C++ Program To Find The Row With Maximum Number Of 1st

Finding the row with the maximum number of ones in a binary matrix is a common problem in computer science, particularly in areas involving data analysis, image processing, and algorithms. This task requires efficiently scanning through a two-dimensional array to identify the row that contains the highest count of the digit one.

In this article, you will learn how to approach this problem using C++, exploring different methods from straightforward iteration to more optimized techniques for specific matrix structures. We will cover the logic, implementation, and analysis of these solutions.

Problem Statement

The core problem is to identify the index of the row in a given binary matrix (a matrix containing only 0s and 1s) that has the highest number of 1s. If multiple rows share the same maximum count of 1s, any one of their indices is considered a valid output.

Consider a scenario in a simple binary image where '1' represents an active pixel and '0' represents an inactive one. You might want to find the row that has the most active pixels to identify a region of interest. Similarly, in a dataset representing features as 0s and 1s, finding the row with the most 1s could indicate the data point with the most active features.

Example

Consider the following 4x4 binary matrix:

0 1 1 1
0 0 1 1
1 1 1 1
0 0 0 0

• Row 0: Three 1s
• Row 1: Two 1s
• Row 2: Four 1s
• Row 3: Zero 1s

The row with the maximum number of 1s is Row 2.

Background & Knowledge Prerequisites

To effectively understand and implement the solutions discussed, readers should be familiar with the following C++ concepts:

• Basic Syntax: Variables, data types, input/output operations.
• Loops: for loops for iteration.
• Arrays: Declaration, initialization, and traversal of one-dimensional and two-dimensional arrays (matrices).
• Functions: Defining and calling functions.

No special libraries or complex data structures are strictly required beyond the standard input/output stream (iostream).

Use Cases or Case Studies

Here are a few practical scenarios where finding the row with the maximum number of 1s can be useful:

• Image Processing: In a binary image, identifying the row with the most 'on' pixels (represented by 1s) could help in feature extraction or region of interest detection.
• Data Analysis: When analyzing datasets where features are represented as binary values (present/absent), finding the row with the most 1s might point to the data entry with the highest number of active attributes.
• Security Logs: If logs are represented as binary matrices (e.g., firewall rules or intrusion detection system flags), a row with many 1s could indicate a particularly active or problematic event sequence.
• Bioinformatics: In genetic sequencing analysis, binary matrices might represent the presence or absence of certain genetic markers. Identifying rows with many 1s could highlight samples with multiple markers.
• Game Development: In tile-based games, a matrix could represent game states. Finding a row with many 1s might indicate a dense area of collectibles or obstacles.

Solution Approaches

We will explore two primary approaches: a straightforward brute-force method and an optimized approach specifically for matrices where rows are sorted (i.e., all 0s appear before all 1s).

1. Brute-Force Row Iteration

This approach involves iterating through each row of the matrix and, for each row, counting the number of 1s. We keep track of the maximum count found so far and the index of the row corresponding to that count.

• Summary: Traverse each row and each column, counting 1s for every row, and update the maximum count and its corresponding row index.
• Code Example:

// Brute-Force Max Ones Row
    #include <iostream>
    #include <vector> // Using std::vector for dynamic arrays
    #include <algorithm> // For std::max, not strictly necessary but useful

    using namespace std;

    // Function to find the row with maximum 1s using brute-force
    int findMaxOnesRowBruteForce(const vector<vector<int>>& matrix) {
        int rows = matrix.size();
        if (rows == 0) {
            return -1; // Handle empty matrix
        }
        int cols = matrix[0].size();
        if (cols == 0) {
            return -1; // Handle empty rows
        }

        int maxOnesCount = 0;
        int maxOnesRowIndex = -1;

        // Step 1: Iterate through each row
        for (int i = 0; i < rows; ++i) {
            int currentOnesCount = 0;
            // Step 2: Iterate through each element in the current row
            for (int j = 0; j < cols; ++j) {
                if (matrix[i][j] == 1) {
                    currentOnesCount++;
                }
            }
            // Step 3: Check if current row has more 1s than the maximum found so far
            if (currentOnesCount > maxOnesCount) {
                maxOnesCount = currentOnesCount;
                maxOnesRowIndex = i;
            }
        }
        return maxOnesRowIndex;
    }

    int main() {
        // Example Matrix 1
        vector<vector<int>> matrix1 = {
            {0, 1, 1, 1},
            {0, 0, 1, 1},
            {1, 1, 1, 1},
            {0, 0, 0, 0}
        };

        // Example Matrix 2 (all zeros)
        vector<vector<int>> matrix2 = {
            {0, 0, 0},
            {0, 0, 0},
            {0, 0, 0}
        };

        // Example Matrix 3 (all ones)
        vector<vector<int>> matrix3 = {
            {1, 1, 1},
            {1, 1, 1},
            {1, 1, 1}
        };

        // Step 1: Test with matrix1
        cout << "Matrix 1:" << endl;
        for (const auto& row : matrix1) {
            for (int val : row) {
                cout << val << " ";
            }
            cout << endl;
        }
        cout << "Row with max 1s (Brute Force): " << findMaxOnesRowBruteForce(matrix1) << endl << endl;

        // Step 2: Test with matrix2
        cout << "Matrix 2:" << endl;
        for (const auto& row : matrix2) {
            for (int val : row) {
                cout << val << " ";
            }
            cout << endl;
        }
        cout << "Row with max 1s (Brute Force): " << findMaxOnesRowBruteForce(matrix2) << endl << endl;

        // Step 3: Test with matrix3
        cout << "Matrix 3:" << endl;
        for (const auto& row : matrix3) {
            for (int val : row) {
                cout << val << " ";
            }
            cout << endl;
        }
        cout << "Row with max 1s (Brute Force): " << findMaxOnesRowBruteForce(matrix3) << endl << endl;

        return 0;
    }

Run

• Sample Output:

Matrix 1:
    0 1 1 1 
    0 0 1 1 
    1 1 1 1 
    0 0 0 0 
    Row with max 1s (Brute Force): 2

    Matrix 2:
    0 0 0 
    0 0 0 
    0 0 0 
    Row with max 1s (Brute Force): -1

    Matrix 3:
    1 1 1 
    1 1 1 
    1 1 1 
    Row with max 1s (Brute Force): 0

• Stepwise Explanation:

1. Initialize maxOnesCount to 0 and maxOnesRowIndex to -1. These variables will store the highest count of 1s found and the index of the row it belongs to.
2. Use an outer for loop to iterate through each row of the matrix, from the first row (index 0) to the last.
3. Inside the outer loop, initialize currentOnesCount to 0 for the current row.
4. Use an inner for loop to iterate through each element in the current row.
5. If an element is 1, increment currentOnesCount.
6. After the inner loop completes (meaning all elements in the current row have been checked), compare currentOnesCount with maxOnesCount.
7. If currentOnesCount is greater, update maxOnesCount with currentOnesCount and maxOnesRowIndex with the current row's index.
8. After the outer loop finishes, maxOnesRowIndex will hold the index of the row with the most 1s.

2. Optimized for Sorted Rows (Two-Pointer Technique)

This optimization applies when each row in the matrix is sorted in non-decreasing order, meaning all 0s appear before all 1s. In such a matrix, the number of 1s in a row is equivalent to cols - index_of_first_one. We can leverage this property with a two-pointer approach, similar to how we search in a sorted matrix.

• Summary: Start from the top-right corner of the matrix. If the current element is a 1, it means all elements to its left in the same row are also 1s (or 0s, but we've found a 1). This row is a candidate, and we move left to find potentially more 1s. If the current element is a 0, we move down to the next row as this row and elements to its left cannot have more 1s than rows below it if we were to move left.
• Code Example:

// Optimized Max Ones Row (Sorted Rows)
    #include <iostream>
    #include <vector>

    using namespace std;

    // Function to find the row with maximum 1s using an optimized two-pointer approach
    // Assumes rows are sorted (all 0s followed by all 1s)
    int findMaxOnesRowOptimized(const vector<vector<int>>& matrix) {
        int rows = matrix.size();
        if (rows == 0) {
            return -1; // Handle empty matrix
        }
        int cols = matrix[0].size();
        if (cols == 0) {
            return -1; // Handle empty rows
        }

        int maxOnesRowIndex = -1;
        int j = cols - 1; // Start from the last column (rightmost)

        // Step 1: Find the first column where a 0 can exist for the first row
        // This effectively finds the first 1 in the first row, or sets j to -1 if all are 0s.
        // Or points to the column index of the first 1 encountered.
        while (j >= 0 && matrix[0][j] == 1) {
            maxOnesRowIndex = 0; // If row 0 has 1s, it's a candidate
            j--;
        }

        // Step 2: Iterate through the remaining rows from the second row
        for (int i = 1; i < rows; ++i) {
            // While current row has 1s at column j or to its right
            // and we haven't gone out of bounds to the left
            while (j >= 0 && matrix[i][j] == 1) {
                maxOnesRowIndex = i; // This row has at least as many 1s as previous candidates, potentially more
                j--; // Move left to find more 1s in this row
            }
        }
        return maxOnesRowIndex;
    }

    int main() {
        // Example Matrix 1 (sorted rows)
        vector<vector<int>> matrix1 = {
            {0, 0, 1, 1},
            {0, 1, 1, 1},
            {0, 0, 0, 1},
            {0, 0, 0, 0}
        };

        // Example Matrix 2 (another sorted matrix)
        vector<vector<int>> matrix2 = {
            {1, 1, 1},
            {0, 0, 0},
            {0, 1, 1}
        };

        // Example Matrix 3 (all zeros)
        vector<vector<int>> matrix3 = {
            {0, 0, 0},
            {0, 0, 0},
            {0, 0, 0}
        };

        // Step 1: Test with matrix1
        cout << "Matrix 1 (Sorted Rows):" << endl;
        for (const auto& row : matrix1) {
            for (int val : row) {
                cout << val << " ";
            }
            cout << endl;
        }
        cout << "Row with max 1s (Optimized): " << findMaxOnesRowOptimized(matrix1) << endl << endl;

        // Step 2: Test with matrix2
        cout << "Matrix 2 (Sorted Rows):" << endl;
        for (const auto& row : matrix2) {
            for (int val : row) {
                cout << val << " ";
            }
            cout << endl;
        }
        cout << "Row with max 1s (Optimized): " << findMaxOnesRowOptimized(matrix2) << endl << endl;

        // Step 3: Test with matrix3
        cout << "Matrix 3 (Sorted Rows):" << endl;
        for (const auto& row : matrix3) {
            for (int val : row) {
                cout << val << " ";
            }
            cout << endl;
        }
        cout << "Row with max 1s (Optimized): " << findMaxOnesRowOptimized(matrix3) << endl << endl;


        return 0;
    }

Run

• Sample Output:

Matrix 1 (Sorted Rows):
    0 0 1 1 
    0 1 1 1 
    0 0 0 1 
    0 0 0 0 
    Row with max 1s (Optimized): 1

    Matrix 2 (Sorted Rows):
    1 1 1 
    0 0 0 
    0 1 1 
    Row with max 1s (Optimized): 0

    Matrix 3 (Sorted Rows):
    0 0 0 
    0 0 0 
    0 0 0 
    Row with max 1s (Optimized): -1

• Stepwise Explanation:

1. Initialize maxOnesRowIndex to -1. This will store the index of the row with the most 1s.
2. Initialize a column pointer j to cols - 1 (the last column). This pointer will always point to the rightmost potential 0 or the first 1 in the current search.
3. Iterate through the first row: while j is valid and matrix[0][j] is 1, set maxOnesRowIndex to 0 and decrement j. This finds the first 1 in row 0, or sets j to -1 if row 0 has no 1s.
4. Iterate through the remaining rows (from i = 1 to rows - 1):

• For each row i, while j is valid (j >= 0) AND the element matrix[i][j] is 1:
• This means the current row i has at least cols - j number of 1s, which is more than or equal to what previous rows could offer (since we moved j to the left).
• Update maxOnesRowIndex to i.
• Decrement j to move further left in the current row, potentially finding even more 1s.
• If matrix[i][j] is 0, it means all elements to the left in this row are also 0s (due to the sorted nature of rows). We cannot find more 1s by moving left in *this* row. The j pointer already points to the position that gave the maximum number of 1s for previous rows. We simply move to the next row (increment i) and continue with the current j.

1. The loop terminates when all rows are processed. maxOnesRowIndex will hold the index of the row with the maximum number of 1s.

This approach is efficient because the column pointer j only moves left, and the row pointer i only moves down. In the worst case, j moves cols times and i moves rows times, leading to an O(rows + cols) time complexity, which is significantly better than O(rows * cols) for the brute-force approach on large matrices.

Conclusion

Finding the row with the maximum number of 1s in a binary matrix can be tackled using different algorithmic strategies. The brute-force approach, while simple to understand and implement, iterates through every element, making it less efficient for very large matrices. However, if the problem guarantees that each row is sorted (all 0s followed by all 1s), a significantly more optimized two-pointer technique can be employed. This optimized method leverages the sorted property to achieve a linear time complexity relative to the dimensions of the matrix. Choosing the right approach depends on the specific constraints and characteristics of the input data.

Summary

• Problem: Identify the row index in a binary matrix that contains the highest count of 1s.
• Brute-Force Approach:
• Iterates through each row and each column.
• Counts 1s for every row.
• Keeps track of the maximum count and corresponding row index.
• Time Complexity: O(rows \* cols).
• Applicable to any binary matrix.
• Optimized Approach (for Sorted Rows):
• Assumes rows are sorted (0s then 1s).
• Uses a two-pointer technique starting from the top-right corner.
• Moves left if a '1' is found (potential new max row), moves down if a '0' is found.
• Time Complexity: O(rows + cols).
• More efficient for large, sorted binary matrices.
• Key Considerations: The structure of the input matrix (sorted or unsorted rows) dictates which approach is more suitable for optimal performance.