LeetCode #519 — MEDIUM

Random Flip Matrix

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

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:

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

Explanation
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.

Constraints:

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Math

Example 1

["Solution","flip","flip","flip","reset","flip"]
[[3,1],[],[],[],[],[]]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #519: Random Flip Matrix
class Solution {
    private int m;
    private int n;
    private int total;
    private Random rand = new Random();
    private Map<Integer, Integer> mp = new HashMap<>();

    public Solution(int m, int n) {
        this.m = m;
        this.n = n;
        this.total = m * n;
    }

    public int[] flip() {
        int x = rand.nextInt(total--);
        int idx = mp.getOrDefault(x, x);
        mp.put(x, mp.getOrDefault(total, total));
        return new int[] {idx / n, idx % n};
    }

    public void reset() {
        total = m * n;
        mp.clear();
    }
}

/**
 * Your Solution object will be instantiated and called as such:
 * Solution obj = new Solution(m, n);
 * int[] param_1 = obj.flip();
 * obj.reset();
 */

// Accepted solution for LeetCode #519: Random Flip Matrix
// Auto-generated Go example from java.
func exampleSolution() {
}
// Reference (java):
// // Accepted solution for LeetCode #519: Random Flip Matrix
// class Solution {
//     private int m;
//     private int n;
//     private int total;
//     private Random rand = new Random();
//     private Map<Integer, Integer> mp = new HashMap<>();
// 
//     public Solution(int m, int n) {
//         this.m = m;
//         this.n = n;
//         this.total = m * n;
//     }
// 
//     public int[] flip() {
//         int x = rand.nextInt(total--);
//         int idx = mp.getOrDefault(x, x);
//         mp.put(x, mp.getOrDefault(total, total));
//         return new int[] {idx / n, idx % n};
//     }
// 
//     public void reset() {
//         total = m * n;
//         mp.clear();
//     }
// }
// 
// /**
//  * Your Solution object will be instantiated and called as such:
//  * Solution obj = new Solution(m, n);
//  * int[] param_1 = obj.flip();
//  * obj.reset();
//  */

# Accepted solution for LeetCode #519: Random Flip Matrix
class Solution:
    def __init__(self, m: int, n: int):
        self.m = m
        self.n = n
        self.total = m * n
        self.mp = {}

    def flip(self) -> List[int]:
        self.total -= 1
        x = random.randint(0, self.total)
        idx = self.mp.get(x, x)
        self.mp[x] = self.mp.get(self.total, self.total)
        return [idx // self.n, idx % self.n]

    def reset(self) -> None:
        self.total = self.m * self.n
        self.mp.clear()


# Your Solution object will be instantiated and called as such:
# obj = Solution(m, n)
# param_1 = obj.flip()
# obj.reset()

// Accepted solution for LeetCode #519: Random Flip Matrix
use rand::prelude::*;
use std::collections::HashMap;

struct Solution {
    size: usize,
    indexes: HashMap<usize, usize>,
    rng: ThreadRng,
    rows: usize,
    cols: usize,
}

impl Solution {
    fn new(n_rows: i32, n_cols: i32) -> Self {
        let rows = n_rows as usize;
        let cols = n_cols as usize;
        let size = rows * cols;
        let indexes = HashMap::new();
        let rng = thread_rng();
        Solution {
            size,
            indexes,
            rng,
            rows,
            cols,
        }
    }

    fn flip(&mut self) -> Vec<i32> {
        let r = self.rng.gen_range(0, self.size);
        let x = *self.indexes.entry(r).or_insert(r);
        self.size -= 1;
        let y = *self.indexes.entry(self.size).or_insert(self.size);
        *self.indexes.entry(r).or_default() = y;
        let col = x % self.cols;
        let row = x / self.cols;
        vec![row as i32, col as i32]
    }

    fn reset(&mut self) {
        self.size = self.rows * self.cols;
        self.indexes = HashMap::new();
    }
}

// Accepted solution for LeetCode #519: Random Flip Matrix
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #519: Random Flip Matrix
// class Solution {
//     private int m;
//     private int n;
//     private int total;
//     private Random rand = new Random();
//     private Map<Integer, Integer> mp = new HashMap<>();
// 
//     public Solution(int m, int n) {
//         this.m = m;
//         this.n = n;
//         this.total = m * n;
//     }
// 
//     public int[] flip() {
//         int x = rand.nextInt(total--);
//         int idx = mp.getOrDefault(x, x);
//         mp.put(x, mp.getOrDefault(total, total));
//         return new int[] {idx / n, idx % n};
//     }
// 
//     public void reset() {
//         total = m * n;
//         mp.clear();
//     }
// }
// 
// /**
//  * Your Solution object will be instantiated and called as such:
//  * Solution obj = new Solution(m, n);
//  * int[] param_1 = obj.flip();
//  * obj.reset();
//  */

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.