# Iterative Segment Tree (Range Maximum Query with Node Update)

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++ 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; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

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

**Time Complexity:** (N * log N)

**Auxiliary Space:** O(N)

## Recommended Posts:

- Segment Tree | Set 2 (Range Maximum Query with Node Update)
- Iterative Segment Tree (Range Minimum Query)
- Segment Tree | Set 2 (Range Minimum Query)
- Difference Array | Range update query in O(1)
- Range and Update Query for Chessboard Pieces
- Binary Indexed Tree : Range Update and Range Queries
- Segment Tree | Set 3 (XOR of given range)
- Segment Tree | Set 1 (Sum of given range)
- Segment Tree | (XOR of a given range )
- Get level of a node in binary tree | iterative approach
- Deepest right leaf node in a binary tree | Iterative approach
- Deepest left leaf node in a binary tree | iterative approach
- Sum of nodes at maximum depth of a Binary Tree | Iterative Approach
- Node having maximum sum of immediate children and itself in n-ary tree
- Get maximum left node in binary tree

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