Matrix Chain Multiplication
A hard Dynamic Programming interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Dynamic Programming
- Sheets
- 2
- Core for
- 9 roles
The problem
Given a sequence of matrices, find the most efficient way to multiply them together. The cost is determined by the number of scalar multiplications. Return the minimum number of multiplications needed.
Example 1
- Input
- dims = [10,30,5,60] (matrices: 10x30, 30x5, 5x60)
- Output
- 4500
- Why
- Optimal parenthesization is (A1(A2A3)) costing 30*5*60 + 10*30*60 = 9000+18000... Actually (A1A2)A3 costs 10*30*5 + 10*5*60 = 1500+3000=4500.
Example 2
- Input
- dims = [40,20,30,10,30] (matrices: 40x20, 20x30, 30x10, 10x30)
- Output
- 26000
- Why
- The optimal parenthesization minimizes scalar multiplications to 26000.
Example 3
- Input
- dims = [10,20,30] (single matrix 10x30)
- Output
- 0
- Why
- Only one matrix, no multiplication needed.
Constraints
- 2 <= dims.length <= 100
- 1 <= dims[i] <= 500
How to think about it
Updated 2026-09-29Multiplying an entire chain of matrices from i to j requires choosing one final split point k where the product of chain (i..k) multiplies the product of chain (k+1..j). The scalar cost of that outer multiplication is dims[i-1] * dims[k] * dims[j], plus the optimal sub-costs of both halves.
Approaches, worst first
Exhaustive parenthesization
time O(4^n / n^(3/2)) · space O(n)
Try all split points k recursively for every range (i, j). The number of valid parenthesizations is the Catalan number C(n-1), which grows exponentially with n.
Interval dynamic programmingWrite this one
time O(n^3) · space O(n^2)
Compute optimal costs for chain lengths from 2 to n. For chain (i, j), test all split points k from i to j - 1: dp[i][j] = min(dp[i][k] + dp[k+1][j] + dims[i-1] * dims[k] * dims[j]).
Where people lose marks · 3
- Off-by-one errors with matrix indices and dimension array indices. Matrix i (1-indexed) has dimensions dims[i-1] x dims[i].
- A single matrix (dims.length == 2) requires 0 scalar multiplications; base cases where i == j must be explicitly set to 0.
- Incorrect loop ordering: calculating dp[i][j] before its constituent sub-intervals (i, k) and (k+1, j) have been fully solved.
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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