LeetCode 1635: Hopper Company Queries I Solution

Master LeetCode problem 1635 (Hopper Company Queries I), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1635. Hopper Company Queries I

Hard

Database

Problem Explanation

Explanation:

To solve this problem, we can use a segment tree data structure. The segment tree will help us efficiently answer queries of finding the maximum value in a range. We will build the segment tree based on the array of weights of the nodes in the tree. Each node in the segment tree will represent a range of the original array.

Algorithmic Steps:

Build the segment tree recursively:
- Initialize the segment tree with an array of size 2*N-1 (where N is the number of weights).
- For each node in the segment tree, calculate the value based on its children.
- The leaf nodes will store the weights of the original array, and the parent nodes will store the maximum weight within their range.
Perform queries:
- For each query, find the range in the segment tree that corresponds to the given node range.
- Query the segment tree to find the maximum weight within that range.

Time Complexity:

Building the segment tree: O(N)
Answering each query: O(log N)

Space Complexity:

Segment tree: O(N)

Solution Code

class HopperCompanyQueries {
    int[] segmentTree;
    int n;

    public HopperCompanyQueries(int[] weights) {
        n = weights.length;
        segmentTree = new int[2 * n - 1];
        buildSegmentTree(weights, 0, 0, n - 1);
    }

    private void buildSegmentTree(int[] weights, int index, int left, int right) {
        if (left == right) {
            segmentTree[index] = weights[left];
        } else {
            int mid = left + (right - left) / 2;
            buildSegmentTree(weights, 2 * index + 1, left, mid);
            buildSegmentTree(weights, 2 * index + 2, mid + 1, right);
            segmentTree[index] = Math.max(segmentTree[2 * index + 1], segmentTree[2 * index + 2]);
        }
    }

    public int query(int start, int end) {
        return queryHelper(0, 0, n - 1, start, end);
    }

    private int queryHelper(int index, int left, int right, int start, int end) {
        if (right < start || left > end) {
            return Integer.MIN_VALUE;
        }
        if (start <= left && right <= end) {
            return segmentTree[index];
        }
        int mid = left + (right - left) / 2;
        int leftMax = queryHelper(2 * index + 1, left, mid, start, end);
        int rightMax = queryHelper(2 * index + 2, mid + 1, right, start, end);
        return Math.max(leftMax, rightMax);
    }
}

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