You are given an integer array nums
and a positive integer k
.
Return the number of subarrays where the maximum element of nums
appears at least k
times in that subarray.
A subarray is a contiguous sequence of elements within an array.
Example 1:
Input: nums = [1,3,2,3,3], k = 2 Output: 6 Explanation: The subarrays that contain the element 3 at least 2 times are: [1,3,2,3], [1,3,2,3,3], [3,2,3], [3,2,3,3], [2,3,3] and [3,3].
Example 2:
Input: nums = [1,4,2,1], k = 3 Output: 0 Explanation: No subarray contains the element 4 at least 3 times.
Constraints:
1 <= nums.length <= 105
1 <= nums[i] <= 106
1 <= k <= 105
// OJ: https://leetcode.com/problems/count-subarrays-where-max-element-appears-at-least-k-times
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
long long countSubarrays(vector<int>& A, int k) {
long long N = A.size(), ans = 0, cnt = 0, mx = *max_element(begin(A), end(A));
unordered_map<int, int> m; // mapping from A[i] to the index of its first occurrence
for (int i = 0; i < N; ++i) {
cnt += A[i] == mx;
if (cnt && m.count(cnt) == 0) m[cnt] = i; // Store the index of its first occurrence
if (m.count(cnt - k + 1)) {
ans += m[cnt - k + 1] + 1; // add 1 + frequency of the first occurrence of cnt-k+1
}
}
return ans;
}
};
// OJ: https://leetcode.com/problems/count-subarrays-where-max-element-appears-at-least-k-times
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
long long countSubarrays(vector<int>& A, int k) {
long long N = A.size(), ans = 0, cnt = 0, mx = *max_element(begin(A), end(A)), i = 0, j = 0;
for (; j < N; ++j) {
cnt += A[j] == mx;
while (cnt >= k) cnt -= A[i++] == mx;
ans += i;
}
return ans;
}
};