[Solved] Why is my string getting repeatedly in this function?

Your for(var j=0; j < accentArray.length; j++) is meaningless. You are not using the j variable anywhere. You just make the inner codes run accentArray.length times. If you want to check if the string includes an accent, you could write if ([…str].some(c => accentArray.includes(c))) There is a better way to remove characters from a string: … Read more

[Solved] Ruby Program to print pattern [closed]

You can use collection, map or each for this data = {a: [1,2,3,4,5,6,7,8,9],b: [1,2,3,4,5,6],c: [2,3,4,5,6,7]} data.map{|k,v| (1..9).map{|a| data[k].include?(a) ? k.to_s.upcase() +a.to_s : ‘ ‘}} 9 solved Ruby Program to print pattern [closed]

[Solved] Check if array Is235

Try this, I tested it and it works, you should use the remainder operator %: public class Array { public static void main(String[] args) { int[] arr = {2, 3, 5, 7, 11}; System.out.println(is235Array(arr)); } public static int is235Array(int[] a) { int countOne = 0; int countTwo = 0; for (int i : a) { … Read more

[Solved] format a 12 digit number in XXXX XXXX XXXX format using javascript

You can use selectionEnd to control the position of cursor var target = document.querySelector(‘.creditCardText’); target.addEventListener(“keyup”, function() { var position = target.selectionStart; var prevVal = target.value; var newVal = prevVal.split(” “).join(“”); // remove spaces if (newVal.length > 0) { newVal = newVal.match(new RegExp(‘.{1,4}’, ‘g’)).join(” “); } target.value = newVal; for (var i = 0; i < … Read more

[Solved] How to transform the digits of a number?

Your logic of result*=devided; is wrong. You need to add the digit(multiplied by it’s 10’s place) to the result. Here is the solution to your problem. int function (int *n, int i){ int temp=n[i]; //tepmorary veraible of the passed number int result=0; int mult=1; int devided; while (temp!=0){ devided=temp%10;//get the last number example 123 = … Read more

[Solved] Sorting Pandas data by hour of the day

after converting to datetime pd.to_datetime(df[‘date’]) you can create a separate column with the hour in it, e.g. df[‘Hour’] = df.date.dt.hour and then sort by it df.sort_values(‘Hour’) EDIT: As you want to sort by time you can instead of using hour, put the timestamp part into a ‘time’ column. In order to get times between 9 … Read more