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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0001_0100.s0074_search_a_2d_matrix;

// #Medium #Top_100_Liked_Questions #Array #Binary_Search #Matrix #Data_Structure_I_Day_5_Array
// #Algorithm_II_Day_1_Binary_Search #Binary_Search_I_Day_8 #Level_2_Day_8_Binary_Search
// #Udemy_2D_Arrays/Matrix #Big_O_Time_O(endRow+endCol)_Space_O(1)
// #2023_08_11_Time_0_ms_(100.00%)_Space_40.9_MB_(71.91%)

/**
 * 74 - Search a 2D Matrix\.
 *
 * Medium
 *
 * Write an efficient algorithm that searches for a value in an `m x n` matrix. This matrix has the following properties:
 *
 * *   Integers in each row are sorted from left to right.
 * *   The first integer of each row is greater than the last integer of the previous row.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/10/05/mat.jpg)
 *
 * **Input:** matrix = \[\[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
 *
 * **Output:** true 
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/10/05/mat2.jpg)
 *
 * **Input:** matrix = \[\[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
 *
 * **Output:** false 
 *
 * **Constraints:**
 *
 * *   `m == matrix.length`
 * *   `n == matrix[i].length`
 * *   `1 <= m, n <= 100`
 * *   -10⁴ <= matrix[i][j], target <= 10⁴
**/
public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int endRow = matrix.length;
        int endCol = matrix[0].length;
        int targetRow = 0;
        boolean result = false;
        for (int i = 0; i < endRow; i++) {
            if (matrix[i][endCol - 1] >= target) {
                targetRow = i;
                break;
            }
        }
        for (int i = 0; i < endCol; i++) {
            if (matrix[targetRow][i] == target) {
                result = true;
                break;
            }
        }
        return result;
    }
}