给你一个整数数组 arr
和一个整数 k
。现需要从数组中恰好移除 k
个元素,请找出移除后数组中不同整数的最少数目。
示例 1:
输入:arr = [5,5,4], k = 1 输出:1 解释:移除 1 个 4 ,数组中只剩下 5 一种整数。
示例 2:
输入:arr = [4,3,1,1,3,3,2], k = 3 输出:2 解释:先移除 4、2 ,然后再移除两个 1 中的任意 1 个或者三个 3 中的任意 1 个,最后剩下 1 和 3 两种整数。
提示:
1 <= arr.length <= 10^5
1 <= arr[i] <= 10^9
0 <= k <= arr.length
class Solution:
def findLeastNumOfUniqueInts(self, arr: List[int], k: int) -> int:
counter = Counter(arr)
t = sorted(counter.items(), key=lambda x: x[1])
for v, cnt in t:
if k >= cnt:
k -= cnt
counter.pop(v)
else:
break
return len(counter)
class Solution {
public int findLeastNumOfUniqueInts(int[] arr, int k) {
Map<Integer, Integer> counter = new HashMap<>();
for (int v : arr) {
counter.put(v, counter.getOrDefault(v, 0) + 1);
}
List<Map.Entry<Integer, Integer>> t = new ArrayList<>(counter.entrySet());
Collections.sort(t, Comparator.comparingInt(Map.Entry::getValue));
for (Map.Entry<Integer, Integer> e : t) {
int v = e.getKey();
int cnt = e.getValue();
if (k >= cnt) {
k -= cnt;
counter.remove(v);
} else {
break;
}
}
return counter.size();
}
}
class Solution {
public:
int findLeastNumOfUniqueInts(vector<int>& arr, int k) {
unordered_map<int, int> counter;
for (int v : arr) ++counter[v];
vector<pair<int, int>> t(counter.begin(), counter.end());
sort(t.begin(), t.end(), [](const auto& a, const auto& b) { return a.second < b.second; });
for (auto [v, cnt] : t) {
if (k >= cnt) {
k -= cnt;
counter.erase(v);
} else
break;
}
return counter.size();
}
};
func findLeastNumOfUniqueInts(arr []int, k int) int {
counter := make(map[int]int)
for _, v := range arr {
counter[v]++
}
var t [][]int
for v, cnt := range counter {
t = append(t, []int{v, cnt})
}
sort.Slice(t, func(i, j int) bool {
return t[i][1] < t[j][1]
})
for _, e := range t {
v, cnt := e[0], e[1]
if k >= cnt {
k -= cnt
delete(counter, v)
} else {
break
}
}
return len(counter)
}