You are given two integers, rows and cols, which represent the number of rows and columns in a 1-based binary matrixA 1-based matrix has an indexing that starts from 1 instead of 0, and a binary matrix contains only binary values, i.e., 0 and 1. Therefore, a 1-based binary matrix is a matrix containing only 0s and 1s, whose indexes (for both rows and columns) start from 1.. In this matrix, each $0$ represents land, and each $1$ represents water.

Initially, on day 0, the whole matrix will just be all 0s, that is, all land. With each passing day, one of the cells of this matrix will get flooded and, therefore, will change to water, that is, from $0$ to $1$. This continues until the entire matrix is flooded. You are given a 1-based array, water_cells, that records which cell will be flooded on each day. Each element $water\_cells[i] = [r_{i},c_{i}]$ indicates the cell present at the ${r_{i}}^{th}$ row and ${c_{i}}^{th}$ column of the matrix that will change from land to water on the $i^{th}$ day.

We can cross any cell of the matrix as long as it’s land. Once it changes to water, we can’t cross it. To cross any cell, we can only move in one of the four cardinal directionsleft, right, up, and down. Given the number of rows and columns of a 1-based binary matrix and a 1-based array, water_cells, you are required to find the last day where you can still cross the matrix, from top to bottom, by walking over the land cells only.

Note: You can start from any cell in the top row, and you need to be able to reach just one cell in the bottom row for it to count as a crossing.

Constraints:

• $2 \leq$ rows$,$ cols $\leq 50$
• $4 \leq$ rows $\times$ cols $\leq 2500$
• water_cells.length $==$ rows $\times$ cols
• $1 \leq r_{i} \leq$ rows
• $1 \leq c_{i} \leq$ cols
• All values of water_cells are unique.

Our aim is to find the maximum number of days where we can cross a 1-based binary matrix, let’s say, matrix, while it is being flooded one cell at a time. We need to cross matrix from top to bottom using a continuous path consisting only of land cells. If, at any time, we encounter a series of connected water cells from the leftmost side of the matrix to its rightmost side that blocks our continuous path of land cells, we won’t be able to cross the matrix.

In our solution, we’ll look for a single component of connected water cells from the left to the right of matrix, on each day. Once encountered, we’ll return this day as the last day where we can still cross matrix.

Note: We are not subtracting 1 from our final answer, because, in the context of this problem, we are starting from day 0, so we don’t need to adjust the value of days.

Since we are required to find a series of connected water cells, we can consider the cells in the matrix to be vertexes in a graph—each vertex connected to its neighbors—and then try to identify connected componentsA connected component, in an undirected graph, is a subgraph in which each pair of nodes is connected with each other via a path. consisting only of water cells using union find. The union find pattern is designed to group elements into sets based on a specified property. The pattern uses a disjoint set data structure, such as an array, to keep track of which set each element belongs to. In our case, there are two properties that define whether any two cells should be grouped ...