思路：

1. 建模：求子区间个数，子区间的要求：存在某数出现过t次以上；
2. 记录某数出现次数用hashmap最佳；牺牲一定的空间复杂度换取时间复杂度。

代码：

import java.util.Collection;

import java.util.HashMap;

import java.util.Map;

import java.util.Scanner;

public class solution {

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();

        int k = sc.nextInt();

        int t = sc.nextInt();

        int[] array = new int[n];

        for (int i = 0; i < n; i++) {

            array[i] = sc.nextInt();

        }

        int result = 0;

        for (int i = 0; i <= array.length - k; i++) {

            HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();

            recordinfo(map, array, i, i + k);

            result += isAppearThanT(map, t);

        }

        System.out.println(result);

    }

    private static void recordinfo(HashMap<Integer, Integer> map, int[] array, int low, int high) {

        for (int i = low; i < high; i++) {

            if (map.containsKey(array[i])) {

                int temp = map.get(array[i]);

                map.put(array[i], temp + 1);

            } else {

                map.put(array[i], 1);

            }

        }

    }

    private static int isAppearThanT(HashMap<Integer, Integer> map, int t) {
        int a = 0;

        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {

            if (entry.getValue() >= t) {

                a++;
            }

        }
        return a;

    }

}