题目
Given an array A of positive integers, call a (contiguous, not necessarily distinct) subarray of A good if the number of different integers in that subarray is exactly K.
(For example, [1,2,3,1,2] has 3 different integers: 1, 2, and 3.)
Return the number of good subarrays of A.
Example 1:
Input: A = [1,2,1,2,3], K = 2
Output: 7
Explanation: Subarrays formed with exactly 2 different integers: [1,2], [2,1], [1,2], [2,3], [1,2,1], [2,1,2], [1,2,1,2].
Example 2:
Input: A = [1,2,1,3,4], K = 3
Output: 3
Explanation: Subarrays formed with exactly 3 different integers: [1,2,1,3], [2,1,3], [1,3,4].
*思路
这个题是思考过程可以从暴力O(n^2)解法evolve到三指针的O(n)解法
暴力解法是首先用一个循环遍历数组的每个元素i
For every starting point i, try to find number of valid subarrays starting from i
例如[1,2,2], K = 2
i = 0时，我们可以得出的valid subarrays有[1,2]和[1,2,2]这2个
Do this for every i in array A, you'll get the answer
暴力解法的缺点是当正在遍历元素 i 时，你每次都得用另一个指针j从i的位置开始遍历后面的所有元素才能知道有哪些valid subarrays。所以时间复杂度是O(n^2)
可以看出每次遍历一个新的元素i，指针j总是得回撤到i的位置然后开始遍历后面的元素
双指针的算法多数都是利用求解答案的某些特性，让我们无需回撤指针从而达到降低一个级别的时间复杂度
例如[1,2,1,2,3], K = 2
i = 0时, valid subarrays有[1,2], [1,2,1], [1,2,1,2] (一共3个)
我们可以有一根指针left让它指向从i开始第一个遇到的valid subarray的末端(用一个dict1 keep track of number of distinct numbers in A[i...left], 如果len(dict1)等于K，则找到了left的位置)
用一根right指针指向最后一个遇到的valid subarray的末端(用一个dict2 keep track of number of distinct numbers in A[i...right], 如果len(dict2)等于K，而且right如果再向右移动就会导致len(dict2) > K)
在我们的这个例子input里，left = 1, right = 3
那么你会发现
number of valid subarrays starting from i is just equal to right - left + 1 = 3 - 1 + 1 = 3
这3个valid subarrays分别是A[i...left], A[i...left+1], A[i...right]
i = 1时
也需要找到相对于i = 1的对应的left和right指针，就可以计算出number of valid subarrays starting from i
当i = 1时，我们就不用考虑A[0]。A[0...left]本来是valid的，但是A[1...left]就不一定valid了。
所以left指针要向右边探，使得A[1...left]也变得valid。同理right指针要向右移动使得A[1...right]变得valid。
当重新计算出相对于i = 1的对应的left和right指针，我们又知道了number of valid subarrays starting from i = 1了
以此类推遍历完i = 0,1,2...n就行了
期间涉及到3个指针(i, left, right)，由于他们全程都只需要向右移动，直到数组结尾，所以时间复杂度是O(n)
答案
class Solution(object):
def subarraysWithKDistinct(self, A, K):
"""
:type A: List[int]
:type K: int
:rtype: int
"""
# window[number] -> occurrences of number
window1 = {}
window2 = {}
ans = 0
left, right = 0, 0
for i in range(len(A)):
# Move left pointer to closet place s.t [i...left] has exactly K distinct chars
# Move right pointer to as far as possible s.t [i...right] has exactly K distinct chars
while left < len(A):
if len(window1) == K:
break
window1[A[left]] = window1.get(A[left], 0) + 1
left += 1
if len(window1) < K:
continue
while right < len(A):
if len(window2) == K:
break
window2[A[right]] = window2.get(A[right], 0) + 1
right += 1
while right < len(A):
if A[right] not in window2:
break
else:
window2[A[right]] = window2.get(A[right], 0) + 1
right += 1
ans += (right - left + 1)
window1[A[i]] -= 1
if window1[A[i]] == 0:
window1.pop(A[i])
window2[A[i]] -= 1
if window2[A[i]] == 0:
window2.pop(A[i])
return ans