# 2537. Count the Number of Good Subarrays
###### tags: `Leetcode` `Medium` `Sliding Window`
Link: https://leetcode.com/problems/count-the-number-of-good-subarrays/description/
## 思路

每次右指针移动 ```currPairs+=map.get(nums[r])```
然后移动左指针直到```currPairs<k``` 那么以右指针为右边界的good subarray数就是l 因为l左边的所有index都可以当左边界
## Code

class Solution:
def countGood(self, nums: List[int], k: int) -> int:
count = Counter()
ans = 0
left, right = 0, 0
pairsCnt = 0
for right in range(len(nums)):
pairsCnt += count[nums[right]]
count[nums[right]] += 1
while pairsCnt>=k:
count[nums[left]] -= 1
pairsCnt -= count[nums[left]]
left += 1
ans += left
return ans

class Solution {
public long countGood(int[] nums, int k) {
int l=0, r=0;
long res=0;
int currPairs = 0;
Map<Integer, Integer> cnt = new HashMap<>();
while(r<nums.length){
currPairs+=cnt.getOrDefault(nums[r], 0);
cnt.put(nums[r], cnt.getOrDefault(nums[r], 0)+1);
while(currPairs>=k){
cnt.put(nums[l], cnt.get(nums[l])-1);
currPairs -= cnt.getOrDefault(nums[l], 0);
l++;
}
res += l;
r++;
}
return res;
}
}