What is the time complexity of Rank Transform of a Matrix?

The optimal solution has a time complexity of O(m*n*alpha(m+n)).

What is the space complexity of Rank Transform of a Matrix?

The space complexity is O(m*n).

What data structures are used to solve Rank Transform of a Matrix?

This problem uses Array, Union Find, Graph, Topological Sort, Sorting, Matrix. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Rank Transform of a Matrix?

Insight 1: The core challenge lies in ensuring that ranks are consistent across rows and columns for elements with different values. Equal values must have the same rank. Insight 2: Union Find is crucial to efficiently determine connected components formed by equal values in the same rows or columns. It ensures that these elements are treated as a single unit when assigning ranks.

Which companies ask LeetCode 1632. Rank Transform of a Matrix in interviews?

This problem is frequently asked at Citadel.

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Rank Transform of a Matrix - LeetCode 1632 Solution

Rank Transform of a Matrix - Complete Solution Guide

Rank Transform of a Matrix is LeetCode problem 1632, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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

Detailed Explanation

The problem asks us to transform a given m x n matrix into a new matrix of the same dimensions, where each element's value represents its rank within the original matrix. The rank is an integer, starting from 1, indicating how large an element is compared to others. The rank must adhere to these rules: if two elements are in the same row or column, their ranks must reflect their relative values (if p < q, then rank(p) < rank(q); if p == q, rank(p) == rank(q); if p > q, rank(p) > rank(q)). Finally, the rank should be as small as possible.

Solution Approach

The provided solutions utilize a combination of sorting and the Union-Find data structure to efficiently determine and assign ranks. The algorithm sorts all unique values present in the matrix. It then iterates through these sorted values, identifying all cells containing the current value. Within each set of cells sharing the same value, it uses Union-Find to group cells that are connected by being in the same row or column. Finally, for each connected component, it determines the maximum rank of any cell in that component's row or column and assigns the next highest rank to all cells in the component.

Step-by-Step Algorithm

Step 1: Group cells based on their values using a hash map. This allows us to efficiently access all cells containing a particular value.
Step 2: Sort the unique values in ascending order. This ensures that we assign ranks starting from the smallest value, making sure ranks are as small as possible.
Step 3: Iterate through the sorted values. For each value:
Step 4: Initialize a Union-Find data structure for all cells with the current value. This allows us to find connected components formed by cells in the same row or column.
Step 5: Iterate through the cells with the current value. For each cell, perform unions with other cells in the same row or column that also have the same value.
Step 6: Identify the connected components. Each component represents a set of cells that must have the same rank.
Step 7: For each connected component, find the maximum rank among all rows and columns that contain a cell in the component.
Step 8: Assign a new rank, which is one greater than the maximum rank found in the previous step, to all cells in the connected component.
Step 9: Update the rank array for the rows and columns affected by the newly assigned ranks.
Step 10: Return the rank matrix.

Key Insights

Insight 1: The core challenge lies in ensuring that ranks are consistent across rows and columns for elements with different values. Equal values must have the same rank.
Insight 2: Union Find is crucial to efficiently determine connected components formed by equal values in the same rows or columns. It ensures that these elements are treated as a single unit when assigning ranks.
Insight 3: Processing elements in sorted order of their values allows us to incrementally build up the rank matrix, guaranteeing that the ranks are as small as possible while adhering to the problem's constraints.

Complexity Analysis

Time Complexity: O(m*n*alpha(m+n))

Space Complexity: O(m*n)

Topics

This problem involves: Array, Union Find, Graph, Topological Sort, Sorting, Matrix.

Companies

Asked at: Citadel.

Rank Transform of a Matrix | LeetCode 1632

Hard

Array Union Find Graph Topological Sort Sorting Matrix

Time: O(m*n*alpha(m+n))Space: O(m*n)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int[][] matrixRankTransform(int[][] matrix) {
        int m = matrix.length;
        int n = matrix[0].length;

        Map<Integer, List<int[]>> groupsByVal = new HashMap<>();
        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                int val = matrix[r][c];
                groupsByVal.computeIfAbsent(val, k -> new ArrayList<>()).add(new int[]{r, c});
            }
        }

        int[][] answer = new int[m][n];
        int[] rank = new int[m + n];

        List<Integer> sortedValues = new ArrayList<>(groupsByVal.keySet());
        Collections.sort(sortedValues);

        for (int val : sortedValues) {
            List<int[]> coords = groupsByVal.get(val);
            int k = coords.size();

            int[] parent = new int[k];
            for (int i = 0; i < k; i++) {
                parent[i] = i;
            }

            Map<Integer, Integer> rowMap = new HashMap<>();
            Map<Integer, Integer> colMap = new HashMap<>();
            for (int i = 0; i < k; i++) {
                int r = coords.get(i)[0];
                int c = coords.get(i)[1];
                if (rowMap.containsKey(r)) {
                    union(parent, i, rowMap.get(r));
                }
                rowMap.put(r, i);

                if (colMap.containsKey(c)) {
                    union(parent, i, colMap.get(c));
                }
                colMap.put(c, i);
            }

            Map<Integer, List<Integer>> components = new HashMap<>();
            for (int i = 0; i < k; i++) {
                int root = find(parent, i);
                components.computeIfAbsent(root, l -> new ArrayList<>()).add(i);
            }

            for (List<Integer> componentIndices : components.values()) {
                int maxRank = 0;
                for (int i : componentIndices) {
                    int r = coords.get(i)[0];
                    int c = coords.get(i)[1];
                    maxRank = Math.max(maxRank, Math.max(rank[r], rank[m + c]));
                }

                int newRank = maxRank + 1;

                for (int i : componentIndices) {
                    int r = coords.get(i)[0];
                    int c = coords.get(i)[1];
                    answer[r][c] = newRank;
                }
            }

            for (int[] coord : coords) {
                int r = coord[0];
                int c = coord[1];
                rank[r] = answer[r][c];
                rank[m + c] = answer[r][c];
            }
        }

        return answer;
    }

    private int find(int[] parent, int i) {
        if (parent[i] == i) {
            return i;
        }
        parent[i] = find(parent, parent[i]);
        return parent[i];
    }

    private void union(int[] parent, int i, int j) {
        int rootI = find(parent, i);
        int rootJ = find(parent, j);
        if (rootI != rootJ) {
            parent[rootJ] = rootI;
        }
    }
}

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Problem Info

Description

Problem 1632: Rank Transform of a Matrix - Detailed explanation and solution approaches.

Topics

Array Union Find Graph Topological Sort Sorting Matrix

Companies

Citadel

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1632

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Group cells based on their values using a hash map. This allows us to efficiently access all cells containing a particular value.
2
Step 2: Sort the unique values in ascending order. This ensures that we assign ranks starting from the smallest value, making sure ranks are as small as possible.
3
Step 3: Iterate through the sorted values. For each value:
4
Step 4: Initialize a Union-Find data structure for all cells with the current value. This allows us to find connected components formed by cells in the same row or column.
5
Step 5: Iterate through the cells with the current value. For each cell, perform unions with other cells in the same row or column that also have the same value.
6
Step 6: Identify the connected components. Each component represents a set of cells that must have the same rank.
7
Step 7: For each connected component, find the maximum rank among all rows and columns that contain a cell in the component.
8
Step 8: Assign a new rank, which is one greater than the maximum rank found in the previous step, to all cells in the connected component.
9
Step 9: Update the rank array for the rows and columns affected by the newly assigned ranks.
10
Step 10: Return the rank matrix.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n*alpha(m+n))

💾

Space Complexity

O(m*n)

📝

Why?

The time complexity is dominated by the Union-Find operations within the loop iterating through the sorted values. The space complexity stems from storing the output matrix and the auxiliary data structures used in Union-Find and the grouping of cells by value.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Topological Sort

One can consider building a directed graph where an edge from node i to j means rank[i] < rank[j]. Then topological sort can be applied to determine ranks, however implementation is complex.

Time: O(m*n*log(m*n)) (Estimation, can vary based on implementation details)Space: O(m*n)

Trade-offs: Topological Sort requires construction of a directed graph which is more complex to implement and reason about compared to Union Find. The time complexity would depend on sort used while performing topological sorting.

✨

Key Insights

1Insight 1: The core challenge lies in ensuring that ranks are consistent across rows and columns for elements with different values. Equal values must have the same rank.
2Insight 2: Union Find is crucial to efficiently determine connected components formed by equal values in the same rows or columns. It ensures that these elements are treated as a single unit when assigning ranks.
3Insight 3: Processing elements in sorted order of their values allows us to incrementally build up the rank matrix, guaranteeing that the ranks are as small as possible while adhering to the problem's constraints.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Graph

Nodes connected by edges. Like a social network or city map!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Matrix with all equal values: In this case, all elements will belong to the same connected component. The initial rank values will be 0, so the new rank assigned to all elements will be 1, fulfilling the requirement that rank should start from 1.
Edge case 2: Matrix with duplicate values in rows or columns: The Union-Find correctly merges the cells with the same value in the same row or column ensuring that all get the same rank.
Edge case 3: Empty Matrix: The code should handle the case where the input matrix is empty (either m or n is zero). This will likely require a check at the beginning to return an empty matrix.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling equal values correctly: Forgetting to use Union Find and assigning different ranks to elements with the same value that are in the same row or column will lead to incorrect results.

✗Common mistake 2: Assigning ranks without considering existing row/column maximums: Forgetting to find the maximum rank in a row or column before assigning a new rank will result in ranks that are not the smallest possible.

✗Common mistake 3: Not sorting the values: If the values are not sorted, then rank assignment can be arbitrary and violate the constraint that ranks should be as small as possible.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Leaderboard Ranking: Ranking players on a leaderboard based on their scores, considering ties and ensuring consistency across different games or seasons.

🏦Real-world scenario 2: Data Analysis: Ranking features or attributes in a dataset based on their importance or relevance, ensuring consistent ranking across different data sources or models.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 13, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Rank Transform of a Matrix"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether union find techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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