Welcome to Subscribe On Youtube

2913. Subarrays Distinct Element Sum of Squares I

Description

You are given a 0-indexed integer array nums.

The distinct count of a subarray of nums is defined as:

  • Let nums[i..j] be a subarray of nums consisting of all the indices from i to j such that 0 <= i <= j < nums.length. Then the number of distinct values in nums[i..j] is called the distinct count of nums[i..j].

Return the sum of the squares of distinct counts of all subarrays of nums.

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

 

Example 1:

Input: nums = [1,2,1]
Output: 15
Explanation: Six possible subarrays are:
[1]: 1 distinct value
[2]: 1 distinct value
[1]: 1 distinct value
[1,2]: 2 distinct values
[2,1]: 2 distinct values
[1,2,1]: 2 distinct values
The sum of the squares of the distinct counts in all subarrays is equal to 12 + 12 + 12 + 22 + 22 + 22 = 15.

Example 2:

Input: nums = [1,1]
Output: 3
Explanation: Three possible subarrays are:
[1]: 1 distinct value
[1]: 1 distinct value
[1,1]: 1 distinct value
The sum of the squares of the distinct counts in all subarrays is equal to 12 + 12 + 12 = 3.

 

Constraints:

  • 1 <= nums.length <= 100
  • 1 <= nums[i] <= 100

Solutions

Solution 1: Enumeration

We can enumerate the left endpoint index $i$ of the subarray, and for each $i$, we enumerate the right endpoint index $j$ in the range $[i, n)$, and calculate the distinct count of $nums[i..j]$ by adding the count of $nums[j]$ to a set $s$, and then taking the square of the size of $s$ as the contribution of $nums[i..j]$ to the answer.

After the enumeration, we return the answer.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

  • class Solution {
    public int sumCounts(List<Integer> nums) {
        int ans = 0;
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            int[] s = new int[101];
            int cnt = 0;
            for (int j = i; j < n; ++j) {
                if (++s[nums.get(j)] == 1) {
                    ++cnt;
                }
                ans += cnt * cnt;
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    int sumCounts(vector<int>& nums) {
        int ans = 0;
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            int s[101]{};
            int cnt = 0;
            for (int j = i; j < n; ++j) {
                if (++s[nums[j]] == 1) {
                    ++cnt;
                }
                ans += cnt * cnt;
            }
        }
        return ans;
    }
};

  • class Solution:
    def sumCounts(self, nums: List[int]) -> int:
        ans, n = 0, len(nums)
        for i in range(n):
            s = set()
            for j in range(i, n):
                s.add(nums[j])
                ans += len(s) * len(s)
        return ans


  • func sumCounts(nums []int) (ans int) {
	for i := range nums {
		s := [101]int{}
		cnt := 0
		for _, x := range nums[i:] {
			s[x]++
			if s[x] == 1 {
				cnt++
			}
			ans += cnt * cnt
		}
	}
	return
}

  • function sumCounts(nums: number[]): number {
    let ans = 0;
    const n = nums.length;
    for (let i = 0; i < n; ++i) {
        const s: number[] = Array(101).fill(0);
        let cnt = 0;
        for (const x of nums.slice(i)) {
            if (++s[x] === 1) {
                ++cnt;
            }
            ans += cnt * cnt;
        }
    }
    return ans;
}

All Problems

All Solutions