LIS using Segment Tree

You are given an array of integers, you need to find the length of the longest increasing sub-sequence.

There can be 4 approaches for solving the problem.

1) Brute force:
In this approach, we try to find all increasing subsequences and then returning the maximum length of longest increasing subsequence. In order to do this, we make use of a recursive function which returns the length of the LIS possible from the current element onwards.
Time Complexity: O(2^n)
Space Complexity: O(n^2)

2) Dynamic Programming
This approach is relying on the fact that LIS up to ith index is independent of the later (n-i+1) elements. Also, the LIS up to (i+1)th element can be calculated by checking the LIS obtained from index 0 to i.
Dynamic Programming approach
Time Complexity: O(n^2)
Space Complexity: O(n)

3) Using Binary Search
The elements are stored in the increasing order in the DP array where the index is determined using binary search. The length of the array gives the length of the LIS.
Time Complexity: O(nlog(n))
Space Complexity: O(n)
Please refer Construction of Longest Increasing Subsequence (N log N) for details.

4) Using Segment Tree
The elements are first sorted in increasing order while retaining their original indices. For strictly increasing LIS, for equal elements, the element with a higher index gets an early spot than the lower. This can be stored in an array of pair.
Now, they are populated in the segment tree. According to their position in their sorted array, they are filled up in the segment tree in the leaves corresponding to their original indices.
Initially, the segment tree was initialised with zeroes. Now, let us assume we have processed ith element in the sorted array. At the (i+1)th iteration, let the original position of the value be j.
Then, it will fill up the jth leaf in the segment tree whose value will be the maximum value of the leaves between 0 to (j-1) +1.
(Length of the LIS formed by the elements lesser than it in the sub array preceding it and +1 for its inclusion)
Arr[] = {5, 1, 3, 9} Indices : {0, 1, 2, 3}
Sorted_Arr[] = {1, 3, 5, 9} Original_Indices : {1, 2, 0, 3} Indices : {0, 1, 2, 3}
9 (2)
9 (4)





// Finding the Longest Increasing Subsequence using 
// Segment Tree
#include <bits/stdc++.h>
using namespace std;
// function to compare two pairs
int compare(pair<int, int> p1, pair<int, int> p2)
     /* For same values, element with the higher 
        index appear earlier in the sorted array.
        This is for strictly increasing subsequence.
        For increasing subsequence, the lower index 
         appears earlier in the sorted array. */
    if (p1.first == p2.first) 
        return p1.second > p2.second;
    // Sorting the array according to their values.
    return p1.first < p2.first;
// Building the entire Segment tree, the root of which
// contains the length of the LIS
void buildTree(int* tree, int pos, int low, int high, 
                                 int index, int value)
    // index is the original index of current element
    // If the index is not present in the given range, 
    // then simply return
    if (index < low || index > high)
    // If low == high then the current position should 
    // be updated to the value
    if (low == high) {
        tree[pos] = value;
    int mid = (high + low) / 2;
    // Recursively call the function on the 
    // child nodes
    buildTree(tree, 2 * pos + 1, low, mid, index, value);
    buildTree(tree, 2 * pos + 2, mid + 1, high, index, value);
    // Assign the current position the max of the 2 child 
    // nodes
    tree[pos] = max(tree[2 * pos + 1], tree[2 * pos + 2]);
// Function to query the Segment tree and return the
// value for a given range
int findMax(int* tree, int pos, int low, int high, 
                               int start, int end)
    // Query: Same as the query function of Segment tree
    // If the current range is totally inside the query 
    // range, return the value of current position
    if (low >= start && high <= end) 
        return tree[pos];
    // If it is out of bound, return the minimum which
    // would be 0 in this case
    if (start > high || end < low) 
        return 0;    
    // Partial overlap
    int mid = (high + low) / 2;
    // Call findMax on child nodes recursively and 
    // return the maximum of the two
    return max(findMax(tree, 2 * pos + 1, low, mid, 
                                        start, end), 
               findMax(tree, 2 * pos + 2, mid + 1, 
                                high, start, end));
int findLIS(int arr[], int n)
    // The array of pairs stores the integers and 
    // indices in p[i]
    pair<int, int> p[n];
    for (int i = 0; i < n; i++) {
        p[i].first = arr[i]; 
        p[i].second = i; 
    // Sorting the array in increasing order
    // of the elements
    sort(p, p + n, compare);
    // Calculating the length of the segment-tree
    int len = pow(2, (int)(ceil(sqrt(n))) + 1) - 1;
    int tree[len];
    // Initializing the tree with zeroes
    memset(tree, 0, sizeof(tree)); 
    // Building the segment-tree, the root node of 
    // which contains the length of LIS for the n
    // elements
    for (int i = 0; i < n; i++) {
        buildTree(tree, 0, 0, n - 1, p[i].second, 
     findMax(tree, 0, 0, n - 1, 0, p[i].second) + 1);
    return tree[0];
// Driver code
int main()
    int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Length of the LIS: " << findLIS(arr, n);
    return 0;


Length of the LIS: 5

Time Complexity: O(nlogn)
Space Complexity: O(nlogn)

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