What is the time complexity of this segment tree algorithm?

Question

class SegTreeNode {
public:
  int start;
  int end;
  int min;
  SegTreeNode *left;
  SegTreeNode *right;
  SegTreeNode(int start, int end) {
    this->start = start;
    this->end = end;
    left = right = NULL;
  }
};

class Solution {
public:
  int largestRectangleArea(vector<int>& heights) {
    if (heights.size() == 0) return 0;
    // first build a segment tree
    SegTreeNode *root = buildSegmentTree(heights, 0, heights.size() - 1);
    // next calculate the maximum area recursively
    return calculateMax(heights, root, 0, heights.size() - 1);
  }
  
  int calculateMax(vector<int>& heights, SegTreeNode* root, int start, int end) {
    if (start > end) {
      return -1;
    }
    if (start == end) {
      return heights[start];
    }
    int minIndex = query(root, heights, start, end);
    int leftMax = calculateMax(heights, root, start, minIndex - 1);
    int rightMax = calculateMax(heights, root, minIndex + 1, end);
    int minMax = heights[minIndex] * (end - start + 1);
    return max( max(leftMax, rightMax), minMax );
  }
  
  SegTreeNode *buildSegmentTree(vector<int>& heights, int start, int end) {
    if (start > end) return NULL;
    SegTreeNode *root = new SegTreeNode(start, end);
    if (start == end) {
        root->min = start;
      return root;
    } else {
      int middle = (start + end) / 2;
      root->left = buildSegmentTree(heights, start, middle);
      root->right = buildSegmentTree(heights, middle + 1, end);
      root->min = heights[root->left->min] < heights[root->right->min] ? root->left->min : root->right->min;
      return root;
    }
  }
  
  int query(SegTreeNode *root, vector<int>& heights, int start, int end) {
    if (root == NULL || end < root->start || start > root->end) return -1;
    if (start <= root->start && end >= root->end) {
      return root->min;
    }
    int leftMin = query(root->left, heights, start, end);
    int rightMin = query(root->right, heights, start, end);
    if (leftMin == -1) return rightMin;
    if (rightMin == -1) return leftMin;
    return heights[leftMin] < heights[rightMin] ? leftMin : rightMin;
  }
};

This is a solution to this leet code problem https://leetcode.com/problems/largest-rectangle-in-histogram/

I believe the following code should have an average time complexity of logN * logN and a worst case time complexity of NlogN

My reasoning is that recursion of calculateMax will take logN on average, and will only deteriorate to N in a worst case scenario where the height of the rectangles in the array are in ascending or descending sorted order.

I just want to know if my logic is correct or not, and would appreciate some type of confirmation.

Thanks :)

Answer 1

Your reasoning is right. But why did you complicate the code so much by using Segment tree. It can be done using stacks also.

Refer this: maximize-the-rectangular-area-under-histogram