# Leetcode 519 - Random flip matrix

Note:

• First, map 2D m x n matrix into a 1D array. How?
• [i, j] -> [i * n + j].
• Explain: There are m * n cells in this matrix, [0, 0] is mapped to 0, and [m - 1, n - 1] is mapped to (m - 1) * n + n - 1 === m * n - 1.
• How should we mark indexes that have been picked?
• Use random * totalLength to get a random num between [0, len - 1].
• If map.get(random) exists, it means the num was picked before, and now it maps to another index.
• If not, it’s a qualified index. So index = this.map.get(random) || random
• How to update the map?
• Map the last element to index ran;
• In case the last element also maps to another index, we need to check map.get(this.len - 1).
• • map.set(random, map.get(len - 1) || len - 1).
• (Total is the length, and that’s why we do total - 1 first);
• Use floor(index / n) to get row, index % n to get col.

Question:

There is an m x n binary grid matrix with all the values set 0 initially. Design an algorithm to randomly pick an index (i, j) where matrix[i][j] == 0 and flips it to 1. All the indices (i, j) where matrix[i][j] == 0 should be equally likely to be returned.

Optimize your algorithm to minimize the number of calls made to the built-in random function of your language and optimize the time and space complexity.

Implement the Solution class:

• Solution(int m, int n) Initializes the object with the size of the binary matrix m and n.
• int[] flip() Returns a random index [i, j] of the matrix where matrix[i][j] == 0 and flips it to 1.
• void reset() Resets all the values of the matrix to be 0.

Example:

Code: