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Paskal uchburchagining dastlabki n ta qatorini toping.

💡 Paskal uchburchagi har bir element o'zidan yuqoridagi 2 ta element yig'indisiga teng. Kiritish: 5
Natija: [[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1]]
📌 Formulasi
Bu masala 42.uz va leetcode.com saytida ham ham uchraydi
DinMuhammad
Paskal uchburchagining dastlabki n ta qatorini toping. 💡 Paskal uchburchagi har bir element o'zidan yuqoridagi 2 ta element yig'indisiga teng. Kiritish: 5 Natija: [[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1]]
def generate(n: int) -> list:
triangle = []

for i in range(n):
row = [1] * (i + 1)
for j in range(1, i):
row[j] = triangle[i-1][j-1] + triangle[i-1][j]
triangle.append(row)

return triangle
Berilgan belgilar ro'yxatini teskari tartibga keltiring.

Xotira murakkabligi O(1) bo'lsin.

Misol 1:
Kiritish: s = ["h","e","l","l","o"]
Natija: ["o","l","l","e","h"]
def letterCombinations(self, digits: str) -> List[str]:
if not digits:
return []

L = {
'2': ['a', 'b', 'c'],
'3': ['d', 'e', 'f'],
'4': ['g', 'h', 'i'],
'5': ['j', 'k', 'l'],
'6': ['m', 'n', 'o'],
'7': ['p', 'q', 'r', 's'],
'8': ['t', 'u', 'v'],
'9': ['w', 'x', 'y', 'z']
}

natija = ['']

for son in digits:
tmp = []
for a in natija:
for letter in L[son]:
tmp.append(a + letter)
natija = tmp

return natija
O’qima, uqi.
O’ylanma, uylan
author: Azim Pulat
4. Ikki tartiblangan massivning medianasi
DinMuhammad
4. Ikki tartiblangan massivning medianasi
Ikki tartiblangan massiv nums1va nums2hajmi mva mos ravishda berilgan holda, ikkita tartiblangan massivning medianasinin qaytaring .

Umumiy ish vaqti murakkabligi bo'lishi kerak O(log (m+n)).
```Belgilangan 32 bitli butun son berilgan bo'lsa , raqamlari teskari bo'lgan holdax qaytaring . Agar teskari harakat qiymatning imzolangan 32 bitli tamsayı oralig'idan tashqariga chiqishiga sabab bo'lsa , uni qaytaring .xx[-231, 231 - 1]0```
Atrof-muhit sizga 64 bitli butun sonlarni (imzolangan yoki imzolanmagan) saqlashga ruxsat bermaydi deb faraz qiling.
1-misol:

Kirish: x = 123
Chiqish: 321
2-misol:

Kirish: x = -123
Chiqish: -321
3-misol:

Kirish: x = 120
Chiqish: 21


Cheklovlar:

-231 <= x <= 231 - 1
yechim
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