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Count number of increasing sub-sequences : O(NlogN)
  • Difficulty Level : Hard
  • Last Updated : 12 May, 2021

Given an array arr[] of length N, the task is to find the number of strictly increasing sub-sequences in the given array.

Examples:  

Input: arr[] = {1, 2, 3} 
Output:
All increasing sub-sequences will be: 
1) {1} 
2) {2} 
3) {3} 
4) {1, 2} 
5) {1, 3} 
6) {2, 3} 
7) {1, 2, 3} 
Thus, answer = 7
Input: arr[] = {3, 2, 1} 
Output: 3  

Approach: An O(N2) approach has already been discussed in this article. Here, an approach with O(NlogN) time using segment tree data structure will be discussed.
In the previous article, the recurrence relation used was:  

dp[i] = 1 + summation(dp[j]), where i <jarr[i] 
 



Due to the fact that the entire sub-array arr[i+1…n-1] was being iterated for each state, an extra O(N) time was being used to solve it. Thus, the complexity was (N2).
The idea is to avoid iterating the O(N) extra loop for each state and reducing its complexity to Log(N).
Assumption: The number of strictly increasing sub-sequences starting from each index ‘i’, where i is greater than a number ‘k’ is known. 
Using this above assumption, the number of increasing sub-sequences starting from index ‘k’ can be found in log(N) time.
Therefore, the summation for all the indexes ‘i’ greater than ‘k’ must be found. But the catch is arr[i] must be greater than arr[k]. To deal with the problem, the following can be done:  

1. For each element of the array, its index is found in the array was sorted. Example – {7, 8, 1, 9, 4} Here, ranks will be: 

7 -> 3 
8 -> 4 
1 -> 1 
9 -> 5 
4 -> 2 
 

2. A segment tree of length ‘N’ is created to answer a range-sum query.

3. To answer the query while solving for index ‘k’, the rank of arr[k] is found first. Let’s say the rank is R. Then, in the segment tree, the range-sum from index {R to N-1} is found.

4. Then, the segment-tree is point updated as segment-(R-1) equals 1 + segtree-query(R, N-1) + segtree-query(R-1, R-1)

Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
#define N 10000
 
// Segment tree array
int seg[3 * N];
 
// Function for point update in segment tree
int update(int in, int l, int r, int up_in, int val)
{
    // Base case
    if (r < up_in || l > up_in)
        return seg[in];
 
    // If l==r==up
    if (l == up_in and r == up_in)
        return seg[in] = val;
 
    // Mid element
    int m = (l + r) / 2;
 
    // Updating the segment tree
    return seg[in] = update(2 * in + 1, l, m, up_in, val)
                     + update(2 * in + 2, m + 1, r, up_in, val);
}
 
// Function for the range sum-query
int query(int in, int l, int r, int l1, int r1)
{
    // Base case
    if (l > r)
        return 0;
    if (r < l1 || l > r1)
        return 0;
    if (l1 <= l and r <= r1)
        return seg[in];
 
    // Mid element
    int m = (l + r) / 2;
 
    // Calling for the left and the right subtree
    return query(2 * in + 1, l, m, l1, r1)
           + query(2 * in + 2, m + 1, r, l1, r1);
}
 
// Function to return the count
int findCnt(int* arr, int n)
{
    // Copying array arr to sort it
    int brr[n];
    for (int i = 0; i < n; i++)
        brr[i] = arr[i];
 
    // Sorting array brr
    sort(brr, brr + n);
 
    // Map to store the rank of each element
    map<int, int> r;
    for (int i = 0; i < n; i++)
        r[brr[i]] = i + 1;
 
    // dp array
    int dp[n] = { 0 };
 
    // To store the final answer
    int ans = 0;
 
    // Updating the dp array
    for (int i = n - 1; i >= 0; i--) {
 
        // Rank of the element
        int rank = r[arr[i]];
 
        // Solving the dp-states using segment tree
        dp[i] = 1 + query(0, 0, n - 1, rank, n - 1);
 
        // Updating the final answer
        ans += dp[i];
 
        // Updating the segment tree
        update(0, 0, n - 1, rank - 1,
               dp[i] + query(0, 0, n - 1, rank - 1, rank - 1));
    }
 
    // Returning the final answer
    return ans;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 10, 9 };
    int n = sizeof(arr) / sizeof(int);
 
    cout << findCnt(arr, n);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
    static final int N = 10000;
 
    // Segment tree array
    static int[] seg = new int[3 * N];
 
    // Function for point update in segment tree
    static int update(int in, int l, int r, int up_in, int val)
    {
        // Base case
        if (r < up_in || l > up_in)
            return seg[in];
 
        // If l==r==up
        if (l == up_in && r == up_in)
            return seg[in] = val;
 
        // Mid element
        int m = (l + r) / 2;
 
        // Updating the segment tree
        return seg[in] = update(2 * in + 1, l, m, up_in, val) +
                update(2 * in + 2, m + 1, r, up_in, val);
    }
 
    // Function for the range sum-query
    static int query(int in, int l, int r, int l1, int r1)
    {
        // Base case
        if (l > r)
            return 0;
        if (r < l1 || l > r1)
            return 0;
        if (l1 <= l && r <= r1)
            return seg[in];
 
        // Mid element
        int m = (l + r) / 2;
 
        // Calling for the left and the right subtree
        return query(2 * in + 1, l, m, l1, r1) +
                query(2 * in + 2, m + 1, r, l1, r1);
    }
 
    // Function to return the count
    static int findCnt(int[] arr, int n)
    {
        // Copying array arr to sort it
        int[] brr = new int[n];
        for (int i = 0; i < n; i++)
            brr[i] = arr[i];
 
        // Sorting array brr
        Arrays.sort(brr);
 
        // Map to store the rank of each element
        HashMap<Integer, Integer> r = new HashMap<Integer, Integer>();
        for (int i = 0; i < n; i++)
            r.put(brr[i], i + 1);
 
        // dp array
        int dp[] = new int[n];
 
        // To store the final answer
        int ans = 0;
 
        // Updating the dp array
        for (int i = n - 1; i >= 0; i--)
        {
 
            // Rank of the element
            int rank = r.get(arr[i]);
 
            // Solving the dp-states using segment tree
            dp[i] = 1 + query(0, 0, n - 1, rank, n - 1);
 
            // Updating the final answer
            ans += dp[i];
 
            // Updating the segment tree
            update(0, 0, n - 1, rank - 1, dp[i] +
                    query(0, 0, n - 1, rank - 1, rank - 1));
        }
 
        // Returning the final answer
        return ans;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 1, 2, 10, 9 };
        int n = arr.length;
 
        System.out.print(findCnt(arr, n));
 
    }
}
 
// This code is contributed by PrinciRaj1992

Python3




# Python3 implementation of the approach
 
N = 10000
 
# Segment tree array
seg = [0] * (3 * N)
 
# Function for point update in segment tree
def update(In, l, r, up_In, val):
     
    # Base case
    if (r < up_In or l > up_In):
        return seg[In]
 
    # If l==r==up
    if (l == up_In and r == up_In):
        seg[In] = val
        return val
 
    # Mid element
    m = (l + r) // 2
 
    # Updating the segment tree
    seg[In] = update(2 * In + 1, l, m, up_In, val) + update(2 * In + 2, m + 1, r, up_In, val)
    return seg[In]
 
 
# Function for the range sum-query
def query(In, l, r, l1, r1):
 
    # Base case
    if (l > r):
        return 0
    if (r < l1 or l > r1):
        return 0
    if (l1 <= l and r <= r1):
        return seg[In]
 
    # Mid element
    m = (l + r) // 2
 
    # CallIng for the left and the right subtree
    return query(2 * In + 1, l, m, l1, r1)+ query(2 * In + 2, m + 1, r, l1, r1)
 
 
# Function to return the count
def fIndCnt(arr, n):
 
    # Copying array arr to sort it
    brr = [0] * n
    for i in range(n):
        brr[i] = arr[i]
 
    # Sorting array brr
    brr = sorted(brr)
 
    # Map to store the rank of each element
    r = dict()
    for i in range(n):
        r[brr[i]] = i + 1
 
    # dp array
    dp = [0] * n
 
    # To store the final answer
    ans = 0
 
    # Updating the dp array
    for i in range(n - 1, -1, -1):
 
        # Rank of the element
        rank = r[arr[i]]
 
        # Solving the dp-states using segment tree
        dp[i] = 1 + query(0, 0, n - 1, rank, n - 1)
 
        # UpdatIng the final answer
        ans += dp[i]
 
        # Updating the segment tree
        update(0, 0, n - 1, rank - 1,dp[i] + query(0, 0, n - 1, rank - 1, rank - 1))
 
    # Returning the final answer
    return ans
 
# Driver code
 
arr = [1, 2, 10, 9]
n = len(arr)
 
print(fIndCnt(arr, n))
 
# This code is contributed by mohit kumar 29

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
    static readonly int N = 10000;
 
    // Segment tree array
    static int[] seg = new int[3 * N];
 
    // Function for point update In segment tree
    static int update(int In, int l, int r,
                        int up_in, int val)
    {
        // Base case
        if (r < up_in || l > up_in)
            return seg[In];
 
        // If l==r==up
        if (l == up_in && r == up_in)
            return seg[In] = val;
 
        // Mid element
        int m = (l + r) / 2;
 
        // Updating the segment tree
        return seg[In] = update(2 * In + 1, l, m, up_in, val) +
                update(2 * In + 2, m + 1, r, up_in, val);
    }
 
    // Function for the range sum-query
    static int query(int In, int l, int r, int l1, int r1)
    {
        // Base case
        if (l > r)
            return 0;
        if (r < l1 || l > r1)
            return 0;
        if (l1 <= l && r <= r1)
            return seg[In];
 
        // Mid element
        int m = (l + r) / 2;
 
        // Calling for the left and the right subtree
        return query(2 * In + 1, l, m, l1, r1) +
                query(2 * In + 2, m + 1, r, l1, r1);
    }
 
    // Function to return the count
    static int findCnt(int[] arr, int n)
    {
        // Copying array arr to sort it
        int[] brr = new int[n];
        for (int i = 0; i < n; i++)
            brr[i] = arr[i];
 
        // Sorting array brr
        Array.Sort(brr);
 
        // Map to store the rank of each element
        Dictionary<int, int> r = new Dictionary<int, int>();
        for (int i = 0; i < n; i++)
            r.Add(brr[i], i + 1);
 
        // dp array
        int []dp = new int[n];
 
        // To store the readonly answer
        int ans = 0;
 
        // Updating the dp array
        for (int i = n - 1; i >= 0; i--)
        {
 
            // Rank of the element
            int rank = r[arr[i]];
 
            // Solving the dp-states using segment tree
            dp[i] = 1 + query(0, 0, n - 1, rank, n - 1);
 
            // Updating the readonly answer
            ans += dp[i];
 
            // Updating the segment tree
            update(0, 0, n - 1, rank - 1, dp[i] +
                    query(0, 0, n - 1, rank - 1, rank - 1));
        }
 
        // Returning the readonly answer
        return ans;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = { 1, 2, 10, 9 };
        int n = arr.Length;
 
        Console.Write(findCnt(arr, n));
 
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
 
// Javascript implementation of the approach
 
var N = 10000;
 
// Segment tree array
var seg = Array(3*N).fill(0);
 
// Function for point update inVal segment tree
function update(inVal, l, r, up_in, val)
{
    // Base case
    if (r < up_in || l > up_in)
        return seg[inVal];
 
    // If l==r==up
    if (l == up_in && r == up_in)
        return seg[inVal] = val;
 
    // Mid element
    var m = parseInt((l + r) / 2);
 
    // Updating the segment tree
    seg[inVal] = update(2 * inVal + 1, l, m, up_in, val)
                     + update(2 * inVal + 2, m + 1, r, up_in, val);
    return seg[inVal];
}
 
// Function for the range sum-query
function query(inVal, l, r, l1, r1)
{
    // Base case
    if (l > r)
        return 0;
    if (r < l1 || l > r1)
        return 0;
    if (l1 <= l && r <= r1)
        return seg[inVal];
 
    // Mid element
    var m = parseInt((l + r) / 2);
 
    // Calling for the left and the right subtree
    return query(2 * inVal + 1, l, m, l1, r1)
           + query(2 * inVal + 2, m + 1, r, l1, r1);
}
 
// Function to return the count
function findCnt(arr, n)
{
    // Copying array arr to sort it
    var brr = Array(n);
    for (var i = 0; i < n; i++)
        brr[i] = arr[i];
 
    // Sorting array brr
    brr.sort((a, b)=> a-b);
 
    // Map to store the rank of each element
    var r = new Map();
    for (var i = 0; i < n; i++)
        r[brr[i]] = i + 1;
 
    // dp array
    var dp = Array(n).fill(0);
 
    // To store the final answer
    var ans = 0;
 
    // Updating the dp array
    for (var i = n - 1; i >= 0; i--) {
 
        // Rank of the element
        var rank = r[arr[i]];
 
        // Solving the dp-states using segment tree
        dp[i] = 1 + query(0, 0, n - 1, rank, n - 1);
 
        // Updating the final answer
        ans += dp[i];
 
        // Updating the segment tree
        update(0, 0, n - 1, rank - 1,
               dp[i] + query(0, 0, n - 1, rank - 1, rank - 1));
    }
 
    // Returning the final answer
    return ans;
}
 
// Driver code
var arr = [1, 2, 10, 9 ];
var n = arr.length;
document.write( findCnt(arr, n));
 
</script>
Output: 
11

 

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