Welcome to Subscribe On Youtube

2762. Continuous Subarrays

Description

You are given a 0-indexed integer array nums. A subarray of nums is called continuous if:

  • Let i, i + 1, ..., j be the indices in the subarray. Then, for each pair of indices i <= i1, i2 <= j, 0 <= |nums[i1] - nums[i2]| <= 2.

Return the total number of continuous subarrays.

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

 

Example 1:

Input: nums = [5,4,2,4]
Output: 8
Explanation: 
Continuous subarray of size 1: [5], [4], [2], [4].
Continuous subarray of size 2: [5,4], [4,2], [2,4].
Continuous subarray of size 3: [4,2,4].
Thereare no subarrys of size 4.
Total continuous subarrays = 4 + 3 + 1 = 8.
It can be shown that there are no more continuous subarrays.

 

Example 2:

Input: nums = [1,2,3]
Output: 6
Explanation: 
Continuous subarray of size 1: [1], [2], [3].
Continuous subarray of size 2: [1,2], [2,3].
Continuous subarray of size 3: [1,2,3].
Total continuous subarrays = 3 + 2 + 1 = 6.

 

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 109

Solutions

Solution 1: Ordered List + Two Pointers

We can use two pointers, $i$ and $j$, to maintain the left and right endpoints of the current subarray, and use an ordered list to maintain all elements in the current subarray.

Iterate through the array $nums$. For the current number $nums[i]$ we’re iterating over, we add it to the ordered list. If the difference between the maximum and minimum values in the ordered list is greater than $2$, we then loop to move the pointer $i$ to the right, continuously removing $nums[i]$ from the ordered list, until the list is empty or the maximum difference between elements in the ordered list is not greater than $2$. At this point, the number of uninterrupted subarrays is $j - i + 1$, which we add to the answer.

After the iteration, return the answer.

The time complexity is $O(n \times \log n)$ and the space complexity is $O(n)$, where $n$ is the length of the array $nums$.

  • class Solution {
    public long continuousSubarrays(int[] nums) {
        long ans = 0;
        int i = 0, n = nums.length;
        TreeMap<Integer, Integer> tm = new TreeMap<>();
        for (int j = 0; j < n; ++j) {
            tm.merge(nums[j], 1, Integer::sum);
            while (tm.lastEntry().getKey() - tm.firstEntry().getKey() > 2) {
                tm.merge(nums[i], -1, Integer::sum);
                if (tm.get(nums[i]) == 0) {
                    tm.remove(nums[i]);
                }
                ++i;
            }
            ans += j - i + 1;
        }
        return ans;
    }
}

  • class Solution {
public:
    long long continuousSubarrays(vector<int>& nums) {
        long long ans = 0;
        int i = 0, n = nums.size();
        multiset<int> s;
        for (int j = 0; j < n; ++j) {
            s.insert(nums[j]);
            while (*s.rbegin() - *s.begin() > 2) {
                s.erase(s.find(nums[i++]));
            }
            ans += j - i + 1;
        }
        return ans;
    }
};

  • from sortedcontainers import SortedList


class Solution:
    def continuousSubarrays(self, nums: List[int]) -> int:
        ans = i = 0
        sl = SortedList()
        for x in nums:
            sl.add(x)
            while sl[-1] - sl[0] > 2:
                sl.remove(nums[i])
                i += 1
            ans += len(sl)
        return ans


  • func continuousSubarrays(nums []int) (ans int64) {
	i := 0
	tm := treemap.NewWithIntComparator()
	for j, x := range nums {
		if v, ok := tm.Get(x); ok {
			tm.Put(x, v.(int)+1)
		} else {
			tm.Put(x, 1)
		}
		for {
			a, _ := tm.Min()
			b, _ := tm.Max()
			if b.(int)-a.(int) > 2 {
				if v, _ := tm.Get(nums[i]); v.(int) == 1 {
					tm.Remove(nums[i])
				} else {
					tm.Put(nums[i], v.(int)-1)
				}
				i++
			} else {
				break
			}
		}
		ans += int64(j - i + 1)
	}
	return
}

All Problems

All Solutions