2762 - Continuous Subarrays

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

Solution 1: Ordered List + Two Pointers

C++

• 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;
      }
  };

JAVA

• 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;
      }
  }

PYTHON

• 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;
      }
  };