LeetCode #1632 — HARD

Rank Transform of a Matrix

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

Given an m x n matrix, return a new matrix answer where answer[row][col] is the rank of matrix[row][col].

The rank is an integer that represents how large an element is compared to other elements. It is calculated using the following rules:

  • The rank is an integer starting from 1.
  • If two elements p and q are in the same row or column, then:
    • If p < q then rank(p) < rank(q)
    • If p == q then rank(p) == rank(q)
    • If p > q then rank(p) > rank(q)
  • The rank should be as small as possible.

The test cases are generated so that answer is unique under the given rules.

Example 1:

Input: matrix = [[1,2],[3,4]]
Output: [[1,2],[2,3]]
Explanation:
The rank of matrix[0][0] is 1 because it is the smallest integer in its row and column.
The rank of matrix[0][1] is 2 because matrix[0][1] > matrix[0][0] and matrix[0][0] is rank 1.
The rank of matrix[1][0] is 2 because matrix[1][0] > matrix[0][0] and matrix[0][0] is rank 1.
The rank of matrix[1][1] is 3 because matrix[1][1] > matrix[0][1], matrix[1][1] > matrix[1][0], and both matrix[0][1] and matrix[1][0] are rank 2.

Example 2:

Input: matrix = [[7,7],[7,7]]
Output: [[1,1],[1,1]]

Example 3:

Input: matrix = [[20,-21,14],[-19,4,19],[22,-47,24],[-19,4,19]]
Output: [[4,2,3],[1,3,4],[5,1,6],[1,3,4]]

Constraints:

  • m == matrix.length
  • n == matrix[i].length
  • 1 <= m, n <= 500
  • -109 <= matrix[row][col] <= 109
Patterns Used
🔗Union-Find→📋Topological Sort→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: Given an m x n matrix, return a new matrix answer where answer[row][col] is the rank of matrix[row][col]. The rank is an integer that represents how large an element is compared to other elements. It is calculated using the following rules: The rank is an integer starting from 1. If two elements p and q are in the same row or column, then: If p < q then rank(p) < rank(q) If p == q then rank(p) == rank(q) If p > q then rank(p) > rank(q) The rank should be as small as possible. The test cases are generated so that answer is unique under the given rules.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Union-Find · Topological Sort

Example 1

[[1,2],[3,4]]

Example 2

[[7,7],[7,7]]

Example 3

[[20,-21,14],[-19,4,19],[22,-47,24],[-19,4,19]]

Related Problems

  • Rank Transform of an Array (rank-transform-of-an-array)
  • GCD Sort of an Array (gcd-sort-of-an-array)
Step 02

Core Insight

What unlocks the optimal approach

  • Sort the cells by value and process them in increasing order.
  • The rank of a cell is the maximum rank in its row and column plus one.
  • Handle the equal cells by treating them as components using a union-find data structure.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1632: Rank Transform of a Matrix
class UnionFind {
    private int[] p;
    private int[] size;

    public UnionFind(int n) {
        p = new int[n];
        size = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
            size[i] = 1;
        }
    }

    public int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }

    public void union(int a, int b) {
        int pa = find(a), pb = find(b);
        if (pa != pb) {
            if (size[pa] > size[pb]) {
                p[pb] = pa;
                size[pa] += size[pb];
            } else {
                p[pa] = pb;
                size[pb] += size[pa];
            }
        }
    }

    public void reset(int x) {
        p[x] = x;
        size[x] = 1;
    }
}

class Solution {
    public int[][] matrixRankTransform(int[][] matrix) {
        int m = matrix.length, n = matrix[0].length;
        TreeMap<Integer, List<int[]>> d = new TreeMap<>();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                d.computeIfAbsent(matrix[i][j], k -> new ArrayList<>()).add(new int[] {i, j});
            }
        }
        int[] rowMax = new int[m];
        int[] colMax = new int[n];
        int[][] ans = new int[m][n];
        UnionFind uf = new UnionFind(m + n);
        int[] rank = new int[m + n];
        for (var ps : d.values()) {
            for (var p : ps) {
                uf.union(p[0], p[1] + m);
            }
            for (var p : ps) {
                int i = p[0], j = p[1];
                rank[uf.find(i)] = Math.max(rank[uf.find(i)], Math.max(rowMax[i], colMax[j]));
            }
            for (var p : ps) {
                int i = p[0], j = p[1];
                ans[i][j] = 1 + rank[uf.find(i)];
                rowMax[i] = ans[i][j];
                colMax[j] = ans[i][j];
            }
            for (var p : ps) {
                uf.reset(p[0]);
                uf.reset(p[1] + m);
            }
        }
        return ans;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(α(n))
Space
O(n)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(n) space

Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.

UNION-FIND
O(α(n)) time
O(n) space

With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.

Shortcut: Union-Find with path compression + rank → O(α(n)) per operation ≈ O(1). Just say “nearly constant.”
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Related Problems
#128Longest Consecutive Sequencemedium→#130Surrounded Regionsmedium→#200Number of Islandsmedium→#329Longest Increasing Path in a Matrixhard→#399Evaluate Divisionmedium→