leetcode:20220213周赛（少点bug就前10%了555）

思路：
1.分奇偶统计数频
2.然后看是否相等
3.相等酌情取下一个

src：

lass Solution:
    def minimumOperations(self, nums: List[int]) -> int:
        num1 = nums[::2]
        num2 = nums[1::2]
        n1 = len(num1)
        n2 = len(num2)
        
        if n1 * n2 == 0:
            return 0
        
        cnt1 = defaultdict(int)
        for num in num1:
            cnt1[num] += 1
        
        cnt2 = defaultdict(int)
        for num in num2:
            cnt2[num] += 1
        
        sort1 = sorted(cnt1.items(),key=lambda x:x[1],reverse=True) #根据词频降序排序
        sort2 = sorted(cnt2.items(),key=lambda x:x[1],reverse=True) #根据词频降序排序
        
        has = 0
        if sort1[0][0] != sort2[0][0]:
            has = sort1[0][1] + sort2[0][1]
        else:
            if len(sort1) == 1 and len(sort2) == 1:
                has = max(sort1[0][1], sort2[0][1])
            elif len(sort1) == 1:
                has = sort1[0][1] + sort2[1][1]
            elif len(sort2) == 1:
                has = sort2[0][1] + sort1[1][1]
            else:
                if sort1[1][1] + sort2[0][1] > sort1[0][1] + sort2[1][1]:
                    has = sort1[1][1] + sort2[0][1]
                else:
                    has = sort1[0][1] + sort2[1][1]
        
        return n1 + n2 - has

思路：
绝对值方程求解，给定x
若大于等于x，则做差；若小于x， 则取该数和0的距离
通过遍历x，可以得到最小的ans
注意，不用每遍历算一次，只需要-一些+一些就好，这就是绝对值函数和遍历混合出来的魅力

src：

class Solution:
    def minimumRemoval(self, beans: List[int]) -> int:
        # 全部变成最小的or最小的变0
        beans.sort()
        n = len(beans)
        #print(beans)
        # 确定一个值x，小于它就全部加上，大于它就取差
        temp = 0
        x = beans[0]
        index = 0
        ans = 0xffffffff
        
        for num in beans:
            temp += num - x
        ans = min(ans, temp)
        
        while index < n:
            nextIndex = index
            while nextIndex < n and beans[nextIndex] == x:
                nextIndex += 1
            if nextIndex >= n:
                break
            temp += x * (nextIndex - index)
            y = beans[nextIndex]
            temp -= (y - x) * (n - nextIndex)
            ans = min(ans, temp)
            
            x = y
            index = nextIndex
        
        return ans

总结：
如果没有bug，就前300了，都怪第二题写错了index