Iterative Segment Tree (Range Minimum Query)

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

Let us consider the following problem understand Segment Trees.

We have an array arr[0 . . . n-1]. We should be able to
1 Find the minimum of elements from index l to r where 0 <= l <= r <= n-1
2 Change value of a specified element of the array to a new value x. We need to do arr[i] = x where 0 <= i <= n-1.



Examples:

Input : 2, 6, 7, 5, 18, 86, 54, 2
        minimum(2, 7)  
        update(3, 4)
        minimum(2, 6) 
Output : Minimum in range 2 to 7 is 2.
         Minimum in range 2 to 6 is 4.

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 minimum 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 minimum 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.

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// CPP Program to implement iterative segment
// tree.
#include <bits/stdc++.h>
#define ll long long
  
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 minimum in a given range */
    for (int i = n - 1; i >= 1; i--)
        segtree[i] = min(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] = min(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 
        minimum at each height. */
    // change the index to leaf node first
    left += n;
    right += n;
  
    // initialize minimum to a very high value
    int mi = (int)1e9;
  
    while (left < right) {
  
        // if left index in odd
        if (left & 1) {
            mi = min(mi, segtree[left]);
  
            // make left index even
            left++;
        }
  
        // if right index in odd
        if (right & 1) {
  
            // make right index even
            right--;
  
            mi = min(mi, segtree[right]);
        }
  
        // move to the next higher level
        left /= 2;
        right /= 2;
    }
    return mi;
}
  
// 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 minimum in the range left to right
    int left = 0, right = 5;
    cout << "Minimum in range " << left << " to " 
         << right << " is "<< range_query(segtree, left, 
                                  right + 1, n) << "\n";
  
    // update the value of index 3 to 1
    int index = 3, value = 1;
  
    // a[3] = 1;
    // Contents of array : {2, 6, 10, 1, 7, 28, 9, 11, 6, 33}
    update(segtree, index, value, n); // point update
  
    // compute minimum in the range left to right
    left = 2, right = 6;
    cout << "Minimum in range " << left << " to "
         << right << " is " << range_query(segtree, 
                      left, right + 1, n) << "\n";
  
    return 0;
}
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Output:
Minimum in range 0 to 5 is 2
Minimum in range 2 to 6 is 1

Time Complexity(n log n)
Auxiliary Space (n)




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