Jul 13, 2022 · Jun 27, 2022 · Jun 28, 2022 · Jun 29, 2022 · Jun 29, 2022 · Jun 29, 2022
diff --git a/1-js/02-first-steps/14-function-basics/1-if-else-required/solution.md b/1-js/02-first-steps/14-function-basics/1-if-else-required/solution.md
 No difference.
 Abi funkcijos veikia vienodai, skirtumų nėra.
diff --git a/1-js/02-first-steps/14-function-basics/1-if-else-required/task.md b/1-js/02-first-steps/14-function-basics/1-if-else-required/task.md

 ---

 #Is "else"required?
 #Ar "else"yra privalomas?

 The following function returns`true` if the parameter`age`is greater than `18`.
 Ši funkcija grąžina`true`, jei parametras`age`yra didesnis nei `18`.

 Otherwise it asks for a confirmation and returns its result:
 Priešingu atveju ji prašo patvirtinimo per `confirm` ir grąžina rezultatą:

 ```js
 function checkAge(age) {
 *!*
  } else {
    // ...
    return confirm('Did parents allow you?');
    return confirm('Ar tėvai leido?');
  }
 */!*
 }
 ```

 Will the function work differently if `else`is removed?
 Ar ši funkcija veiks kitaip, jei `else`bus pašalinta?

 ```js
 function checkAge(age) {
  }
 *!*
  // ...
  return confirm('Did parents allow you?');
  return confirm('Ar tėvai leido?');
 */!*
 }
 ```

 Is there any difference in the behavior of these two variants?
 Ar yra nors vienas šio varianto elgsenos skirtumas?
diff --git a/1-js/02-first-steps/14-function-basics/2-rewrite-function-question-or/solution.md b/1-js/02-first-steps/14-function-basics/2-rewrite-function-question-or/solution.md
 Using a question mark operator `'?'`:
 Naudojant operatorių `?`:

 ```js
 function checkAge(age) {
  return (age > 18) ? true : confirm('Did parents allow you?');
  return (age > 18) ? true : confirm('Ar tėvai leido?');
 }
 ```

 Using OR `||` (the shortest variant):
 Naudojant operatorių `||` (trumpiausias variantas):

 ```js
 function checkAge(age) {
  return (age > 18) || confirm('Did parents allow you?');
  return (age > 18) || confirm('Ar tėvai leido?');
 }
 ```

 Note that the parentheses around `age > 18`are not required here. They exist for better readabilty.
 Atkreipkite dėmesį, kad skliaustai aplink `age > 18`yra neprivalomi. Jie skirti geresniam kodo skaitomumui užtikrinti.
diff --git a/1-js/02-first-steps/14-function-basics/2-rewrite-function-question-or/task.md b/1-js/02-first-steps/14-function-basics/2-rewrite-function-question-or/task.md

 ---

 #Rewrite the function using '?'or '||'
 #Perrašykite funkciją naudodami operatorių '?'arba '||'

 The following function returns`true` if the parameter`age`is greater than `18`.
 Ši funkcija grąžina`true`, jei parametras`age`yra didesnis nei `18`.

 Otherwise it asks for a confirmation and returns its result.
 Priešingu atveju klausiama `confirm' ir grąžinamas rezultatas.

 ```js
 function checkAge(age) {
  if (age > 18) {
    return true;
  } else {
    return confirm('Did parents allow you?');
    return confirm('Ar tėvai leido?');
  }
 }
 ```

 Rewrite it, to perform the same, but without `if`,in a single line.
 Perrašykite funkciją taip, kad ji darytų tą patį, bet be `if`,vienoje eilutėje.

 Make two variants of`checkAge`:
 Sukurkite dvi funkcijos`checkAge` variantus:

 1.Using a question mark operator `?`
 2.Using OR `||`
 1.Naudojant operatorių `?`
 2.Naudojant operatorių `||`
diff --git a/1-js/02-first-steps/14-function-basics/3-min/solution.md b/1-js/02-first-steps/14-function-basics/3-min/solution.md
 A solution using `if`:
 Sprendimo variantas naudojant `if`:

 ```js
 function min(a, b) {
 }
 ```

 A solution with a question mark operator `'?'`:
 Sprendimo variantas su operatoriumi `?`:

 ```js
 function min(a, b) {
  return a < b ? a : b;
 }
 ```

 P.S.In the case of an equality`a == b`it does not matter what to return.
 P.S.Lygybės`a == b`atveju nesvarbu, ką grąžinti.
diff --git a/1-js/02-first-steps/14-function-basics/3-min/task.md b/1-js/02-first-steps/14-function-basics/3-min/task.md

 ---

 #Function min(a, b)
 #Funkcija min(a, b)

 Write a function`min(a,b)` which returns the least of two numbers`a`and `b`.
 Parašykite funkciją`min(a,b)`, kuri grąžina mažesnį iš skaičių`a`ir `b`.

 For instance:
 Pavyzdžiui:

 ```js
 min(2, 5) == 2
diff --git a/1-js/02-first-steps/14-function-basics/4-pow/solution.md b/1-js/02-first-steps/14-function-basics/4-pow/solution.md
 let n = prompt("n?", '');

 if (n < 1) {
  alert(`Power${n}is not supported, use a positive integer`);
  alert(`${n}laipsnis nepalaikomas, naudokite natūralų skaičių`);
 } else {
  alert( pow(x, n) );
 }
diff --git a/1-js/02-first-steps/14-function-basics/4-pow/task.md b/1-js/02-first-steps/14-function-basics/4-pow/task.md

 ---

 #Function pow(x,n)
 #Funkcija pow(x,n)

 Write a function`pow(x,n)` that returns `x`in power`n`. Or, in other words, multiplies `x`by itself `n`times and returns the result.
 Sukurkite funkciją`pow(x,n)`, kuri grąžina `x`iki`n` laipsnio. Kitaip tariant, padaugina `x`iš savęs `n`kartų ir grąžina rezultatą.

 ```js
 pow(3, 2) = 3 * 3 = 9
 pow(3, 3) = 3 * 3 * 3 = 27
 pow(1, 100) = 1 * 1 * ...* 1 = 1
 ```

 Create a web-page that prompts for`x`and `n`, and then shows the result of `pow(x,n)`.
 Sukurkite svetainę, kuri pateikia užklausas`x`ir `n` ir išveda rezultatą `pow(x,n)`.

 [demo]

 P.S.In this task the function should support only natural values of`n`: integers up from `1`.
 P.S.Šioje užduotyje reikalaujama, kad funkcija palaikytų tik natūraliąsias`n` vertės, t. y. sveikuosius skaičius nuo `1`.
Original file line number	Diff line number	Diff line change
		@@ -1 +1 @@
		No difference.
		Abi funkcijos veikia vienodai, skirtumų nėra.
Original file line number	Diff line number	Diff line change
Expand Up		@@ -2,11 +2,11 @@ importance: 4

		---

		#Is "else"required?
		#Ar "else"yra privalomas?

		The following function returns`true` if the parameter`age`is greater than `18`.
		Ši funkcija grąžina`true`, jei parametras`age`yra didesnis nei `18`.

		Otherwise it asks for a confirmation and returns its result:
		Priešingu atveju ji prašo patvirtinimo per `confirm` ir grąžina rezultatą:

		```js
		function checkAge(age) {
Expand All		@@ -15,13 +15,13 @@ function checkAge(age) {
		!
		} else {
		// ...
		return confirm('Did parents allow you?');
		return confirm('Ar tėvai leido?');
		}
		/!
		}
		```

		Will the function work differently if `else`is removed?
		Ar ši funkcija veiks kitaip, jei `else`bus pašalinta?

		```js
		function checkAge(age) {
Expand All		@@ -30,9 +30,9 @@ function checkAge(age) {
		}
		!
		// ...
		return confirm('Did parents allow you?');
		return confirm('Ar tėvai leido?');
		/!
		}
		```

		Is there any difference in the behavior of these two variants?
		Ar yra nors vienas šio varianto elgsenos skirtumas?
Original file line number	Diff line number	Diff line change
		@@ -1,17 +1,17 @@
		Using a question mark operator `'?'`:
		Naudojant operatorių `?`:

		```js
		function checkAge(age) {
		return (age > 18) ? true : confirm('Did parents allow you?');
		return (age > 18) ? true : confirm('Ar tėvai leido?');
		}
		```

		Using OR `\|\|` (the shortest variant):
		Naudojant operatorių `\|\|` (trumpiausias variantas):

		```js
		function checkAge(age) {
		return (age > 18) \|\| confirm('Did parents allow you?');
		return (age > 18) \|\| confirm('Ar tėvai leido?');
		}
		```

		Note that the parentheses around `age > 18`are not required here. They exist for better readabilty.
		Atkreipkite dėmesį, kad skliaustai aplink `age > 18`yra neprivalomi. Jie skirti geresniam kodo skaitomumui užtikrinti.
Original file line number	Diff line number	Diff line change
Expand Up		@@ -2,25 +2,25 @@ importance: 4

		---

		#Rewrite the function using '?'or '\|\|'
		#Perrašykite funkciją naudodami operatorių '?'arba '\|\|'

		The following function returns`true` if the parameter`age`is greater than `18`.
		Ši funkcija grąžina`true`, jei parametras`age`yra didesnis nei `18`.

		Otherwise it asks for a confirmation and returns its result.
		Priešingu atveju klausiama `confirm' ir grąžinamas rezultatas.

		```js
		function checkAge(age) {
		if (age > 18) {
		return true;
		} else {
		return confirm('Did parents allow you?');
		return confirm('Ar tėvai leido?');
		}
		}
		```

		Rewrite it, to perform the same, but without `if`,in a single line.
		Perrašykite funkciją taip, kad ji darytų tą patį, bet be `if`,vienoje eilutėje.

		Make two variants of`checkAge`:
		Sukurkite dvi funkcijos`checkAge` variantus:

		1.Using a question mark operator `?`
		2.Using OR `\|\|`
		1.Naudojant operatorių `?`
		2.Naudojant operatorių `\|\|`
Original file line number	Diff line number	Diff line change
		@@ -1,4 +1,4 @@
		A solution using `if`:
		Sprendimo variantas naudojant `if`:

		```js
		function min(a, b) {
Expand All		@@ -10,12 +10,12 @@ function min(a, b) {
		}
		```

		A solution with a question mark operator `'?'`:
		Sprendimo variantas su operatoriumi `?`:

		```js
		function min(a, b) {
		return a < b ? a : b;
		}
		```

		P.S.In the case of an equality`a == b`it does not matter what to return.
		P.S.Lygybės`a == b`atveju nesvarbu, ką grąžinti.
Original file line number	Diff line number	Diff line change
Expand Up		@@ -2,11 +2,11 @@ importance: 1

		---

		#Function min(a, b)
		#Funkcija min(a, b)

		Write a function`min(a,b)` which returns the least of two numbers`a`and `b`.
		Parašykite funkciją`min(a,b)`, kuri grąžina mažesnį iš skaičių`a`ir `b`.

		For instance:
		Pavyzdžiui:

		```js
		min(2, 5) == 2
Expand Down
Original file line number	Diff line number	Diff line change
Expand Up		@@ -14,7 +14,7 @@ let x = prompt("x?", '');
		let n = prompt("n?", '');

		if (n < 1) {
		alert(`Power${n}is not supported, use a positive integer`);
		alert(`${n}laipsnis nepalaikomas, naudokite natūralų skaičių`);
		} else {
		alert( pow(x, n) );
		}
Expand Down
Original file line number	Diff line number	Diff line change
Expand Up		@@ -2,18 +2,18 @@ importance: 4

		---

		#Function pow(x,n)
		#Funkcija pow(x,n)

		Write a function`pow(x,n)` that returns `x`in power`n`. Or, in other words, multiplies `x`by itself `n`times and returns the result.
		Sukurkite funkciją`pow(x,n)`, kuri grąžina `x`iki`n` laipsnio. Kitaip tariant, padaugina `x`iš savęs `n`kartų ir grąžina rezultatą.

		```js
		pow(3, 2) = 3 * 3 = 9
		pow(3, 3) = 3 * 3 * 3 = 27
		pow(1, 100) = 1 * 1 * ...* 1 = 1
		```

		Create a web-page that prompts for`x`and `n`, and then shows the result of `pow(x,n)`.
		Sukurkite svetainę, kuri pateikia užklausas`x`ir `n` ir išveda rezultatą `pow(x,n)`.

		[demo]

		P.S.In this task the function should support only natural values of`n`: integers up from `1`.
		P.S.Šioje užduotyje reikalaujama, kad funkcija palaikytų tik natūraliąsias`n` vertės, t. y. sveikuosius skaičius nuo `1`.