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
- Find the minimum 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. 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.
Implementation:
C++
// 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 is odd if (left & 1) { mi = min(mi, segtree[left]); // make left index even left++; } // if right index is 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; } |
Java
// Java Program to implement iterative segment // tree. import java.io.*; import java.util.*; class GFG { static void construct_segment_tree( int [] segtree, 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] = Math.min(segtree[ 2 * i], segtree[ 2 * i + 1 ]); } static void update( 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] = Math.min(segtree[ 2 * pos], segtree[ 2 * pos + 1 ]); } } static int range_query( 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 is odd if ((left & 1 ) == 1 ) { mi = Math.min(mi, segtree[left]); // make left index even left++; } // if right index is odd if ((right & 1 ) == 1 ) { // make right index even right--; mi = Math.min(mi, segtree[right]); } // move to the next higher level left /= 2 ; right /= 2 ; } return mi; } // Driver Code public static void main(String[] args) { int [] a = { 2 , 6 , 10 , 4 , 7 , 28 , 9 , 11 , 6 , 33 }; int n = a.length; /* * Construct the segment tree by assigning * the values to the internal nodes */ int [] segtree = new int [ 2 * n]; construct_segment_tree(segtree, a, n); // compute minimum in the range left to right int left = 0 , right = 5 ; System.out.printf( "Minimum in range %d to %d is %d\n" , left, right, range_query(segtree, left, right + 1 , 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 ; System.out.printf( "Minimum in range %d to %d is %d\n" , left, right, range_query(segtree, left, right + 1 , n)); } } // This code is contributed by // sanjeev2552 |
Python3
# Python3 program to implement # iterative segment tree. def construct_segment_tree(segtree, a, n): # assign values to leaves of # the segment tree for i in range (n): segtree[n + i] = a[i]; # assign values to remaining nodes # to compute minimum in a given range for i in range (n - 1 , 0 , - 1 ): segtree[i] = min (segtree[ 2 * i], segtree[ 2 * i + 1 ]) def range_query(segtree, left, right, n): left + = n 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 minimum at each height change the index to leaf node first """ mi = 1e9 # initialize minimum to a very high value while (left < right): if (left & 1 ): # if left index is odd mi = min (mi, segtree[left]) left = left + 1 if (right & 1 ): # if right index is odd right - = 1 mi = min (mi, segtree[right]) # move to the next higher level left = left / / 2 right = right / / 2 return mi def 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] = min (segtree[ 2 * pos], segtree[ 2 * pos + 1 ]) # Driver Code # Elements in list 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 for i in range ( 2 * n)] construct_segment_tree(segtree, a, n); left = 0 right = 5 #compute minimum in the range left to right print ( "Minimum in range" , left, "to" , right, "is" , range_query(segtree, left, right + 1 , n)) # update the value of index 3 to 1 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 left = 2 right = 6 # compute minimum in the range left to right print ( "Minimum in range" , left, "to" , right, "is" , range_query(segtree, left, right + 1 , n)) # This code is contributed by sarthak Raghuwanshi |
C#
// C# Program to implement iterative segment // tree. using System; class GFG { static void construct_segment_tree( int [] segtree, 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] = Math.Min(segtree[2 * i], segtree[2 * i + 1]); } static void update( 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] = Math.Min(segtree[2 * pos], segtree[2 * pos + 1]); } } static int range_query( 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 is odd if ((left & 1) == 1) { mi = Math.Min(mi, segtree[left]); // make left index even left++; } // if right index is odd if ((right & 1) == 1) { // make right index even right--; mi = Math.Min(mi, segtree[right]); } // move to the next higher level left /= 2; right /= 2; } return mi; } // Driver Code public static void Main(String[] args) { int [] a = {2, 6, 10, 4, 7, 28, 9, 11, 6, 33}; int n = a.Length; /* * Construct the segment tree by assigning * the values to the internal nodes */ int [] segtree = new int [2 * n]; construct_segment_tree(segtree, a, n); // compute minimum in the range left to right int left = 0, right = 5; Console.Write( "Minimum in range {0} to {1} is {2}\n" , left, right, range_query(segtree, left, right + 1, 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; Console.Write( "Minimum in range {0} to {1} is {2}\n" , left, right, range_query(segtree, left, right + 1, n)); } } // This code is contributed by Rajput-Ji |
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 ( var i = 0; i < n; i++) segtree[n + i] = a[i]; /* * assign values to internal nodes * to compute minimum in a given range */ for ( var i = n - 1; i >= 1; i--) segtree[i] = Math.min(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.min(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 * minimum at each height. */ // change the index to leaf node first left += n; right += n; // initialize minimum to a very high value var mi = 1000000000; while (left < right) { // if left index is odd if ((left & 1) == 1) { mi = Math.min(mi, segtree[left]); // make left index even left++; } // if right index is odd if ((right & 1) == 1) { // make right index even right--; mi = Math.min(mi, segtree[right]); } // move to the next higher level left /= 2; right /= 2; } return mi; } // Driver Code var a = [2, 6, 10, 4, 7, 28, 9, 11, 6, 33]; var n = a.length; /* * Construct the segment tree by assigning * the values to the internal nodes */ var segtree = Array(2*n).fill(0); construct_segment_tree(segtree, a, n); // compute minimum in the range left to right var left = 0, right = 5; document.write(`Minimum in range ${left} to ${right} is ${range_query(segtree,left, right + 1, n)}<br>`) // update the value of index 3 to 1 var 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; document.write(`Minimum in range ${left} to ${right} is ${range_query(segtree, left, right + 1, n)}<br>`); // This code is contributed by rutvik_56. </script> |
Minimum in range 0 to 5 is 2 Minimum in range 2 to 6 is 1
Complexity Analysis:
- Time Complexity : (n log n)
- Auxiliary Space : (n)