293.使数组元素互不相同所需的最小操作次数

3396. 使数组元素互不相同所需的最少操作次数 - 力扣（LeetCode）

class Solution {
    public int minimumOperations(int[] nums) {
        HashSet<Integer> hs=new HashSet<>();
        int le=0;
        for(int i=nums.length-1;i>=0;i--){
            if(hs.contains(nums[i])){
                break;
            }
            le++;
            hs.add(nums[i]);
        }
        return (nums.length-le)%3==0?(nums.length-le)/3:(nums.length-le)/3+1;
    }
}

class Solution:
    def minimumOperations(self, nums):
        hs = set()
        le = 0
        for i in range(len(nums)-1, -1, -1):
            if nums[i] in hs:
                break
            le += 1
            hs.add(nums[i])
        return (len(nums) - le) // 3 if (len(nums) - le) % 3 == 0 else (len(nums) - le) // 3 + 1