();++i)
{
vec[s[i]-'a']++;
}
int odd_count=0,flag=0;
for(int i=0;i<vec.size();++i)
{
if(vec[i]%2!=0)
odd_count++;
if(odd_count>1)
{
flag=1;
cout<<"Can't be palindrome"<<endl;
break;
}
}
if(flag==0)
cout<<"Yes can be palindrome"<<endl;
My code check if can it is palindrome or can be manipulated to Palindrome
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
//Tested on windows 64 bit arhc by using cygwin64 and GCC
bool isPalindrome (char *text);
int main()
{
char text[100]; // it could be N with defining N
bool isPal,isPosPal = false;
printf("Give me a string to test if it is Anagram of Palindrome\n");
gets(text);
isPal = isPalindrome(text);
isPosPal = isAnagramOfPalindrome(text);
if(isPal == false)
{
printf("Not a palindrome.\n");
}
else
{
printf("Palindrome.\n");
}
if(isPosPal == false)
{
printf("Not Anagram of Palindrome\n");
}
else
{
printf("Anagram of Palindrome\n");
}
return 0;
}
bool isPalindrome (char *text) {
int begin, middle, end, length = 0;
length = getLength(text);
end = length - 1;
middle = length/2;
for (begin = 0; begin < middle; begin++)
{
if (text[begin] != text[end])
{
return false;
}
end--;
}
if (begin == middle)
return true;
}
int getLength (char *text) {
int length = 0;
while (text[length] != '\0')
length++;
printf("length: %d\n",length);
return length;
}
int isAnagramOfPalindrome (char *text) {
int length = getLength(text);
int i = 0,j=0;
bool arr[26] = {false};
int counter = 0;
//char string[100]="neveroddoreven";
int a;
for (i = 0; i < length; i++)
{
a = text[i];
a = a-97;
if(arr[a])
{
arr[a] = false;
}
else
{
arr[a] = true;
}
}
for(j = 0; j < 27 ; j++)
{
if (arr[a] == true)
{
counter++;
}
}
printf("counter: %d\n",counter);
if(counter > 1)
{
return false;
}
else if(counter == 1)
{
if(length % 2 == 0)
return false;
else
return true;
}
else if(counter == 0)
{
return true;
}
}
as others have posted, the idea is to have each character occur an even number of times for an even length string, and one character an odd number of times for an odd length string.
The reason the books suggest using a hash table is due to execution time. It is an O(1) operation to insert into / retrieve from a hash map. Yes a list can be used but the execution time will be slightly slower as the sorting of the list will be O(N log N) time.
Pseudo code for a list implementation would be:
sortedList = unsortedList.sort;
bool oddCharFound = false;
//if language does not permit nullable char then initialise
//current char to first element, initialise count to 1 and loop from i=1
currentChar = null;
currentCharCount = 0;
for (int i=0; i <= sortedList.Length; i++) //start from first element go one past end of list
{
if(i == sortedList.Length
|| sortedList[i] != currentChar)
{
if(currentCharCount % 2 = 1)
{
//check if breaks rule
if((sortedList.Length % 2 = 1 && oddCharFound)
|| oddCharFound)
{
return false;
}
else
{
oddCharFound = true;
}
}
if(i!= sortedList.Length)
{
currentCharCount = 1;
currentChar = sortedList[i];
}
}
else
{
currentCharCount++;
}
}
return true;
Here is a simple solution using an array; no sort needed
#include <stdio.h>
#include <stdlib.h>
int main()
{
int a[256] = { 0 };
unsigned char i[] = {"aaBcBccc"};
unsigned char *p = &i[0];
int c = 0;
int j;
int flag = 0;
while (*p != 0)
{
a[*p]++;
p++;
}
for(j=0; j<256; j++)
{
if(a[j] & 1)
{
c++;
if(c > 1)
{
flag = 1;
break;
}
}
}
if(flag)
printf("Nope\n");
else
printf("yup\n");
return 0;
}
C#:
bool ok = s.GroupBy(c => c).Select(g => g.Count()).Where(c => c == 1).Count() < 2;
This solution, however, does use hashing.
Assuming all input characters are lower case letters.
#include<stdio.h>
int main(){
char *str;
char arr[27];
int j;
int a;
j = 0;
printf("Enter the string : ");
scanf("%s", str);
while (*str != '\0'){
a = *str;
a = a%27;
if(arr[a] == *str){
arr[a]=0;
j--;
}else{
arr[a] = *str;
j++;
}
*str++;
}
if(j==0 || j== -1 || j==1){
printf ("\nThe string can be a palindrome\n");
}
}
I have the method below, which in my Blackjack app will get the value of the hand which is an NSMutableArray. The problem here is that when 2 Ace's are in a hand, it should be a 12, but because it counts Ace's as 11, it results in being 22, which then makes lowValue returned.
How can I make it so that I can check and see if the for loop has already found an Ace and if it finds another, makes the next Ace worth only 1, not 11?
Thanks!
int getHandValue(NSMutableArray *hand) {
int lowValue = 0;
int highValue = 0;
for (KCCard *aCard in hand) {
if (aCard.value == Ace) {
lowValue+= 1;
highValue+= 11;
} else if (aCard.value == Jack || aCard.value == Queen || aCard.value == King) {
lowValue += 10;
highValue += 10;
} else {
lowValue += aCard.value;
highValue += aCard.value;
}
}
return (highValue > 21) ? lowValue : highValue;
}
Perhaps you can add a boolean value before the for loop setting it initially to NO. When an Ace is found then you can break from the for loop after setting the boolean to YES, that way if you encounter another Ace && your boolean value == YES you can handle the case accordingly.
int getHandValue(NSMutableArray *hand) {
int lowValue = 0;
int highValue = 0;
BOOL isFoundAce = NO;
for (KCCard *aCard in hand) {
if (aCard.value == Ace) {
if (isFoundAce) {
lowValue+= 1;
highValue+= 1;
}
else {
lowValue+= 1;
highValue+= 11;
isFoundAce= YES;
}
} else if (aCard.value == Jack || aCard.value == Queen || aCard.value == King) {
lowValue += 10;
highValue += 10;
} else {
lowValue += aCard.value;
highValue += aCard.value;
}
}
return (highValue > 21) ? lowValue : highValue;
}
My example without a redundant code from zsxwing's example:
int getHandValue(NSMutableArray *hand) {
int cardValue = 0;
int aceCount = 0;
for (KCCard *aCard in hand) {
if (aCard.value == Ace) {
aceCount++;
cardValue += 11;
} else if (aCard.value == Jack || aCard.value == Queen || aCard.value == King) {
cardValue += 10;
} else {
cardValue += aCard.value;
}
}
while ((cardValue > 21) && (aceCount > 0)) {
cardValue -= 10;
aceCount--;
}
return cardValue;
}