数组范围查询频率与C程序中的值相同的元素?
在这里,我们将看到一个有趣的问题。我们有一个包含N个元素的数组。我们必须执行一个查询Q,如下所示:
Q(开始,结束)表示从开始到结束的次数恰好是“p”次,出现次数“p”。
因此,如果数组类似于:{1、5、2、3、1、3、5、7、3、9、8},而查询为-
Q(1,8)-这里的1出现一次,而3出现3次。所以答案是2
Q(0,2)-这里1出现一次。所以答案是1
算法
查询(s,e)-
Begin get the elements and count the frequency of each element ‘e’ into one map count := count + 1 for each key-value pair p, do if p.key = p.value, then count := count + 1 done return count; End
示例
#include <iostream> #include <map> using namespace std; int query(int start, int end, int arr[]) { map<int, int> freq; for (int i = start; i <= end; i++) //get element and store frequency freq[arr[i]]++; int count = 0; for (auto x : freq) if (x.first == x.second) //when the frequencies are same, increase count count++; return count; } int main() { int A[] = {1, 5, 2, 3, 1, 3, 5, 7, 3, 9, 8}; int n = sizeof(A) / sizeof(A[0]); int queries[][3] = {{ 0, 1 }, { 1, 8 }, { 0, 2 }, { 1, 6 }, { 3, 5 }, { 7, 9 } }; int query_count = sizeof(queries) / sizeof(queries[0]); for (int i = 0; i < query_count; i++) { int start = queries[i][0]; int end = queries[i][1]; cout << "Answer for Query " << (i + 1) << " = " << query(start, end, A) << endl; } }
输出结果
Answer for Query 1 = 1 Answer for Query 2 = 2 Answer for Query 3 = 1 Answer for Query 4 = 1 Answer for Query 5 = 1 Answer for Query 6 = 0