Range Max Query

Problem

You are tasked with implementing a segment tree using Java and Python to calculate the maximum value in a given range [left, right] of an array. A segment tree allows both querying and updating efficiently.

The main operations are:

Build: Construct the segment tree from a given array.
Update: Modify a specific index in the array and update the segment tree accordingly.
Query: Find the maximum value in a given range [left, right].

Examples

Example 1

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Input: [2, 6, 10, 4, 7, 28, 9, 11, 6, 33]
Query Range: [0, 5]
Output: 28
Explanation: The maximum value in range `[0, 5]` (elements: `[2, 6, 10, 4, 7, 28]`) is 28.

Example 2

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Input: [2, 6, 10, 4, 7, 28, 9, 11, 6, 33]
Change index `3` to `40`
Query Range: [2, 6]
Output: 40
Explanation: After updating the element at index `3` to `40`, the maximum value in range `[2, 6]` (elements: `[10, 40, 7, 28, 9]`) is 40.

Solution

Method 1 - Using the segment tree

A segment tree is an efficient way to perform range queries such as maximum, minimum, or summation operations. The tree is built such that:

Each leaf node contains an element of the array.
Internal nodes represent aggregate information over their child nodes (in this case, the maximum value).

Key insight:

A query for the maximum in a range [left, right] moves through specialised paths divided into segments within the tree hierarchy.

Approach

Tree Representation:
- Use a size of 2 * n to store both leaves and internal nodes.
- The leaves of the segment tree contain the original array values.
Build:
- Populate the leaves with array values.
- Compute the maximum values for internal nodes, moving from the bottom up.
Update:
- Update values at the leaf node corresponding to the array index.
- Propagate the changes upwards, recalculating maximums along the way.
Query:
- Traverse the range [left, right] within the tree.
- Aggregate the maximum values as you traverse segments.

Code

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public class SegmentTreeMaxQuery {

    private final int n;
    private final int[] segTree;

    public SegmentTreeMaxQuery(int n) {
        this.n = n;
        segTree = new int[2 * n];
    }

    void build(int[] a) {
        for (int i = 0; i < n; i++) {
            segTree[n + i] = a[i];
        }

        for (int i = n - 1; i >= 1; i--) {
            segTree[i] = Math.max(segTree[2 * i], segTree[2 * i + 1]);
        }
    }

    void update(int pos, int value) {
        pos += n;
        segTree[pos] = value;

        while (pos > 1) {
            pos >>= 1;
            segTree[pos] = Math.max(segTree[2 * pos], segTree[2 * pos + 1]);
        }
    }

    int query(int left, int right) {
        left += n;
        right += n;
        int maxVal = Integer.MIN_VALUE;

        while (left < right) {
            if ((left & 1) == 1) {
                maxVal = Math.max(maxVal, segTree[left++]);
            }
            if ((right & 1) == 1) {
                maxVal = Math.max(maxVal, segTree[--right]);
            }
            left >>= 1;
            right >>= 1;
        }
        return maxVal;
    }

    public static void main(String[] args) {
        int[] a = { 2, 6, 10, 4, 7, 28, 9, 11, 6, 33 };
        SegmentTreeMaxQuery st = new SegmentTreeMaxQuery(a.length);

        st.build(a);

        System.out.printf(
            "Maximum in range %d to %d is %d\n",
            0,
            5,
            st.query(0, 6)
        );

        st.update(3, 40);

        System.out.printf(
            "Maximum in range %d to %d is %d\n",
            2,
            6,
            st.query(2, 7)
        );
    }
}

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class SegmentTreeMaxQuery:
    def __init__(self, n):
        self.n = n
        self.seg_tree = [0] * (2 * n)

    def build(self, a):
        for i in range(self.n):
            self.seg_tree[self.n + i] = a[i]
        
        for i in range(self.n - 1, 0, -1):
            self.seg_tree[i] = max(self.seg_tree[2 * i], self.seg_tree[2 * i + 1])

    def update(self, pos, value):
        pos += self.n
        self.seg_tree[pos] = value

        while pos > 1:
            pos //= 2
            self.seg_tree[pos] = max(self.seg_tree[2 * pos], self.seg_tree[2 * pos + 1])

    def query(self, left, right):
        left += self.n
        right += self.n
        max_val = float('-inf')

        while left < right:
            if left % 2 == 1:
                max_val = max(max_val, self.seg_tree[left])
                left += 1
            if right % 2 == 1:
                right -= 1
                max_val = max(max_val, self.seg_tree[right])
            
            left //= 2
            right //= 2
        return max_val

# Driver Code
a = [2, 6, 10, 4, 7, 28, 9, 11, 6, 33]
st = SegmentTreeMaxQuery(len(a))

st.build(a)
print(f"Maximum in range 0 to 5 is {st.query(0, 6)}")

st.update(3, 40)
print(f"Maximum in range 2 to 6 is {st.query(2, 7)}")

Complexity

⏰ Time complexity
- Tree Build: O(n) (populate leaves and compute internal nodes).
- Update: O(log n) (move up the tree for updates).
- Query: O(log n) (traverse log levels in the tree).
🧺 Space complexity: O(n)

Problem#

Examples#

Example 1#

Example 2#

Solution#

Method 1 - Using the segment tree#

Approach#

Code#

Complexity#