Iterative Segment Tree (Range Maximum Query with Node Update)

Last Updated : 16 Jan, 2023

Given an array arr[0 . . . n-1]. The task is to perform the following operation:

Find the maximum of elements from index l to r where 0 <= l <= r <= n-1.
Change value of a specified element of the array to a new value x. Given i and x, change A[i] to x, 0 <= i <= n-1.

Examples:

Input: a[] = {2, 6, 7, 5, 18, 86, 54, 2}
Query1: maximum(2, 7)
Query2: update(3, 90)
Query3: maximum(2, 6)

Output:
Maximum in range 2 to 7 is 86.
Maximum in range 2 to 6 is 90.

We have discussed Recursive segment tree implementation. In this post, iterative implementation is discussed.

The iterative version of the segment tree basically uses the fact, that for an index i, left child = 2 * i and right child = 2 * i + 1 in the tree. The parent for an index i in the segment tree array can be found by parent = i / 2. Thus we can easily travel up and down through the levels of the tree one by one. At first we compute the maximum in the ranges while constructing the tree starting from the leaf nodes and climbing up through the levels one by one. We use the same concept while processing the queries for finding the maximum in a range.

Since there are (log n) levels in the worst case, so querying takes log n time. For update of a particular index to a given value we start updating the segment tree starting from the leaf nodes and update all those nodes which are affected by the updation of the current node by gradually moving up through the levels at every iteration. Updation also takes log n time because there we have to update all the levels starting from the leaf node where we update the exact value at the exact index given by the user.

Below is the implementation of the above approach.

C++

// C++ Program to implement
// iterative segment tree.
#include <bits/stdc++.h>
using namespace std;

void construct_segment_tree(vector<int>& segtree,
                            vector<int>& a, int n)
{
    // assign values to leaves of the segment tree
    for (int i = 0; i < n; i++)
        segtree[n + i] = a[i];

    /* assign values to internal nodes
    to compute maximum in a given range */
    for (int i = n - 1; i >= 1; i--)
        segtree[i] = max(segtree[2 * i],
                         segtree[2 * i + 1]);
}

void update(vector<int>& segtree, int pos, int value,
            int n)
{
    // change the index to leaf node first
    pos += n;

    // update the value at the leaf node
    // at the exact index
    segtree[pos] = value;

    while (pos > 1) {

        // move up one level at a time in the tree
        pos >>= 1;

        // update the values in the nodes in
        // the next higher level
        segtree[pos] = max(segtree[2 * pos],
                           segtree[2 * pos + 1]);
    }
}

int range_query(vector<int>& segtree, int left, int
                                                    right,
                int n)
{
    /* Basically the left and right indices will move
        towards right and left respectively and with
        every each next higher level and compute the 
        maximum at each height. */
    // change the index to leaf node first
    left += n;
    right += n;

    // initialize maximum to a very low value
    int ma = INT_MIN;

    while (left < right) {

        // if left index in odd
        if (left & 1) {
            ma = max(ma, segtree[left]);

            // make left index even
            left++;
        }

        // if right index in odd
        if (right & 1) {

            // make right index even
            right--;

            ma = max(ma, segtree[right]);
        }

        // move to the next higher level
        left /= 2;
        right /= 2;
    }
    return ma;
}

// Driver code
int main()
{
    vector<int> a = { 2, 6, 10, 4, 7, 28, 9, 11, 6, 33 };
    int n = a.size();

    /* Construct the segment tree by assigning 
    the values to the internal nodes*/
    vector<int> segtree(2 * n);
    construct_segment_tree(segtree, a, n);

    // compute maximum in the range left to right
    int left = 1, right = 5;
    cout << "Maximum in range " << left << " to "
         << right << " is " << range_query(segtree, left,
                                           right + 1, n)
         << "\n";

    // update the value of index 5 to 32
    int index = 5, value = 32;

    // a[5] = 32;
    // Contents of array : {2, 6, 10, 4, 7, 32, 9, 11, 6, 33}
    update(segtree, index, value, n);

    // compute maximum in the range left to right
    left = 2, right = 8;
    cout << "Maximum in range " << left << " to "
         << right << " is " << range_query(segtree,
                                           left, right + 1, n)
         << "\n";

    return 0;
}

Java

// Java Program for the above approach
import java.util.Arrays;

public class IterativeSegmentTree {
  private int[] segtree;
  private int n;

  public IterativeSegmentTree(int[] a)
  {
    n = a.length;
    segtree = new int[2 * n];
    Arrays.fill(segtree, Integer.MIN_VALUE);
    constructSegmentTree(a);
  }

  private void constructSegmentTree(int[] a)
  {
    // assign values to leaves of the segment tree
    for (int i = 0; i < n; i++) {
      segtree[n + i] = a[i];
    }

    /* assign values to internal nodes
        to compute maximum in a given range */
    for (int i = n - 1; i >= 1; i--) {
      segtree[i] = Math.max(segtree[2 * i],
                            segtree[2 * i + 1]);
    }
  }

  public void update(int pos, int value)
  {
    // change the index to leaf node first
    pos += n;

    // update the value at the leaf node
    // at the exact index
    segtree[pos] = value;

    while (pos > 1) {
      // move up one level at a time in the tree
      pos >>= 1;

      // update the values in the nodes in
      // the next higher level
      segtree[pos] = Math.max(segtree[2 * pos],
                              segtree[2 * pos + 1]);
    }
  }

  public int rangeQuery(int left, int right)
  {
    /* Basically the left and right indices will move
        towards right and left respectively and with
        every each next higher level and compute the
        maximum at each height. */
    // change the index to leaf node first
    left += n;
    right += n;

    // initialize maximum to a very low value
    int ma = Integer.MIN_VALUE;

    while (left < right) {
      // if left index in odd
      if ((left & 1) == 1) {
        ma = Math.max(ma, segtree[left]);

        // make left index even
        left++;
      }

      // if right index in odd
      if ((right & 1) == 1) {
        // make right index even
        right--;

        ma = Math.max(ma, segtree[right]);
      }

      // move to the next higher level
      left /= 2;
      right /= 2;
    }
    return ma;
  }
  //Driver code
  public static void main(String[] args)
  {
    int[] a = { 2, 6, 10, 4, 7, 28, 9, 11, 6, 33 };
    IterativeSegmentTree tree
      = new IterativeSegmentTree(a);

    // compute maximum in the range left to right
    int left = 1, right = 5;
    System.out.println(
      "Maximum in range " + left + " to " + right
      + " is " + tree.rangeQuery(left, right + 1));

    // update the value of index 5 to 32
    int index = 5, value = 32;

    // a[5] = 32;
    // Contents of array : {2, 6, 10, 4, 7, 32, 9, 11,
    // 6, 33}
    tree.update(index, value);

    // compute maximum in the range left to right
    left = 2;
    right = 8;
    System.out.println(
      "Maximum in range " + left + " to " + right
      + " is " + tree.rangeQuery(left, right + 1));
  }
}

// This code is contributed by lokeshpotta20.

Python3

# Python Program to implement
# iterative segment tree.

from sys import maxsize
INT_MIN = -maxsize

def construct_segment_tree(a: list, n: int):
    global segtree

    # assign values to leaves of the segment tree
    for i in range(n):
        segtree[n + i] = a[i]

    # assign values to internal nodes
    # to compute maximum in a given range */
    for i in range(n - 1, 0, -1):
        segtree[i] = max(segtree[2 * i], segtree[2 * i + 1])

def update(pos: int, value: int, n: int):
    global segtree

    # change the index to leaf node first
    pos += n

    # update the value at the leaf node
    # at the exact index
    segtree[pos] = value

    while pos > 1:

        # move up one level at a time in the tree
        pos //= 2

        # update the values in the nodes in
        # the next higher level
        segtree[pos] = max(segtree[2 * pos], segtree[2 * pos + 1])

def range_query(left: int, right: int, n: int) -> int:
    global segtree

    # Basically the left and right indices will move
    # towards right and left respectively and with
    # every each next higher level and compute the
    # maximum at each height.
    # change the index to leaf node first
    left += n
    right += n

    # initialize maximum to a very low value
    ma = INT_MIN
    while left < right:

        # if left index in odd
        if left & 1:
            ma = max(ma, segtree[left])

            # make left index even
            left += 1

        # if right index in odd
        if right & 1:

            # make right index even
            right -= 1

            ma = max(ma, segtree[right])

        # move to the next higher level
        left //= 2
        right //= 2
    return ma


# Driver Code
if __name__ == "__main__":
    a = [2, 6, 10, 4, 7, 28, 9, 11, 6, 33]
    n = len(a)

    # Construct the segment tree by assigning
    # the values to the internal nodes
    segtree = [0] * (2 * n)
    construct_segment_tree(a, n)

    # compute maximum in the range left to right
    left = 1
    right = 5
    print("Maximum in range %d to %d is %d" %
        (left, right, range_query(left, right + 1, n)))

    # update the value of index 5 to 32
    index = 5
    value = 32

    # a[5] = 32;
    # Contents of array : {2, 6, 10, 4, 7, 32, 9, 11, 6, 33}
    update(index, value, n)

    # compute maximum in the range left to right
    left = 2
    right = 8
    print("Maximum in range %d to %d is %d" %
        (left, right, range_query(left, right + 1, n)))

# This code is contributed by
# sanjeev2552

// C# Program for the above approach
using System;

public class IterativeSegmentTree {
static int[] segtree;
static int n;

public IterativeSegmentTree(int[] a)
{
    n = a.Length;
    segtree = new int[2 * n];
    for(int i=0;i<2*n;i++)
    {
        segtree[i]= Int32.MinValue;
    }
    
    constructSegmentTree(a);
}

private void constructSegmentTree(int[] a)
{
    // assign values to leaves of the segment tree
    for (int i = 0; i < n; i++) {
    segtree[n + i] = a[i];
    }

    /* assign values to internal nodes
        to compute maximum in a given range */
    for (int i = n - 1; i >= 1; i--) {
    segtree[i] = Math.Max(segtree[2 * i],
                            segtree[2 * i + 1]);
    }
}

public void update(int pos, int value)
{
    // change the index to leaf node first
    pos += n;

    // update the value at the leaf node
    // at the exact index
    segtree[pos] = value;

    while (pos > 1) {
    // move up one level at a time in the tree
    pos >>= 1;

    // update the values in the nodes in
    // the next higher level
    segtree[pos] = Math.Max(segtree[2 * pos],
                            segtree[2 * pos + 1]);
    }
}

public int rangeQuery(int left, int right)
{
    /* Basically the left and right indices will move
        towards right and left respectively and with
        every each next higher level and compute the
        maximum at each height. */
    // change the index to leaf node first
    left += n;
    right += n;

    // initialize maximum to a very low value
    int ma = Int32.MinValue;

    while (left < right) {
    // if left index in odd
    if ((left & 1) == 1) {
        ma = Math.Max(ma, segtree[left]);

        // make left index even
        left++;
    }

    // if right index in odd
    if ((right & 1) == 1) {
        // make right index even
        right--;

        ma = Math.Max(ma, segtree[right]);
    }

    // move to the next higher level
    left /= 2;
    right /= 2;
    }
    return ma;
}
//Driver code
static public void Main ()
{
    int[] a = { 2, 6, 10, 4, 7, 28, 9, 11, 6, 33 };
    IterativeSegmentTree tree
    = new IterativeSegmentTree(a);

    // compute maximum in the range left to right
    int left = 1, right = 5;
    Console.WriteLine(
    "Maximum in range " + left + " to " + right
    + " is " + tree.rangeQuery(left, right + 1));

    // update the value of index 5 to 32
    int index = 5, value = 32;

    // a[5] = 32;
    // Contents of array : {2, 6, 10, 4, 7, 32, 9, 11,
    // 6, 33}
    tree.update(index, value);

    // compute maximum in the range left to right
    left = 2;
    right = 8;
    Console.WriteLine(
    "Maximum in range " + left + " to " + right
    + " is " + tree.rangeQuery(left, right + 1));
}
}

// This code is contributed by Pushpesh Raj

JavaScript

<script>

// Javascript program to implement 
// iterative segment tree.
function construct_segment_tree(segtree, a, n)
{
    
    // Assign values to leaves of the segment tree
    for(let i = 0; i < n; i++)
        segtree[n + i] = a[i];

    // Assign values to internal nodes
    // to compute maximum in a given range 
    for(let i = n - 1; i >= 1; i--)
        segtree[i] = Math.max(segtree[2 * i],
                              segtree[2 * i + 1]);
}

function update(segtree, pos, value, n)
{
    
    // Change the index to leaf node first
    pos += n;

    // Update the value at the leaf node
    // at the exact index
    segtree[pos] = value;

    while (pos > 1) 
    {
        
        // Move up one level at a time in the tree
        pos >>= 1;

        // Update the values in the nodes in
        // the next higher level
        segtree[pos] = Math.max(segtree[2 * pos],
                                segtree[2 * pos + 1]);
    }
}

function range_query(segtree, left, right, n)
{
    
    /* Basically the left and right indices will move
        towards right and left respectively and with
        every each next higher level and compute the 
        maximum at each height. */
    // change the index to leaf node first
    left += n;
    right += n;

    // Initialize maximum to a very low value
    let ma = Number.MIN_VALUE;

    while (left < right)
    {
        
        // If left index in odd
        if ((left & 1) != 0)
        {
            ma = Math.max(ma, segtree[left]);

            // Make left index even
            left++;
        }

        // If right index in odd
        if ((right & 1) > 0) 
        {
            
            // Make right index even
            right--;

            ma = Math.max(ma, segtree[right]);
        }

        // Move to the next higher level
        left = parseInt(left / 2, 10);
        right = parseInt(right / 2, 10);
    }
    return ma;
}

// Driver code
let a = [ 2, 6, 10, 4, 7, 28, 9, 11, 6, 33 ];
let n = a.length;

// Construct the segment tree by assigning 
// the values to the internal nodes
let segtree = new Array(2 * n);
construct_segment_tree(segtree, a, n);

// Compute maximum in the range left to right
let left = 1, right = 5;
document.write("Maximum in range " + left + 
               " to " + right + " is " + 
               range_query(segtree, left, right + 1, n) + "</br>");

// Update the value of index 5 to 32
let index = 5, value = 32;

// a[5] = 32;
// Contents of array : {2, 6, 10, 4, 7, 32, 9, 11, 6, 33}
update(segtree, index, value, n);

// Compute maximum in the range left to right
left = 2, right = 8;
document.write("Maximum in range " + left + 
               " to " + right + " is " + 
               range_query(segtree, left, right + 1, n) + "</br>");

// This code is contributed by divyesh072019

</script>

Output

Maximum in range 1 to 5 is 28
Maximum in range 2 to 8 is 32

Complexity Analysis:

Time Complexity: (N * log N)
Auxiliary Space: O(N)

Iterative Segment Tree (Range Maximum Query with Node Update)

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