share ur solution on the comment section or in the group.

``` class Solution:
def search(self, nums: List[int], target: int) -> int:
left = 0
right = len(nums)-1
while(left<=right):
mid = (left+right)//2
if nums[mid] == target:
return mid
elif nums[mid] >target:
right = mid -1
else:
left = mid+1
return -1```

answer:

744. find the smallest that greater than the target letter #binary_search #leet_code_Q6 #Easy You are given an array of characters letters that is sorted in non-decreasing order, and a character target. There are at least two different characters in letters.

Return the smallest character in letters that is lexicographically greater than target. If such a character does not exist, return the first character in letters.

Example 1:

Input: letters = ["c","f","j"], target = "a"
Output: "c"
Explanation: The smallest character that is lexicographically greater than 'a' in letters is 'c'.
Example 2:

Input: letters = ["c","f","j"], target = "c"
Output: "f"
Explanation: The smallest character that is lexicographically greater than 'c' in letters is 'f'.
Example 3:

Input: letters = ["x","x","y","y"], target = "z"
Output: "x"
Explanation: There are no characters in letters that is lexicographically greater than 'z' so we return letters[0]

Constraints:

2 <= letters.length <= 104
letters[i] is a lowercase English letter.
letters is sorted in non-decreasing order.
letters contains at least two different characters.
target is a lowercase English letter.

answer:

class Solution:
def nextGreatestLetter(self, letters: List[str], target: str) -> str:
left = 0
right = len(letters)-1
while(left<=right):
mid = (left+right)//2
if(letters[mid] > target )and (letters[mid-1] <= target or mid==left):
return letters[mid]
elif letters[mid] > target:
right = mid -1
else:
left = mid +1
return letters[0]

let's submit our answer:

it's 91% faster.😎

የማብድ ቢመስለኝ ፣ ድንገት አንቺን ሳጣ
አረቄ ቤት ሔጄ ፣ አረቄ ስጠጣ
አንድ ግጥም ሰማሁ
ፍቅርሽን ከልቤ ፣ ነቅሎ የሚያወጣ፡፡
ያውም መንገድ በሚል...
ለመንገደኛ ሰው ፣ ሰካራም የፃፈው
እዛ ጋ ቁጭ ብሎ...
"ከዳችኝ" እያለ ፣ የሚለፈልፈው
መንገድ የሚል ግጥሙ ፣ጆሮዬን ገረፈው፡፡
ጆሮዬ ሲገረፍ ፣
ጠባሳ ጣለብኝ ፣ ግጥሙን እንዳልረሳ
"መንገድ አያደርስም
መንገድ አይመልስም ፣ እግር ካልተነሳ!!!"
እያለ ይገጥማል...
ደጋግሞ ደጋግሞ ፣ ሌላ አይናገርም
እሱም ልክ እንደኔ ...
አንዷን በመሔዷ ፣ ሳያጣት አይቀርም፡፡
ብቻ ሰክሪያለሁ
ለመንገደኛ ሴት ፣
የተፃፈ ግጥምን ፣ ጆሮዬ ያደሞጣል
መሔድሽን ያየ
እንደሌሌሽኝ ሲያውቅ ፣ ሊኖረኝ ይመጣል
ይህ ነው መንገድ ማለት!
የሰካራም ግጥምን ሰክሬ ስረዳው
ማፍቀረ ሳይሆን ለኔ ፣ መርሳት ነው ሚጎዳው!!!

ሰለወደድኩት ነው

let's takcle leet code medium question for the first time

#binary_search #Medium #leet_code_Q7 852. Peak Index in a Mountain Array

An array arr is a mountain if the following properties hold:

arr.length >= 3
There exists some i with 0 < i < arr.length - 1 such that:
arr[0] < arr[1] < ... < arr[i - 1] < arr[i]
arr[i] > arr[i + 1] > ... > arr[arr.length - 1]
Given a mountain array arr, return the index i such that arr[0] < arr[1] < ... < arr[i - 1] < arr[i] > arr[i + 1] > ... > arr[arr.length - 1].

You must solve it in O(log(arr.length)) time complexity.

Example 1:

Input: arr = [0,1,0]
Output: 1
Example 2:

Input: arr = [0,2,1,0]
Output: 1
Example 3:

Input: arr = [0,10,5,2]
Output: 1

Constraints:

3 <= arr.length <= 105
0 <= arr[i] <= 106
arr is guaranteed to be a mountain array.

How do we solve this? Any ideas?

any one who wants to share ur answer .

join the discussion group and share ur ideas leetcoders

Let's solve the peak index problem in mountain arrays. A peak index in a mountain array refers to the index i where:

Elements to the left of index i are strictly increasing: arr[0] < arr[1] < ... < arr[i-1].
Elements to the right of index i are strictly decreasing: arr[i] > arr[i+1] > ... > arr[arr.length - 1].
In simpler terms, the element at the peak index is the largest element in the array, with elements increasing to its left and decreasing to its right.

Here's a common approach to find the peak index in a mountain array:

1 We can use a binary search algorithm.
2 Start with the middle element of the array.
3 If the middle element is greater than its right neighbor, then the peak must be in the left half of the array. Move to the left half and go to step 2 .
4 If the middle element is less than its left neighbor, then the peak must be in the right half of the array. Move to the right half and go to step 2.
5 If the middle element is greater than both its neighbors, then it's the peak!

Answer:

class Solution:
def peakIndexInMountainArray(self, arr: List[int]) -> int:
left = 0
right = len(arr)-1
while(left<=right):
mid = (left+right)//2
if arr[mid] > arr[mid-1] and arr[mid] > arr[mid+1]:
return mid
elif arr[mid] > arr[mid+1]:
right = mid -1
else:
left = mid +1