Shortest Common Supersequence
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
Given two strings, find the length of their shortest common supersequence. A supersequence contains both strings as subsequences.
Example 1
- Input
- str1 = "abac", str2 = "cab"
- Output
- 5
- Why
- The shortest common supersequence is "cabac", length 5.
Example 2
- Input
- str1 = "aaaaaaaa", str2 = "aaaaaaaa"
- Output
- 8
- Why
- Both strings are identical, the supersequence is the string itself.
Example 3
- Input
- str1 = "abc", str2 = "def"
- Output
- 6
- Why
- The shortest common supersequence is "abcdef", length 6.
Constraints
- 1 <= str1.length, str2.length <= 1000
- str1 and str2 consist of lowercase English letters
How to think about it
Updated 2026-09-29A common supersequence must contain all characters of both strings, but characters appearing in their Longest Common Subsequence can be shared and included once instead of twice. The shortest supersequence length is therefore len(str1) + len(str2) - len(LCS).
Approaches, worst first
Recursive supersequence alignment
time O(2^(m+n)) · space O(m + n)
Step through both strings. If str1[i] == str2[j], consume both with 1 character added to length. If they differ, branch into consuming str1[i] or str2[j] and take 1 + min. Explores exponential paths.
2D dynamic programming table
time O(m * n) · space O(m * n)
Build dp[i][j] tracking SCS length for prefixes. If str1[i-1] == str2[j-1], dp[i][j] = 1 + dp[i-1][j-1]; otherwise dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1]).
LCS subtraction formulaWrite this one
time O(m * n) · space O(min(m, n))
Compute the length of the Longest Common Subsequence of str1 and str2 using standard 1D space optimized LCS. Return m + n - lcs.
Where people lose marks · 3
- Off-by-one errors when using LCS subtraction formula: make sure to subtract the LCS length once, not twice.
- Reconstructing the supersequence string when only length is requested (or vice versa). Check problem return type.
- Strings with no overlapping letters (e.g. 'abc' and 'def'): LCS is 0, so length is m + n.
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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