Introduction

In combinatorics, the problem of determining the number of increasing subsequences of a given length $k$ from a sequence is a fundamental question with significant implications in algorithm design, bioinformatics, and coding theory. This article examines both foundational concepts and practical techniques for solving this problem efficiently.

Understanding Increasing Subsequences

An increasing subsequence of a sequence is a subsequence where each element is strictly larger than the preceding one. For example, given the sequence ${1, 3, 2, 4}$, possible increasing subsequences include ${1, 3}$, ${1, 2}$, ${1, 4}$, ${3, 4}$, ${2, 4}$, ${1, 3, 4}$, ${1, 2, 4}$, and the single elements ${1}$, ${3}$, ${2}$, ${4}$.

Problem Statement

Given a sequence of $n$ distinct integers, find the number of increasing subsequences of length exactly $k$. This problem can be generalized to include sequences with repeated numbers by using non-strict inequality, but we focus on the strictly increasing case here.

Dynamic Programming Approach

A common way to solve this problem is through dynamic programming. The approach involves building a table where each entry $dp[i][j]$ represents the number of increasing subsequences of length $j$ that end at element $i$ (using 1-based indexing for the sequence).

Initialization

For length 1, every single element forms an increasing subsequence by itself:

$$dp[i][1] = 1 \quad \text{for all } i \in {1, 2, \ldots, n}$$

Transition

For every element at position $i$ and subsequence length $j$, count those that can be formed by preceding elements smaller than $a_i$:

$$dp[i][j] = \sum_{\substack{m=1 \\ a_m < a_i}}^{i-1} dp[m][j-1]$$

In words, to count increasing subsequences of length $j$ ending at position $i$, look at all earlier positions $m$ where $a_m < a_i$ and sum up the count of increasing subsequences of length $j-1$ ending at those positions.

Final Result

The total number of increasing subsequences of length $k$ is the sum over all ending positions:

$$\text{answer} = \sum_{i=1}^{n} dp[i][k]$$

Example

Consider the sequence ${2, 5, 3, 7, 4}$ and $k = 3$.

Step 1: Initialize $dp[i][1] = 1$ for all $i$.

Step 2: Fill the table for $j = 2$:

• $dp[2][2] = dp[1][1] = 1$ (subsequence ${2, 5}$)
• $dp[3][2] = dp[1][1] = 1$ (subsequence ${2, 3}$)
• $dp[4][2] = dp[1][1] + dp[2][1] + dp[3][1] = 3$ (subsequences ${2,7}, {5,7}, {3,7}$)
• $dp[5][2] = dp[1][1] + dp[3][1] = 2$ (subsequences ${2,4}, {3,4}$)

Step 3: Fill for $j = 3$:

• $dp[4][3] = dp[2][2] + dp[3][2] = 1 + 1 = 2$ (subsequences ${2,5,7}, {2,3,7}$)
• $dp[5][3] = dp[3][2] = 1$ (subsequence ${2,3,4}$)

Answer: $dp[4][3] + dp[5][3] = 2 + 1 = 3$

Complexity Analysis

The time complexity of this basic algorithm is $O(n^2 \cdot k)$ due to the nested loops iterating over possible subsequences for each element and subsequence length. The space complexity is $O(n \cdot k)$ for the DP table.

Optimization with BIT/Fenwick Tree

The inner summation over all preceding elements can be accelerated using a Binary Indexed Tree (BIT). By processing elements in a sorted order and querying prefix sums, each transition step can be done in $O(\log n)$ instead of $O(n)$, bringing the total time complexity down to $O(n \cdot k \cdot \log n)$. This optimization is important for large sequences.

Summary Table

Conclusion

Computing the number of increasing subsequences of length $k$ is both a classical and practically relevant problem. The dynamic programming approach provides a clear and correct solution, while data structure optimizations like BIT reduce the time complexity for larger inputs. Understanding this problem also builds intuition for related challenges like the Longest Increasing Subsequence (LIS) and patience sorting.