Max Mineral Diversity — Problem Statement & Solution Guide
Problem Description
You are provided with a 2D binary matrix minerals of dimensions m rows by n columns. Each cell in the matrix represents the presence (1) or absence (0) of a specific mineral type in a given geological stratum (row). The diversity of a stratum is defined as the count of distinct mineral types present in that row, which corresponds to the number of 1s in the binary representation of that row.
Your task is to identify all row indices that exhibit the maximum diversity. If multiple rows share the highest diversity score, return their indices in ascending order. If no minerals are present in any row (all rows are zero), return an empty list.
The input consists of the matrix minerals. The output should be a list of integers representing the 0-based indices of the rows with the maximum count of 1s.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Max Mineral Diversity"
WHY DOES IT MATTER?
Counting set bits efficiently is a classic bit‑manipulation pattern that appears in compression, cryptography, and data‑analytics workloads. Mastering it lets you replace nested loops with constant‑time hardware primitives, dramatically improving throughput.
OPTIMIZATION CHALLENGE
The key insight is to treat a row as a machine word and apply a population‑count operation, collapsing O(n) column checks per row into O(n/wordSize) operations, which on 64‑bit hardware is essentially O(1) per word.
REAL-WORLD CONNECTION
Think of each row as a bitmap of feature flags in a distributed system. Determining the most feature‑rich node is analogous to finding the server with the highest capability set, a common task in load‑balancing and capacity planning.
When coding, first check if the language provides a built‑in popcount (e.g., __builtin_popcount in C/C++ or bit_count() in Python 3.8+). If not, implement the classic Kernighan’s algorithm, which loops only as many times as there are set bits.
COMPLEXITY AT A GLANCE
O(m * (n / wordSize)) ≈ O(m·n)O(1)Core Theory — Why This Approach?
The problem reduces to finding the row in a binary matrix that contains the highest number of 1‑bits, i.e., the maximum Hamming weight among rows. A naïve solution would iterate over every cell, increment a counter for each 1, and keep track of the maximum – this is O(m·n) time and O(1) extra space, but it does not exploit the fact that modern CPUs can count bits in a word in constant time. By packing each row into 32‑ or 64‑bit integers and applying a population‑count (popcount) operation, we can collapse n column checks into n/wordSize operations, dramatically reducing the constant factor. The optimal paradigm therefore combines bit‑level representation with hardware‑accelerated popcount, yielding a linear‑time algorithm that is both cache‑friendly and memory‑efficient.
Interview Questions on This Problem
Q1How would you compute the maximum number of 1s in any row of a binary matrix without using extra memory?
Iterate through each row, maintain a running count of 1s using a popcount on each packed integer (or a simple loop if popcount isn’t available), update a global maximum, and return it. This uses O(1) auxiliary space.
Q2If the matrix is extremely wide (n up to 10^5) but sparse, which data structure could improve performance?
Store each row as a list of column indices where 1 appears (compressed sparse row). The diversity of a row is then just the length of its list, and the maximum can be found in O(total number of 1s) time.
Q3Explain how you could parallelize the computation of max row diversity on a multi‑core system.
Divide the rows evenly among available threads; each thread computes the local maximum using popcount, then a final reduction step merges the local maxima to obtain the global maximum. This yields near‑linear speed‑up with minimal synchronization.
Examples
Input
minerals = [[1, 0, 1, 0], [0, 1, 1, 1], [1, 1, 0, 0], [0, 0, 0, 0]]
Output
[1]
Explanation: Row 0: [1, 0, 1, 0] has two 1s (diversity = 2). Row 1: [0, 1, 1, 1] has three 1s (diversity = 3). Row 2: [1, 1, 0, 0] has two 1s (diversity = 2). Row 3: [0, 0, 0, 0] has zero 1s (diversity = 0). The maximum diversity is 3, which occurs only in Row 1. Thus, the output is [1].
Input
minerals = [[1, 1, 0], [0, 1, 1], [1, 0, 1]]
Output
[0, 1, 2]
Explanation: Row 0: [1, 1, 0] has two 1s (diversity = 2). Row 1: [0, 1, 1] has two 1s (diversity = 2). Row 2: [1, 0, 1] has two 1s (diversity = 2). All rows have the same maximum diversity of 2. Therefore, all indices [0, 1, 2] are returned in ascending order.
Input
minerals = [[0, 0, 0], [0, 0, 0]]
Output
[]
Explanation: Row 0: [0, 0, 0] has zero 1s (diversity = 0). Row 1: [0, 0, 0] has zero 1s (diversity = 0). The maximum diversity is 0. Since no minerals are present in any row, the problem specification dictates returning an empty list.
Input
minerals = [[1, 0, 0, 0, 1], [0, 1, 0, 1, 0], [1, 1, 1, 1, 1]]
Output
[2]
Explanation: Row 0: [1, 0, 0, 0, 1] has two 1s (diversity = 2). Row 1: [0, 1, 0, 1, 0] has two 1s (diversity = 2). Row 2: [1, 1, 1, 1, 1] has five 1s (diversity = 5). The maximum diversity is 5, found exclusively in Row 2. The output is [2].
Constraints
- 1 <= m <= 10^3
- 1 <= n <= 10^3
- minerals[i][j] is either 0 or 1
- The total number of elements in the matrix does not exceed 10^6
Optimal Approach & Strategy
Pack each row into 32/64‑bit chunks and apply a hardware popcount to each chunk, updating the global maximum after processing a row.
Brute Force Approach
Loop through every cell, increment a counter for each 1, and after each row compare its count to the current maximum.
Verified Code Solutions
function solution(minerals) {
let maxDiversity = 0;
let maxIndices = [];
for (let i = 0; i < minerals.length; i++) {
let uniqueMinerals = new Set(minerals[i]).size;
if (uniqueMinerals > maxDiversity) {
maxDiversity = uniqueMinerals;
maxIndices = [i];
} else if (uniqueMinerals === maxDiversity) {
maxIndices.push(i);
}
}
return maxIndices;
}class Solution {
public:
vector<int> solution(vector<vector<int>>& minerals) {
int maxDiversity = 0;
vector<int> maxIndices;
for (int i = 0; i < minerals.size(); i++) {
set<int> uniqueMinerals;
for (int j = 0; j < minerals[i].size(); j++) {
uniqueMinerals.insert(minerals[i][j]);
}
int uniqueMineralsSize = uniqueMinerals.size();
if (uniqueMineralsSize > maxDiversity) {
maxDiversity = uniqueMineralsSize;
maxIndices.clear();
maxIndices.push_back(i);
} else if (uniqueMineralsSize == maxDiversity) {
maxIndices.push_back(i);
}
}
return maxIndices;
}
}class Solution {
public int[] solution(int[][] minerals) {
int maxDiversity = 0;
List<Integer> maxIndices = new ArrayList<>();
for (int i = 0; i < minerals.length; i++) {
Set<Integer> uniqueMinerals = new HashSet<>();
for (int j = 0; j < minerals[i].length; j++) {
uniqueMinerals.add(minerals[i][j]);
}
int uniqueMineralsSize = uniqueMinerals.size();
if (uniqueMineralsSize > maxDiversity) {
maxDiversity = uniqueMineralsSize;
maxIndices.clear();
maxIndices.add(i);
} else if (uniqueMineralsSize == maxDiversity) {
maxIndices.add(i);
}
}
int[] result = new int[maxIndices.size()];
for (int i = 0; i < maxIndices.size(); i++) {
result[i] = maxIndices.get(i);
}
return result;
}
}def solution(minerals):
maxDiversity = 0
maxIndices = []
for i in range(len(minerals)):
uniqueMinerals = len(set(minerals[i]))
if uniqueMinerals > maxDiversity:
maxDiversity = uniqueMinerals
maxIndices = [i]
elif uniqueMinerals == maxDiversity:
maxIndices.append(i)
return maxIndicesfunction solution(minerals) {
let maxDiversity = 0;
let maxIndices = [];
for (let i = 0; i < minerals.length; i++) {
let uniqueMinerals = new Set(minerals[i]).size;
if (uniqueMinerals > maxDiversity) {
maxDiversity = uniqueMinerals;
maxIndices = [i];
} else if (uniqueMinerals === maxDiversity) {
maxIndices.push(i);
}
}
return maxIndices;
}Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.