Matrix Chain Visualization for Coding Interviews
Optimal parenthesization; interval DP with split-point k.
Concept
Matrix Chain Multiplication asks: given matrices A1, A2, ..., An with compatible dimensions, in what order should you multiply them to minimize the total number of scalar multiplications? Matrix multiplication is associative, so the result is the same regardless of parenthesization, but the cost is not — multiplying a 10x100 by a 100x5 and then a 5x50 takes 10*100*5 + 10*5*50 = 7500 scalar mults, while the other grouping costs 100*5*50 + 10*100*50 = 75000. Over long chains the factor can be 1000x. The problem is the archetypal example of interval DP: every decision is 'where to split an interval into two halves', and the answer is the minimum over all split points. It also appears in optimal binary search tree construction and SQL query join ordering.
How it works
Let dims[] be the dimension vector such that matrix Ai has shape dims[i-1] x dims[i]. Let dp[i][j] be the minimum scalar multiplications needed to compute the product Ai * A(i+1) * ... * Aj. For length-1 chains (i == j) the cost is zero. For longer chains, we try every possible split point k where i <= k < j: the cost of splitting here is dp[i][k] + dp[k+1][j] + dims[i-1] * dims[k] * dims[j], meaning the cost of computing each half plus the cost of multiplying the two resulting matrices of shape dims[i-1] x dims[k] and dims[k] x dims[j]. Take the minimum over k. Fill the table by increasing chain length: length 2 first, then 3, up to n. To reconstruct the parenthesization, store the split point that achieved the minimum and recurse: for the chain (i, j), split at s[i][j], emit '(' + parens(i, s[i][j]) + ' x ' + parens(s[i][j] + 1, j) + ')'. Single matrices are emitted as 'Ai'. The algorithm is O(n^3) time and O(n^2) space, which handles chains up to a few thousand matrices comfortably.
Implementation
public final class MatrixChain {
public static final class Result {
public final long minCost;
public final String parens;
public Result(long minCost, String parens) { this.minCost = minCost; this.parens = parens; }
}
private MatrixChain() {}
public static Result matrixChainOrder(int[] dims) {
int n = dims.length - 1;
if (n <= 0) return new Result(0, "");
long[][] dp = new long[n + 1][n + 1];
int[][] split = new int[n + 1][n + 1];
for (int len = 2; len <= n; len++) {
for (int i = 1; i + len - 1 <= n; i++) {
int j = i + len - 1;
dp[i][j] = Long.MAX_VALUE;
for (int k = i; k < j; k++) {
long cost = dp[i][k] + dp[k + 1][j]
+ (long) dims[i - 1] * dims[k] * dims[j];
if (cost < dp[i][j]) {
dp[i][j] = cost;
split[i][j] = k;
}
}
}
}
String parens = buildParens(split, 1, n);
return new Result(dp[1][n], parens);
}
private static String buildParens(int[][] split, int i, int j) {
if (i == j) return "A" + i;
int k = split[i][j];
return "(" + buildParens(split, i, k) + " x " + buildParens(split, k + 1, j) + ")";
}
}
Complexity
- Time:
O(n^3) - Space:
O(n^2)
Common pitfalls
- Indexing dims off by one — matrix Ai has shape dims[i-1] x dims[i], not dims[i] x dims[i+1].
- Using int for cost — scalar multiplication counts easily overflow 32 bits on long chains.
- Iterating by i and j instead of by chain length; many cells get filled before their dependencies exist.
- Forgetting base case dp[i][i] = 0 — default zero works in Java but not in every language.
- Reporting the shape of the result instead of the number of multiplications.
Key design decisions & trade-offs
- Recursive memoized vs iterative bottom-up — Chosen: Iterative for cache locality. Bottom-up iterates the table in order; memoization adds recursion overhead.
- Store split table vs reconstruct on demand — Chosen: Store the split table. O(n^2) extra memory in exchange for O(n) parenthesization reconstruction.
- Hu-Shing O(n log n) — Chosen: Only for very large n. Complex to implement; the O(n^3) DP handles n = 1000 in milliseconds.
Practice problems
Interview follow-ups
- Adapt the DP to find the optimal order for a chain of database joins.
- Implement optimal binary search tree construction — same interval DP pattern.
- Solve the Burst Balloons problem (LeetCode 312) using the same split structure.
- Extend to weighted costs where different pairs have different per-mult costs.
Why it matters
Matrix Chain is a core part of the Dynamic Programming toolkit. If you are preparing for a technical interview at a top-tier company or teaching a course, a working visualization is the fastest path from "I read about it" to "I understand it".