519. Random Flip Matrix LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones

Random Flip Matrix 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 1:

["Solution", "flip", "flip", "flip", "reset", "flip"]
[[3, 1], [], [], [], [], []]
[null, [1, 0], [2, 0], [0, 0], null, [2, 0]]

Solution solution = new Solution(3, 1);
solution.flip();  // return [1, 0], [0,0], [1,0], and [2,0] should be equally likely to be returned.
solution.flip();  // return [2, 0], Since [1,0] was returned, [2,0] and [0,0]
solution.flip();  // return [0, 0], Based on the previously returned indices, only [0,0] can be returned.
solution.reset(); // All the values are reset to 0 and can be returned.
solution.flip();  // return [2, 0], [0,0], [1,0], and [2,0] should be equally likely to be returned.


  • 1 <= m, n <= 104
  • There will be at least one free cell for each call to flip.
  • At most 1000 calls will be made to flip and reset.

Random Flip Matrix Solutions

Time: O(n)
Space: O(n)


 Will be updated Soon


 Will be updated Soon


  Will be updated Soon

Watch Tutorial

Checkout more Solutions here

Leave a Comment

Your email address will not be published. Required fields are marked *