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2104. Sum of Subarray Ranges

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https://leetcode.com/problems/sum-of-subarray-ranges/


You are given an integer array nums. The range of a subarray of nums is the difference between the largest and smallest element in the subarray.

Return the sum of all subarray ranges of nums.

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

 

Example 1:

Input: nums = [1,2,3]
Output: 4
Explanation: The 6 subarrays of nums are the following:
[1], range = largest - smallest = 1 - 1 = 0 
[2], range = 2 - 2 = 0
[3], range = 3 - 3 = 0
[1,2], range = 2 - 1 = 1
[2,3], range = 3 - 2 = 1
[1,2,3], range = 3 - 1 = 2
So the sum of all ranges is 0 + 0 + 0 + 1 + 1 + 2 = 4.

Example 2:

Input: nums = [1,3,3]
Output: 4
Explanation: The 6 subarrays of nums are the following:
[1], range = largest - smallest = 1 - 1 = 0
[3], range = 3 - 3 = 0
[3], range = 3 - 3 = 0
[1,3], range = 3 - 1 = 2
[3,3], range = 3 - 3 = 0
[1,3,3], range = 3 - 1 = 2
So the sum of all ranges is 0 + 0 + 0 + 2 + 0 + 2 = 4.

Example 3:

Input: nums = [4,-2,-3,4,1]
Output: 59
Explanation: The sum of all subarray ranges of nums is 59.

 

Constraints:

  • 1 <= nums.length <= 1000
  • -109 <= nums[i] <= 109

 

Follow-up: Could you find a solution with O(n) time complexity?

관련자료

댓글 2

Coffee님의 댓글

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class Solution {
    public long subArrayRanges(int[] nums) {
        
        // find max
        // find min
        int min = nums[0];
        int max = nums[0];
        
        long accSum = 0;
        
        for(int i=0; i<nums.length; i++){    
            min = nums[i];
            max = min;
            for(int j=i+1; j<nums.length; j++){
                max = Math.max(max, nums[j]);
                min = Math.min(min, nums[j]);
                accSum += (max - min);
            }   
        }
        return accSum;
        
    }
}

austin님의 댓글

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O(n) linear solution
class Solution {
public:
    long long subArrayRanges(vector<int>& nums) {
        auto check = [&nums](auto &s, auto r, auto cmp) {
            long long ret = 0;
            while(!s.empty() && (r == nums.size() || cmp(nums[s.top()], nums[r]))) {
                auto m = s.top(); s.pop();
                ret += (m - (s.empty() ? -1 :s.top())) * (r - m) * nums[m];                
            }  
            s.emplace(r);
            return ret;
        };        
        long long ret = 0LL;
        stack<long long> ms, ls;
        for(long long r = 0; r <= nums.size(); ++r) {
            ret -= check(ls, r, greater<int>());
            ret += check(ms, r, less<int>());
        }
        return ret;
    }
};
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