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Formatted question description: https://leetcode.ca/all/2537.html

# 2537. Count the Number of Good Subarrays

## Description

Given an integer array nums and an integer k, return the number of good subarrays of nums.

A subarray arr is good if it there are at least k pairs of indices (i, j) such that i < j and arr[i] == arr[j].

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [1,1,1,1,1], k = 10
Output: 1
Explanation: The only good subarray is the array nums itself.


Example 2:

Input: nums = [3,1,4,3,2,2,4], k = 2
Output: 4
Explanation: There are 4 different good subarrays:
- [3,1,4,3,2,2] that has 2 pairs.
- [3,1,4,3,2,2,4] that has 3 pairs.
- [1,4,3,2,2,4] that has 2 pairs.
- [4,3,2,2,4] that has 2 pairs.


Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i], k <= 109

## Solutions

• class Solution {
public long countGood(int[] nums, int k) {
Map<Integer, Integer> cnt = new HashMap<>();
long ans = 0, cur = 0;
int i = 0;
for (int x : nums) {
cur += cnt.getOrDefault(x, 0);
cnt.merge(x, 1, Integer::sum);
while (cur - cnt.get(nums[i]) + 1 >= k) {
cur -= cnt.merge(nums[i++], -1, Integer::sum);
}
if (cur >= k) {
ans += i + 1;
}
}
return ans;
}
}

• class Solution {
public:
long long countGood(vector<int>& nums, int k) {
unordered_map<int, int> cnt;
long long ans = 0;
long long cur = 0;
int i = 0;
for (int& x : nums) {
cur += cnt[x]++;
while (cur - cnt[nums[i]] + 1 >= k) {
cur -= --cnt[nums[i++]];
}
if (cur >= k) {
ans += i + 1;
}
}
return ans;
}
};


• class Solution:
def countGood(self, nums: List[int], k: int) -> int:
cnt = Counter()
ans = cur = 0
i = 0
for x in nums:
cur += cnt[x]
cnt[x] += 1
while cur - cnt[nums[i]] + 1 >= k:
cnt[nums[i]] -= 1
cur -= cnt[nums[i]]
i += 1
if cur >= k:
ans += i + 1
return ans


• func countGood(nums []int, k int) int64 {
cnt := map[int]int{}
ans, cur := 0, 0
i := 0
for _, x := range nums {
cur += cnt[x]
cnt[x]++
for cur-cnt[nums[i]]+1 >= k {
cnt[nums[i]]--
cur -= cnt[nums[i]]
i++
}
if cur >= k {
ans += i + 1
}
}
return int64(ans)
}