Massive Algorithms: Segment Tree | Set 1 (Sum of given range)

Segment Tree | Set 1 (Sum of given range) - GeeksforGeeks

Segment Tree | Set 1 (Sum of given range) - GeeksforGeeks
a segment tree is a tree data structure for storing intervals, or segments. It allows querying which of the stored segments contain a given point.

Segment Trees is a Tree data structure for storing intervals, or segments, It allows querying which of the stored segments contain a given point. It is, in principle, a static structure; that is, its content cannot be modified once the structure is built. It only uses O(N lg(N)) storage.
A similar data structure is the interval tree.

Order of growth of segment trees operations
build_tree: O(N lg(N))
update_tree: O(lg(N + k))
query_tree: O(lg(N + k))

We have an array arr[0 . . . n-1]. We should be able to
1 Find the sum of elements from index l to r where 0 <= l <= r <= n-1
2 Change value of a specified element of the array arr[i] = x where 0 <= i <= n-1.
Java code:
http://www.codebytes.in/2015/01/segment-tree-for-finding-sum-of-given.html
Lintcode https://codesolutiony.wordpress.com/2015/05/12/lintcode-interval-sum-ii/

Given an integer array in the construct method, implement two methods query(start, end) and modify(index, value):

For query(start, end), return the sum from index start to index end in the given array.
For modify(index, value), modify the number in the given index to value
Have you met this question in a real interview? Yes
Example
Given array A = [1,2,7,8,5].

query(0, 2), return 10.
modify(0, 4), change A[0] from 1 to 4.
query(0, 1), return 6.
modify(2, 1), change A[2] from 7 to 1.
query(2, 4), return 14.
Note
We suggest you finish problem Segment Tree Build, Segment Tree Query and Segment Tree Modify first.

public class Solution {

    class SumTreeNode {

        SumTreeNode left;

        SumTreeNode right;

        long sum;

        int start;

        int end;

}

    SumTreeNode root = null;

    public Solution(int[] A) {

        root = constructTree(A, 0, A.length - 1);

}

    private SumTreeNode constructTree(int[] a, int start, int end) {

        if (start > end) {

            return null;

}

        if (start == end) {

            return new SumTreeNode(start, start, (long)a[start]);

}

        int mid = (start + end) / 2;

        SumTreeNode root = new SumTreeNode(start, end);

        root.left = constructTree(a, start, mid);

        root.right = constructTree(a, mid + 1, end);

        long sum = 0;

        if (root.left != null) {

            sum += root.left.sum;

}

        if (root.right != null) {

            sum += root.right.sum;

}

        root.sum = sum;

        return root;

}

/**

     * @param start, end: Indices

     * @return: The sum from start to end

*/

    public long query(int start, int end) {

        return subQuery(root, start, end);

}

    public long subQuery(SumTreeNode root, int start, int end) {

        if (root == null || root.end < start || root.start > end) {

            return 0;

}

        if (root.start >= start && root.end <= end) {

            return root.sum;

}

        return subQuery(root.left, start, end) + subQuery(root.right, start, end);

}

/**

     * @param index, value: modify A[index] to value.

*/

    public void modify(int index, int value) {

        subModify(root, index, value);

}

    private void subModify(SumTreeNode root, int index, int value) {

        if (root == null || root.end < index || root.start > index) {

            return;

}

        if (root.start == root.end) {

            root.sum = value;

            return;

}

        if ((root.start + root.end) / 2 >= index) {

            subModify(root.left, index, value);

        } else {

            subModify(root.right, index, value);

}

        long sum = 0;

        if (root.left != null) {

            sum += root.left.sum;

}

        if (root.right != null) {

            sum += root.right.sum;

}

        root.sum = sum;

}

}

What if the number of query and updates are equal? Can we perform both the operations in O(log n) time once given the array? We can use a Segment Tree to do both operations in O(Logn) time.

Representation of Segment trees
1. Leaf Nodes are the elements of the input array.
2. Each internal node represents some merging of the leaf nodes. The merging may be different for different problems. For this problem, merging is sum of leaves under a node.

An array representation of tree is used to represent Segment Trees. For each node at index i, the left child is at index 2*i+1, right child at 2*i+2 and the parent is at (i-1)/2

Construction of Segment Tree from given array

We start with a segment arr[0 . . . n-1]. and every time we divide the current segment into two halves(if it has not yet become a segment of length 1), and then call the same procedure on both halves, and for each such segment we store the sum in corresponding node.

All levels of the constructed segment tree will be completely filled except the last level. Also, the tree will be a Full Binary Tree because we always divide segments in two halves at every level. Since the constructed tree is always full binary tree with n leaves, there will be n-1 internal nodes. So total number of nodes will be 2*n – 1.

Height of the segment tree will be $\lceil \log_2{n} \rceil$ . Since the tree is represented using array and relation between parent and child indexes must be maintained, size of memory allocated for segment tree will be $2 * 2 ^{\lceil \log_2{n} \rceil} - 1$ .

The first node will hold the information for the interval [i, j]
If i<j the left and right son will hold the information for the intervals [i, (i+j)/2] and [(i+j)/2+1, j]

Query for Sum of given range
Once the tree is constructed, how to get the sum using the constructed segment tree. Following is algorithm to get the sum of elements.

int getSum(node, l, r) 
{
   if range of node is within l and r
        return value in node
   else if range of node is completely outside l and r
        return 0
   else
    return getSum(node's left child, l, r) + 
           getSum(node's right child, l, r)
}

Update a value
Like tree construction and query operations, update can also be done recursively. We are given an index which needs to updated. Let diff be the value to be added. We start from root of the segment tree, and add diff to all nodes which have given index in their range. If a node doesn’t have given index in its range, we don’t make any changes to that node.

class SegmentTree 

{

    int st[]; // The array that stores segment tree nodes

    /* Constructor to construct segment tree from given array. This

       constructor  allocates memory for segment tree and calls

       constructSTUtil() to  fill the allocated memory */

    SegmentTree(int arr[], int n)

{

        // Allocate memory for segment tree

        //Height of segment tree

        int x = (int) (Math.ceil(Math.log(n) / Math.log(2)));

        //Maximum size of segment tree

        int max_size = 2 * (int) Math.pow(2, x) - 1;

        st = new int[max_size]; // Memory allocation

        constructSTUtil(arr, 0, n - 1, 0);

}

    // A utility function to get the middle index from corner indexes.

    int getMid(int s, int e) {

        return s + (e - s) / 2;

}

    /*  A recursive function to get the sum of values in given range

        of the array.  The following are parameters for this function.

      st    --> Pointer to segment tree

      si    --> Index of current node in the segment tree. Initially

                0 is passed as root is always at index 0

      ss & se  --> Starting and ending indexes of the segment represented

                    by current node, i.e., st[si]

      qs & qe  --> Starting and ending indexes of query range */

    int getSumUtil(int ss, int se, int qs, int qe, int si)

{

        // If segment of this node is a part of given range, then return

        // the sum of the segment

        if (qs <= ss && qe >= se)

            return st[si];

        // If segment of this node is outside the given range

        if (se < qs || ss > qe)

            return 0;

        // If a part of this segment overlaps with the given range

        int mid = getMid(ss, se);

        return getSumUtil(ss, mid, qs, qe, 2 * si + 1) +

                getSumUtil(mid + 1, se, qs, qe, 2 * si + 2);

}

    /* A recursive function to update the nodes which have the given 

       index in their range. The following are parameters

        st, si, ss and se are same as getSumUtil()

        i    --> index of the element to be updated. This index is in

                 input array.

       diff --> Value to be added to all nodes which have i in range */

    void updateValueUtil(int ss, int se, int i, int diff, int si)

{

        // Base Case: If the input index lies outside the range of 

        // this segment

        if (i < ss || i > se)

            return;

        // If the input index is in range of this node, then update the

        // value of the node and its children

        st[si] = st[si] + diff;

        if (se != ss) {

            int mid = getMid(ss, se);

            updateValueUtil(ss, mid, i, diff, 2 * si + 1);

            updateValueUtil(mid + 1, se, i, diff, 2 * si + 2);

}

}

    // The function to update a value in input array and segment tree.

   // It uses updateValueUtil() to update the value in segment tree

    void updateValue(int arr[], int n, int i, int new_val)

{

        // Check for erroneous input index

        if (i < 0 || i > n - 1) {

            System.out.println("Invalid Input");

            return;

}

        // Get the difference between new value and old value

        int diff = new_val - arr[i];

        // Update the value in array

        arr[i] = new_val;

        // Update the values of nodes in segment tree

        updateValueUtil(0, n - 1, i, diff, 0);

}

    // Return sum of elements in range from index qs (quey start) to

   // qe (query end).  It mainly uses getSumUtil()

    int getSum(int n, int qs, int qe)

{

        // Check for erroneous input values

        if (qs < 0 || qe > n - 1 || qs > qe) {

            System.out.println("Invalid Input");

            return -1;

}

        return getSumUtil(0, n - 1, qs, qe, 0);

}

    // A recursive function that constructs Segment Tree for array[ss..se].

    // si is index of current node in segment tree st

    int constructSTUtil(int arr[], int ss, int se, int si)

{

        // If there is one element in array, store it in current node of

        // segment tree and return

        if (ss == se) {

            st[si] = arr[ss];

            return arr[ss];

}

        // If there are more than one elements, then recur for left and

        // right subtrees and store the sum of values in this node

        int mid = getMid(ss, se);

        st[si] = constructSTUtil(arr, ss, mid, si * 2 + 1) +

                 constructSTUtil(arr, mid + 1, se, si * 2 + 2);

        return st[si];

}

}

Java Program to Implement Segment Tree
package org.codeexample.algorithms.collected.geometry;

public class SegmentTree {
private int[] tree;
private int maxsize;
private int height;

private final int STARTINDEX = 0;
private final int ENDINDEX;
private final int ROOT = 0;

public SegmentTree(int size) {
height = (int) (Math.ceil(Math.log(size) / Math.log(2)));
maxsize = 2 * (int) Math.pow(2, height) - 1;
tree = new int[maxsize];
ENDINDEX = size - 1;
}

private int leftchild(int pos) {
return 2 * pos + 1;
}

private int rightchild(int pos) {
return 2 * pos + 2;
}

private int mid(int start, int end) {
return (start + (end - start) / 2);
}

private int getSumUtil(int startIndex, int endIndex, int queryStart,
int queryEnd, int current) {
if (queryStart <= startIndex && queryEnd >= endIndex) {
return tree[current];
}
if (endIndex < queryStart || startIndex > queryEnd) {
return 0;
}
int mid = mid(startIndex, endIndex);
return getSumUtil(startIndex, mid, queryStart, queryEnd,
leftchild(current))
+ getSumUtil(mid + 1, endIndex, queryStart, queryEnd,
rightchild(current));
}

public int getSum(int queryStart, int queryEnd) {
if (queryStart < 0 || queryEnd > tree.length) {
return -1;
}
return getSumUtil(STARTINDEX, ENDINDEX, queryStart, queryEnd, ROOT);
}

private int constructSegmentTreeUtil(int[] elements, int startIndex,
int endIndex, int current) {
if (startIndex == endIndex) {
tree[current] = elements[startIndex];
return tree[current];
}
int mid = mid(startIndex, endIndex);
tree[current] = constructSegmentTreeUtil(elements, startIndex, mid,
leftchild(current))
+ constructSegmentTreeUtil(elements, mid + 1, endIndex,
rightchild(current));
return tree[current];
}

public void constructSegmentTree(int[] elements) {
constructSegmentTreeUtil(elements, STARTINDEX, ENDINDEX, ROOT);
}

private void updateTreeUtil(int startIndex, int endIndex, int updatePos,
int update, int current) {
if (updatePos < startIndex || updatePos > endIndex) {
return;
}
tree[current] = tree[current] + update;
if (startIndex != endIndex) {
int mid = mid(startIndex, endIndex);
updateTreeUtil(startIndex, mid, updatePos, update,
leftchild(current));
updateTreeUtil(mid + 1, endIndex, updatePos, update,
rightchild(current));
}
}

public void update(int update, int updatePos, int[] elements) {
int updatediff = update - elements[updatePos];
elements[updatePos] = update;
updateTreeUtil(STARTINDEX, ENDINDEX, updatePos, updatediff, ROOT);
}
}

Another solution is to create another array and store sum from start to i at the ith index in this array. Sum of a given range can now be calculated in O(1) time, but update operation takes O(n) time now. This works well if the number of query operations are large and very few updates.

Have a separate driver method, which will handle corner case, pre optimization(exit early), set init value for parameters etc.

Iterative Version:

http://ideone.com/zWJ0pq

http://se7so.blogspot.com/2012/12/segment-trees-and-lazy-propagation.html

https://www.youtube.com/watch?v=ZBHKZF5w4YU

http://letuskode.blogspot.com/2013/01/segtrees.html
Read full article from Segment Tree | Set 1 (Sum of given range) - GeeksforGeeks