i am using image picker package in flutter , so i want to mention the max size of image selected should be 5 MB, in the package there is a parameter named imagequality which take 0-100, but don't equivalent for 5 mb
getImageFromGallery() async {
try {
var pickedfiles = await imagePicker.pickMultiImage(imageQuality: 5);
if (pickedfiles.isNotEmpty) {
for (int i = 0; i < pickedfiles.length; i++) {
previewList.add(pickedfiles[i].path);
}
setState(() {});
}
} catch (e) {
log(e.toString());
}
}
Try this, I hope this helps
selectImagesFromImagePicker() async{
// e.g
final knowSize = await defindSize(fileSelected, 2);
if (double.parse(knowSize.toString()) > 5.0) {
SomeSnackBar("File too large", "Limit is 5 mb below");
} else {
log("$knowSize mb", name: "FILE SIZE");
filesChosen.add(File(x!));
}
}
// Function to know the size of a certain file
defindSize(String? path, int decimals) {
final file = File(path!);
int bytes = file.lengthSync();
log("lengthSync $bytes", name: "FILE SIZE");
double sizeMb = bytes / (1000 * 1000);
// the return is mb default
return sizeMb.toStringAsFixed(decimals);
// below if you wanted to return suffix just uncomment them
// const suffixes = ["b", "kb", "mb", "gb", "tb"];
// var i = (m.log(bytes) / m.log(1024)).floor();
// return ((bytes / m.pow(1024, i)).toStringAsFixed(decimals)) + suffixes[i];
}
I am sending some values from a microcontroller to a flutter app in the form of Uint8List then I convert it to. When I send values individually, I have no issue at all. however, when I send 100 values together or any number of values continuously, I get some values correct and others divided in multiple lines. I will be providing a screenshot for the result because its hard to represent it with typing.
This is the code tor receiving:
Future<void> _onDataReceived( Uint8List data ) async {
var fixedList = new List<String>.filled(6, '', growable: true);
// var fixedList1 = <Uint8List>[];
var fixed1 = new List<int>.filled(7, 0, growable: true);
// fixedList1.add(data);
if (data.length >= 7) {
for (int k = 0; k < 7; k++) {
// int x=0;
fixed1[k] = data[k+1];
l++;
}
if (l > 6) {
for (int i = 0; i <= 5; i++) {
fixedList[i] = String.fromCharCode(fixed1[i]);
fixed1[i]=0;
}
l = 0;
fixed1.clear();
print(fixedList); //fixedList1.clear();
}
}
}
Anyone knows what is going on here?
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 :)
When i read data from a big txt file block by block ,I got the error as blow:
Unfinished UTF-8 octet sequence (at offset 4096)
code:
File file = File(path!);
RandomAccessFile _raf = await file.open();
_raf.setPositionSync(skip ?? 0);
var data = _raf.readSync(block);// block = 64*64
content.value = utf8.decode(data.toList());
UTF*8 is variable length encoding.
The error come from data not align to UTF8 boundary
Alternative way is to trim data byte on left and right before call utf.decode
This will lost first and last character. You may read and add more bytes to cover last character and align with utf8 boundary
bool isDataByte(int i) {
return i & 0xc0 == 0x80;
}
Future<void> main(List<String> arguments) async {
var _raf = await File('utf8.txt').open();
_raf.setPositionSync(skip);
var data = _raf.readSync(8 * 8);
var utfData = data.toList();
int l, r;
for (l = 0; isDataByte(utfData[l]) && l < utfData.length; l++) {}
for (r = utfData.length - 1; isDataByte(utfData[r]) && r > l; r--) {}
var value = utf8.decode(utfData.sublist(l, r));
print(value);
}
Optional read more 4 bytes and expand to cover last character
bool isDataByte(int i) {
return i & 0xc0 == 0x80;
}
Future<void> main(List<String> arguments) async {
var _raf = await File('utf8.txt').open();
_raf.setPositionSync(skip);
var block = 8 * 8;
var data = _raf.readSync(block + 4);
var utfData = data.toList();
int l, r;
for (l = 0; isDataByte(utfData[l]) && l < block; l++) {}
for (r = block; isDataByte(utfData[r]) && r < block + 4; r++) {}
var value = utf8.decode(utfData.sublist(l, r));
print(value);
}
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");
}
}