Distinct Subsequences
A hard Dynamic Programming interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Dynamic Programming
- Sheets
- 1
- Core for
- 9 roles
The problem
Count how many different ways you can delete characters from s, without reordering the ones that stay, so that what remains spells exactly t. Two ways are different when they keep different positions of s.
Example 1
- Input
- s = "rabbbit", t = "rabbit"
- Output
- 3
- Why
- Three ways to remove characters from "rabbbit" to get "rabbit".
Example 2
- Input
- s = "babgbag", t = "bag"
- Output
- 5
- Why
- Five distinct subsequences of "babgbag" equal "bag".
Example 3
- Input
- s = "abc", t = "abc"
- Output
- 1
- Why
- Only one way: the strings are identical.
Constraints
- 1 <= s.length, t.length <= 1000
- s and t consist of English letters
How to think about it
Updated 2026-09-29When matching character t[j-1] against s[i-1], you always have the option to ignore s[i-1] and inherit ways from s[0..i-2]. If the characters match, you can additionally add the ways to form t[0..j-2] from s[0..i-2]. Summing both possibilities gives the total distinct subsequences.
Approaches, worst first
Recursive match branching
time O(2^m) · space O(m)
Branch on whether to consume matching characters or skip the current character in s. Without memoization, explores overlapping string prefixes exponentially.
2D dynamic programming grid
time O(m * n) · space O(m * n)
dp[i][j] is the number of ways s[0..i-1] forms t[0..j-1]. dp[i][0] = 1 for all i. For matching characters, dp[i][j] = dp[i-1][j] + dp[i-1][j-1]; otherwise dp[i][j] = dp[i-1][j].
1D backwards accumulation arrayWrite this one
time O(m * n) · space O(n)
Allocate 1D array dp of size t.length + 1 with dp[0] = 1. For each char c in s, iterate j from t.length down to 1: if c == t[j-1], add dp[j-1] to dp[j].
Where people lose marks · 3
- Iterating forward in 1D array: updating j from 1 up to t.length uses the same character of s multiple times to match multiple positions of t.
- Base case failure: an empty target t can always be formed in exactly 1 way (by deleting all characters of s).
- Integer overflow: counts can exceed standard 32-bit signed integers on long strings with duplicate characters.
The theory behind it
Dynamic Programming — the ground this problem stands on. All Dynamic Programming problems
What Dynamic Programming is
Dynamic programming is a method for solving a complex problem by breaking it into overlapping subproblems, solving each subproblem only once, and remembering the answers in a lookup table. Instead of recalculating identical questions over and over, future steps look up previous answers directly. By assembling these saved pieces from the bottom up or storing them during recursion, a task that would take billions of steps finishes in a fraction of a second.
When to reach for it
Reach for dynamic programming when questions ask for the maximum profit, minimum cost, total number of distinct ways to achieve a goal, or whether a target can be formed. Signals include overlapping choices where making a choice now affects what choices remain later, but greedy picking fails to find the true global optimum. If drawing a recursive decision tree reveals the same subproblem states repeating across branches, dynamic programming is needed.
How the pattern works
Identify the state variables that uniquely describe a subproblem, such as an array index and remaining capacity. Write the base cases first, representing states whose answers are known without calculation. Next, write the recurrence relation that expresses the current state using previously solved states, taking the minimum, maximum, or sum among your options. Build the solution either top-down by caching recursive returns in a memo table, or bottom-up by filling an array in topological dependency order. When each state depends only on the previous row, compress storage down to a single array.
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