LeetCode 1635: Hopper Company Queries I
LeetCode 1635 Solution 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)
:
LeetCode 1635 Solutions in Java, C++, Python
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);
}
}Interactive Code Editor for LeetCode 1635
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