Python list/dict functions for leetcode

List

• len(array) : length of array
  • list.index(obj)
• enumerate(array) : adds counter at the beginning.
  • for count, item in enumerate(['x','y','z']) # 0x 1y 2z
• range(0,len(mylist)1,2)
  • range(0,3), is 0,1,2
  • for i in reversed(range(len(A)))
• nums.append(val), nums.remove(val), nums.reverse()
  • nums.extend(nums2)
  • list.insert(index, obj)
  • list.pop() default pop index is -1, remove the last one
• nums[:]
  • copy the value within function: b[:]=a
    • .copy is equals to b[:]
    • temp = self.original[:]
    • copy.deepcopy could copy nested list
    • import copy , copy.deepcopy(a)
• min(a,b)
• nums.sort()
  • list.sort() change in place
  • sorted() return a new sorted list, leaving original unaffected
    • a.sort(key=lambda: x: str(x), reverse = False)
• zip(alist,blist) : return a zip object which is an iterator of tuples, paired together. Lengths depends on the shorter input list.
  • usually use tuple(), list(), set() on result.
  • list(zip(['x','y','z'], [3,4,5])) = [('x', 3), ('y', 4), ('z', 5)];
  • letter, number = zip(*result_list) # unzip
• matrix
  • test = [[0 for i in range(m)] for j in range(n)]
    • watch out , 32, mn, then the corner is test[n-1][m-1]
    • lee62
  • make one: dp = [[0] \* n for \_ in range(n)]
  • transpose a two dimension list
    • Trans = [[row[i] for row in board] for i in range(len(board[0]))]

dict

• for k, v in dic.items()
  • dict = {'a': 1, 'b': 2, 'b': '3'}
  • dic1 == dic2
  • dic.values()
• d = dict(collections.Counter(s)) : s 3 times, a 1 time
  • c.most_common(3), according to frequency, first 3 value,count 's list
  • c.items() , a list with value,count
  • c.elements: ['a', 'a', 'a', 'a', 'b', 'b']
• collections.defaultdict(int) # default 0
  • senti_dic = collections.defaultdict(lambda: 0)
  • defaultdict(lambda: 'Vanilla')
  • dict(collections.Counter(nums1)): key is item in num1, value is the count.

e.g.1 with zip, enumerate

134 !(2019-05-28) Gas station 2 #array #2lee

There are N gas stations along a circular route, where the amount of gas at station i is gas[i]. You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations. Return the starting gas station’s index if you can travel around the circuit once in the clockwise direction, otherwise return -1.

Note: If there exists a solution, it is guaranteed to be unique. Both input arrays are non-empty and have the same length. Each element in the input arrays is a non-negative integer.

Example 1:

Input: gas = [1,2,3,4,5] cost = [3,4,5,1,2] Output: 3

Explanation: Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4 Travel to station 4. Your tank = 4 - 1 + 5 = 8 Travel to station 0. Your tank = 8 - 2 + 1 = 7 Travel to station 1. Your tank = 7 - 3 + 2 = 6 Travel to station 2. Your tank = 6 - 4 + 3 = 5 Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3. Therefore, return 3 as the starting index.

Code:

 def canCompleteCircuit2(self, gas: List[int], cost: List[int]) -> int:
     tmp = [x - y for x, y in zip(gas, cost)]
     print(tmp+tmp)
     cur = 0
     ans = 0
     for i, val in enumerate(tmp + tmp):
         cur += val
         if cur < 0:
             ans = i + 1
             if ans >= len(gas):
                 return -1
             cur = 0
     return ans
            
          

e.g.2 with dict(collection.Counter(array))

Given a non-empty array of integers, every element appears twice except for one. Find that single one. Note: Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? Example 1: Input: [2,2,1] Output: 1

class Solution:
    def singleNumber(self, nums: List[int]) -> int:
        dic = dict(collections.Counter(nums))
        for i in dic:
            if dic[i] == 1:
                return i