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

let's submit our answer:

How to be a creative thinker | Carnegie Mellon University Po-Shen Loh

441. Arranging Coins #binary_search #Easy #leet_code_Q8 #441

You have n coins and you want to build a staircase with these coins. The staircase consists of k rows where the ith row has exactly i coins. The last row of the staircase may be incomplete.

Given the integer n, return the number of complete rows of the staircase you will build.

Input: n = 5
Output: 2
Explanation: Because the 3rd row is incomplete, we return 2.

Input: n = 8
Output: 3
Explanation: Because the 4th row is incomplete, we return 3

You can contact me if you have any comments, ideas, questions, or anything to talk about https://news.1rj.ru/str/zprogramming_bot

let's solve the question.

First, how can we determine the number of coins that are enough for building an n-staircase? Let's look at an example: to build 1 stair, we need 1 coin; for 2 stairs, we need 3 coins; for 3 stairs, we need 5 coins; and for 4 stairs, we need 10 coins. What do you notice from this pattern? The number of stairs and the number of coins have a relationship, so we can use the formula (n+1 * n) / 2 to calculate the number of coins needed based on the number of stairs.

Let's say n is the number of stairs and c is the number of coins. Based on this, we can iterate from 1 up to n, where (n+1 * n) / 2 > c. When we find the first n that fulfills this condition, we subtract 1 so that n is the maximum number of stairs that we can build with the given number of coins.