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Formatted question description: https://leetcode.ca/all/1343.html

1343. Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold (Medium)

Given an array of integers arr and two integers k and threshold.

Return the number of sub-arrays of size k and average greater than or equal to threshold.

 

Example 1:

Input: arr = [2,2,2,2,5,5,5,8], k = 3, threshold = 4
Output: 3
Explanation: Sub-arrays [2,5,5],[5,5,5] and [5,5,8] have averages 4, 5 and 6 respectively. All other sub-arrays of size 3 have averages less than 4 (the threshold).

Example 2:

Input: arr = [1,1,1,1,1], k = 1, threshold = 0
Output: 5

Example 3:

Input: arr = [11,13,17,23,29,31,7,5,2,3], k = 3, threshold = 5
Output: 6
Explanation: The first 6 sub-arrays of size 3 have averages greater than 5. Note that averages are not integers.

Example 4:

Input: arr = [7,7,7,7,7,7,7], k = 7, threshold = 7
Output: 1

Example 5:

Input: arr = [4,4,4,4], k = 4, threshold = 1
Output: 1

 

Constraints:

  • 1 <= arr.length <= 10^5
  • 1 <= arr[i] <= 10^4
  • 1 <= k <= arr.length
  • 0 <= threshold <= 10^4

Related Topics:
Array

Solution 1.

// OJ: https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int numOfSubarrays(vector<int>& arr, int k, int threshold) {
        int sum = 0, cnt = 0;
        for (int i = 0; i < arr.size(); ++i) {
            sum += arr[i];
            if (i >= k) sum -= arr[i - k];
            if (i + 1 >= k && sum / k >= threshold) ++cnt;
        }
        return cnt;
    }
};

  • class Solution {
    public int numOfSubarrays(int[] arr, int k, int threshold) {
        int count = 0;
        int sumThreshold = threshold * k;
        int sum = 0;
        for (int i = 0; i < k; i++)
            sum += arr[i];
        if (sum >= sumThreshold)
            count++;
        int length = arr.length;
        for (int i = k; i < length; i++) {
            sum -= arr[i - k];
            sum += arr[i];
            if (sum >= sumThreshold)
                count++;
        }
        return count;
    }
}

############

class Solution {
    public int numOfSubarrays(int[] arr, int k, int threshold) {
        int s = 0;
        for (int i = 0; i < k; ++i) {
            s += arr[i];
        }
        int ans = s / k >= threshold ? 1 : 0;
        for (int i = k; i < arr.length; ++i) {
            s += arr[i] - arr[i - k];
            ans += s / k >= threshold ? 1 : 0;
        }
        return ans;
    }
}

  • // OJ: https://leetcode.com/problems/number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int numOfSubarrays(vector<int>& arr, int k, int threshold) {
        int sum = 0, cnt = 0;
        for (int i = 0; i < arr.size(); ++i) {
            sum += arr[i];
            if (i >= k) sum -= arr[i - k];
            if (i + 1 >= k && sum / k >= threshold) ++cnt;
        }
        return cnt;
    }
};

  • class Solution:
    def numOfSubarrays(self, arr: List[int], k: int, threshold: int) -> int:
        s = sum(arr[:k])
        ans = int(s / k >= threshold)
        for i in range(k, len(arr)):
            s += arr[i]
            s -= arr[i - k]
            ans += int(s / k >= threshold)
        return ans


  • func numOfSubarrays(arr []int, k int, threshold int) (ans int) {
	s := 0
	for _, x := range arr[:k] {
		s += x
	}
	if s/k >= threshold {
		ans++
	}
	for i := k; i < len(arr); i++ {
		s += arr[i] - arr[i-k]
		if s/k >= threshold {
			ans++
		}
	}
	return
}

  • function numOfSubarrays(arr: number[], k: number, threshold: number): number {
    let s = arr.slice(0, k).reduce((acc, cur) => acc + cur, 0);
    let ans = s >= k * threshold ? 1 : 0;
    for (let i = k; i < arr.length; ++i) {
        s += arr[i] - arr[i - k];
        ans += s >= k * threshold ? 1 : 0;
    }
    return ans;
}

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