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Day31 - Greedy-Part1.md

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Day31 - Greedy Part 1

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455. Assign Cookies

Easy

Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie.

Each child i has a greed factor g[i], which is the minimum size of a cookie that the child will be content with; and each cookie j has a size s[j]. If s[j] >= g[i], we can assign the cookie j to the child i, and the child i will be content. Your goal is to maximize the number of your content children and output the maximum number.

 

Example 1:

Input: g = [1,2,3], s = [1,1]
Output: 1
Explanation: You have 3 children and 2 cookies. The greed factors of 3 children are 1, 2, 3. 
And even though you have 2 cookies, since their size is both 1, you could only make the child whose greed factor is 1 content.
You need to output 1.

Example 2:

Input: g = [1,2], s = [1,2,3]
Output: 2
Explanation: You have 2 children and 3 cookies. The greed factors of 2 children are 1, 2. 
You have 3 cookies and their sizes are big enough to gratify all of the children, 
You need to output 2.

 

Constraints:

  • 1 <= g.length <= 3 * 104
  • 0 <= s.length <= 3 * 104
  • 1 <= g[i], s[j] <= 231 - 1
### Breakdown and Thought Process:

My Solution 1：Greedy Large

class Solution:
    def findContentChildren(self, g: List[int], s: List[int]) -> int:
        # 按降序分配最大的饼干给胃口最大的孩子
        g.sort()
        s.sort()
        s_index = len(s) - 1
        count = 0
        # 注意这里的循环顺序不能颠倒
        # 例如 g[1,2,7,10] s[1,3,5,9]
        for i in range(len(g)-1, -1, -1):
            if s_index>=0 and s[s_index] >= g[i]:
                s_index -= 1
                count += 1
  
        return count

Complexity Analysis:

  • Time Complexity:
    O(nlogn + mlogm) , where nlogn for sorting s and mlogm for sorting g.

  • Space Complexity:
    O(1)


Dividing Line

376. Wiggle Subsequence

Medium

A wiggle sequence is a sequence where the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with one element and a sequence with two non-equal elements are trivially wiggle sequences.

  • For example, [1, 7, 4, 9, 2, 5] is a wiggle sequence because the differences (6, -3, 5, -7, 3) alternate between positive and negative.
  • In contrast, [1, 4, 7, 2, 5] and [1, 7, 4, 5, 5] are not wiggle sequences. The first is not because its first two differences are positive, and the second is not because its last difference is zero.

A subsequence is obtained by deleting some elements (possibly zero) from the original sequence, leaving the remaining elements in their original order.

Given an integer array nums, return the length of the longest wiggle subsequence of nums.

 

Example 1:

Input: nums = [1,7,4,9,2,5]
Output: 6
Explanation: The entire sequence is a wiggle sequence with differences (6, -3, 5, -7, 3).

Example 2:

Input: nums = [1,17,5,10,13,15,10,5,16,8]
Output: 7
Explanation: There are several subsequences that achieve this length.
One is [1, 17, 10, 13, 10, 16, 8] with differences (16, -7, 3, -3, 6, -8).

Example 3:

Input: nums = [1,2,3,4,5,6,7,8,9]
Output: 2

 

Constraints:

  • 1 <= nums.length <= 1000
  • 0 <= nums[i] <= 1000

 

Follow up: Could you solve this in O(n) time?


Solving approach 1:

q376-thoughts

My Solution 1：Greedy

class Solution:
    def wiggleMaxLength(self, nums: List[int]) -> int:
        pre_diff = 0
        cur_diff = 0

        result = 1
        # 处理三种情况 1.上下有平坡
        # 2. 首位元素
        # 单调坡度平坡


        for i in range(len(nums)-1):
            cur_diff = nums[i+1] - nums[i]
            if (pre_diff >= 0 and cur_diff < 0) or (pre_diff <= 0 and cur_diff > 0):
                result += 1
                pre_diff = cur_diff
            #pre_diff = cur_diff 放在这里没有考虑单调平坡情况

        return result

Complexity Analysis:

  • Time Complexity:
    O(n)

  • Space Complexity:
    O(1)


Dividing Line

53. Maximum Subarray

Medium

Given an integer array nums, find the subarray with the largest sum, and return its sum.

 

Example 1:

Input: nums = [-2,1,-3,4,-1,2,1,-5,4]
Output: 6
Explanation: The subarray [4,-1,2,1] has the largest sum 6.

Example 2:

Input: nums = [1]
Output: 1
Explanation: The subarray [1] has the largest sum 1.

Example 3:

Input: nums = [5,4,-1,7,8]
Output: 23
Explanation: The subarray [5,4,-1,7,8] has the largest sum 23.

 

Constraints:

  • 1 <= nums.length <= 105
  • -104 <= nums[i] <= 104

 

Follow up: If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

My Solution 1：Kadane greedy

class Solution:
    def maxSubArray(self, nums: List[int]) -> int:


        res, count = float('-inf'),float('-inf')
        for num in nums: 
            #现在的数值是取求和，或是重新以现在的数值开始
            count = num + max(count, 0)
            
            res = max(res, count)
        
        return res

Complexity Analysis:

  • Time Complexity:
    O(n)
  • Space Complexity:
    O(1)

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