This game uses C++ I wrote two questions , Used after the game python I have gained a lot from the supplementary questions , I learned a little about python Built in functions for . Method learning comes from other excellent bloggers .

6120. How many pairs can an array form （A topic ）

Ideas

Method 1 ： I wrote my own algorithm , First pair nums Sort , Then traverse the array to find whether the adjacent positions are the same .
Method 2 ： Count the number of times each number appears nums, And then nums Divide 2 Add up and get , Is the first answer , Unmatched use nums Subtract twice the matching logarithm from the length of .

Code

class Solution:
    def numberOfPairs(self, nums: List[int]) -> List[int]:
        cnt = Counter(nums)
        pairs = sum(num // 2 for num in cnt.values())
        return [pairs, len(nums) - pairs * 2]

6164. The maximum sum of digits and equal pairs （B topic ）

Ideas

1. Group by digits , Key memory digits and , Value store itself
2. Maintain the first two numbers with a large root heap

Code

class Solution:
    def maximumSum(self, nums: List[int]) -> int:
        groups = defaultdict(list)
        for num in nums:
            s = 0
            n = num
            while n:
                s += n % 10
                n //= 10
            if len(groups[s]) < 2 :
                heappush(groups[s], num)
            else:
                heappushpop(groups[s], num) #  First pop up and press in 
        ans = -1
        for g in groups.values():
            if len(g) > 1:
                ans = max(ans, g[-1] + g[-2])
        return ans

6121. Query the number K Small numbers （C topic ）

Ideas

Use the principle of Radix sorting to sort .
First we need to use zip() Function to take iteratable objects as parameters , Package the corresponding elements in the object into tuples , And then return the objects made up of these tuples .
And then use zip Function will nums and queries Save the subscript of the array , Sort , You have to use it here sorted Stable sequencing .
Then use the idea of cardinality to sort queries Of trim Sort ,nums Sort from the last column , Then find the corresponding trim, Take second place k The following table of small numbers .
Finally, save these subscripts to ans Array returns .

Code

class Solution:
    def smallestTrimmedNumbers(self, nums: List[str], queries: List[List[int]]) -> List[int]:
        # 
        qs = sorted(zip(queries, range(len(queries))), key=lambda q: q[0][1]) #  Sort according to the columns of the two-dimensional array 
        a = sorted(zip(nums, range(len(nums))), key=lambda t: t[0][-1]) #  Sort by the last character 
        j = 2 # j Record the current row of string 
        ans = [0] * len(queries) #  Define an answer array 
        for (k, trim),i in qs:
            while j <= trim: #  Rank at trim individual 
                a.sort(key=lambda t:t[0][-j]) #  Countdown j Sort columns 
                j += 1 #  Next time 
            ans[i] = a[k - 1][1] #  Record the... After sorting k Small subscript 
        return ans

6122. The minimum number of deletions that make the array divisible （D topic ）

Ideas

First seek numsDivide Maximum common divisor of g, And then in nums Find the smallest one that can be g Divisible number , recorded , The number of deletions is nums Middle ratio g Small number .
We need to use reduce function ,reduce function , Put the first two operations in the first set , And then calculate the result with the third one .

Code

class Solution:
    def minOperations(self, nums: List[int], numsDivide: List[int]) -> int:
        g = reduce(gcd, numsDivide)
        mn = min((num for num in nums if g % num == 0), default = 0)
        if mn == 0:
            return -1
        return sum(num < mn for num in nums)