[Solved] Count Subarrays With Score Less Than K LeetCode Contest Problem

Count Subarrays With Score Less Than K: The score of an array is defined as the product of its sum and its length.

• For example, the score of [1, 2, 3, 4, 5] is (1 + 2 + 3 + 4 + 5) * 5 = 75.

Given a positive integer array nums and an integer k, return the number of non-empty subarrays of nums whose score is strictly less than k.

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

Example 1:

Input: nums = [2,1,4,3,5], k = 10
Output: 6
Explanation:
The 6 subarrays having scores less than 10 are:
- [2] with score 2 * 1 = 2.
- [1] with score 1 * 1 = 1.
- [4] with score 4 * 1 = 4.
- [3] with score 3 * 1 = 3. 
- [5] with score 5 * 1 = 5.
- [2,1] with score (2 + 1) * 2 = 6.
Note that subarrays such as [1,4] and [4,3,5] are not considered because their scores are 10 and 36 respectively, while we need scores strictly less than 10.

Example 2:

Input: nums = [1,1,1], k = 5
Output: 5
Explanation:
Every subarray except [1,1,1] has a score less than 5.
[1,1,1] has a score (1 + 1 + 1) * 3 = 9, which is greater than 5.
Thus, there are 5 subarrays having scores less than 5.

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 105
• 1 <= k <= 1015

Solution:

Subarrays + Less Than K = Slide Window

Explanation

Maintain a sliding window from A[i] to A[j], with a score less than k.
cur counts the current sum of elements between the windows.

Iterate j from 0 to A.length - 1,
each round we firstly add A[j] to the window,
update cur += A[j].

The current sum is cur and length is j - i + 1,
If the score cur * (j - i + 1) >= k,
the window is too big,
we remove A[i] and update i++.
We continue doing this until the score is less than k.

Then we find the smallest i where subarray A[i] to A[j] has scored less than k.
For any subarray ending at A[j] with a shorter length, they have scored less than k.
There are j – i + 1 subarrays in total, so we update the result res += j - i + 1

Complexity

Time O(n)
Space O(1)

Runtime: 3 ms, faster than 100.00% of Java online submissions for Count Subarrays With Score Less Than K.

Memory Usage: 52 MB, less than 83.33% of Java online submissions for Count Subarrays With Score Less Than K.

    public long countSubarrays(int[] A, long k) {
        long res = 0, cur = 0;
        for (int j = 0, i = 0; j < A.length; ++j) {
            cur += A[j];
            while (cur * (j - i + 1) >= k)
                cur -= A[i++];
            res += j - i + 1;
        }
        return res;
    }

    long long countSubarrays(vector<int>& A, long long k) {
        long long res = 0, cur = 0;
        for (int j = 0, i = 0; j < A.size(); ++j) {
            cur += A[j];
            while (cur * (j - i + 1) >= k)
                cur -= A[i++];
            res += j - i + 1;
        }
        return res;
    }

    def countSubarrays(self, A, k):
        res = cur = i = 0
        for j in range(len(A)):
            cur += A[j]
            while cur * (j - i + 1) >= k:
                cur -= A[i]
                i += 1
            res += j - i + 1
        return res

class Solution {
    typedef long long ll;
    typedef long double ld;
    typedef vector<ll> vi;
    typedef pair<ll, ll> pi;
#define endl '\n'
    static const ll mod = 1e9;
public:
    long long countSubarrays(vector<int>& v, long long k) {
        ll sz = v.size(), res = 0;
        vi pre(sz);
        pre[0] = v[0];
        for (ll i = 1;i < sz;i++)
            pre[i] = pre[i - 1] + v[i];
        for (ll i = 0;i < sz;++i) {
            ll l = i,
                r = sz - 1,
                ind = -1;
            while (l <= r) {
                ll mid = l + ((r - l) >> 1);
                ll score = pre[mid] - (i > 0 ? pre[i - 1] : 0);
                score *= mid - i + 1;
                if (score < k) {
                    ind = mid;
                    l = mid + 1;
                    continue;
                }
                r = mid - 1;
            }
            if (ind != -1)
                res += ind - i + 1;
        }
        return res;
    }
};

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