Bläddra i källkod

Another problem done

master
Lachlan Jacob 5 år sedan
förälder
incheckning
b32ac0f7e2
3 ändrade filer med 41 tillägg och 0 borttagningar
  1. 15
    0
      852/main.py
  2. 23
    0
      852/problem.txt
  3. 3
    0
      852/run.sh

+ 15
- 0
852/main.py Visa fil

@@ -0,0 +1,15 @@
class Solution:
def peakIndexInMountainArray(self, A):
# return the peak of the mountain... i.e Where the array starts going down in value
last = -1
for ind, n in enumerate(A):
if n < last:
return ind - 1
last = n

# This should never happen
return -1

s = Solution()
print("Expected: 1")
print("Got:", s.peakIndexInMountainArray([0, 2, 1, 0]))

+ 23
- 0
852/problem.txt Visa fil

@@ -0,0 +1,23 @@
Let's call an array A a mountain if the following properties hold:

A.length >= 3
There exists some 0 < i < A.length - 1 such that A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]

Given an array that is definitely a mountain, return any i such that A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1].

Example 1:

Input: [0,1,0]
Output: 1

Example 2:

Input: [0,2,1,0]
Output: 1

Note:

3 <= A.length <= 10000
0 <= A[i] <= 10^6
A is a mountain, as defined above.


+ 3
- 0
852/run.sh Visa fil

@@ -0,0 +1,3 @@
#!/bin/bash

python3 main.py

Laddar…
Avbryt
Spara