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2080. 区间内查询数字的频率

English Version

题目描述

请你设计一个数据结构，它能求出给定子数组内一个给定值的 频率 。

子数组中一个值的 频率 指的是这个子数组中这个值的出现次数。

请你实现 RangeFreqQuery 类：

  • RangeFreqQuery(int[] arr) 用下标从 0 开始的整数数组 arr 构造一个类的实例。
  • int query(int left, int right, int value) 返回子数组 arr[left...right] 中 value 的 频率 。

一个 子数组 指的是数组中一段连续的元素。arr[left...right] 指的是 nums 中包含下标 left 和 right 在内 的中间一段连续元素。

 

示例 1：

输入：
["RangeFreqQuery", "query", "query"]
[[[12, 33, 4, 56, 22, 2, 34, 33, 22, 12, 34, 56]], [1, 2, 4], [0, 11, 33]]
输出：
[null, 1, 2]

解释：
RangeFreqQuery rangeFreqQuery = new RangeFreqQuery([12, 33, 4, 56, 22, 2, 34, 33, 22, 12, 34, 56]);
rangeFreqQuery.query(1, 2, 4); // 返回 1 。4 在子数组 [33, 4] 中出现 1 次。
rangeFreqQuery.query(0, 11, 33); // 返回 2 。33 在整个子数组中出现 2 次。

 

提示：

  • 1 <= arr.length <= 105
  • 1 <= arr[i], value <= 104
  • 0 <= left <= right < arr.length
  • 调用 query 不超过 105 次。

解法

方法一

class RangeFreqQuery:
    def __init__(self, arr: List[int]):
        self.mp = defaultdict(list)
        for i, x in enumerate(arr):
            self.mp[x].append(i)

    def query(self, left: int, right: int, value: int) -> int:
        if value not in self.mp:
            return 0
        arr = self.mp[value]
        l, r = bisect_right(arr, left - 1), bisect_right(arr, right)
        return r - l


# Your RangeFreqQuery object will be instantiated and called as such:
# obj = RangeFreqQuery(arr)
# param_1 = obj.query(left,right,value)
class RangeFreqQuery {
    private Map<Integer, List<Integer>> mp = new HashMap<>();

    public RangeFreqQuery(int[] arr) {
        for (int i = 0; i < arr.length; ++i) {
            mp.computeIfAbsent(arr[i], k -> new ArrayList<>()).add(i);
        }
    }

    public int query(int left, int right, int value) {
        if (!mp.containsKey(value)) {
            return 0;
        }
        List<Integer> arr = mp.get(value);
        int l = search(arr, left - 1);
        int r = search(arr, right);
        return r - l;
    }

    private int search(List<Integer> arr, int val) {
        int left = 0, right = arr.size();
        while (left < right) {
            int mid = (left + right) >> 1;
            if (arr.get(mid) > val) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

/**
 * Your RangeFreqQuery object will be instantiated and called as such:
 * RangeFreqQuery obj = new RangeFreqQuery(arr);
 * int param_1 = obj.query(left,right,value);
 */
class RangeFreqQuery {
public:
    unordered_map<int, vector<int>> mp;
    RangeFreqQuery(vector<int>& arr) {
        for (int i = 0; i < arr.size(); ++i)
            mp[arr[i]].push_back(i);
    }

    int query(int left, int right, int value) {
        if (!mp.count(value)) return 0;
        auto& arr = mp[value];
        auto l = upper_bound(arr.begin(), arr.end(), left - 1);
        auto r = upper_bound(arr.begin(), arr.end(), right);
        return r - l;
    }
};

/**
 * Your RangeFreqQuery object will be instantiated and called as such:
 * RangeFreqQuery* obj = new RangeFreqQuery(arr);
 * int param_1 = obj->query(left,right,value);
 */
type RangeFreqQuery struct {
	mp map[int][]int
}

func Constructor(arr []int) RangeFreqQuery {
	mp := make(map[int][]int)
	for i, v := range arr {
		mp[v] = append(mp[v], i)
	}
	return RangeFreqQuery{mp}
}

func (this *RangeFreqQuery) Query(left int, right int, value int) int {
	arr := this.mp[value]
	l := sort.SearchInts(arr, left)
	r := sort.SearchInts(arr, right+1)
	return r - l
}

/**
 * Your RangeFreqQuery object will be instantiated and called as such:
 * obj := Constructor(arr);
 * param_1 := obj.Query(left,right,value);
 */