Description
Given an array of integers nums
and an integer k
. A continuous subarray is called nice if there are k
odd numbers on it.
Return the number of nice sub-arrays.
Example 1:
Input: nums = [1,1,2,1,1], k = 3 Output: 2 Explanation: The only sub-arrays with 3 odd numbers are [1,1,2,1] and [1,2,1,1].
Example 2:
Input: nums = [2,4,6], k = 1 Output: 0 Explanation: There is no odd numbers in the array.
Example 3:
Input: nums = [2,2,2,1,2,2,1,2,2,2], k = 2 Output: 16
Constraints:
1 <= nums.length <= 50000
1 <= nums[i] <= 10^5
1 <= k <= nums.length
Code
The atMost Trick
和 Binary Subarrays With Sum 、Subarrays with K Different Integers、Count Vowel Substrings of a String 一樣,要求得剛好有 k
個 odd number 的 sub-arrays,我們可以用至多 k
個 odd number 的 sub-arrays 的數量去減掉至多 k - 1
個 odd number 的 sub-arrays 的數量。
Time Complexity: , Space Complexity:
class Solution {
public:
int numberOfSubarrays(vector<int>& nums, int k) {
return atMost(nums, k) - atMost(nums, k - 1);
}
int atMost(vector<int>& nums, int k) {
int i = 0, res = 0;
int oddCount = 0;
for(int j = 0; j < nums.size(); j++) {
if(nums[j] % 2 != 0) oddCount++;
while(oddCount > k) {
if(nums[i++] % 2 != 0) oddCount--;
}
res += j - i + 1;
}
return res;
}
};