Example: Efficient Pointer Max Ones Row in C

// Efficient Pointer Max Ones Row
#include <stdio.h>

#define ROWS 4
#define COLS 4

int main() {
    int matrix[ROWS][COLS] = {
        {0, 0, 0, 1},
        {0, 0, 1, 1},
        {0, 1, 1, 1},
        {0, 0, 0, 0}
    };

    int maxOnesRow = -1; // Initialize to -1 to indicate no 1s found yet
    int i = 0;           // Current row index, starts at the top
    int j = COLS - 1;    // Current column index, starts at the rightmost column

    // Step 1: Traverse the matrix starting from the top-right corner
    // The loop continues as long as we are within matrix bounds
    while (i < ROWS && j >= 0) {
        if (matrix[i][j] == 1) {
            // Step 2: If the current element is '1'
            // This row (i) is a candidate for having max ones.
            // Since rows are sorted, all elements to the left (matrix[i][0] to matrix[i][j])
            // are also 1s or 0s. The '1's extend at least up to column 'j'.
            // Update maxOnesRow and try to find '1's extending further left in this row
            maxOnesRow = i; // Mark current row as the best candidate found so far
            j--;            // Move left to find if current row has even more 1s
        } else {
            // Step 3: If the current element is '0'
            // This entire column 'j' and anything to its right in the current row 'i'
            // are 0s. So, this row cannot have more 1s by moving left.
            // Move to the next row (downwards) to find a row with '1's.
            i++;
        }
    }

    printf("Row with maximum 1s: %d (0-indexed)\n", maxOnesRow);
    return 0;
}