Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using array strategy.
Given an integer array nums, design an algorithm to randomly shuffle the array. All permutations of the array should be equally likely as a result of the shuffling.
Implement the Solution class:
Solution(int[] nums) Initializes the object with the integer array nums.int[] reset() Resets the array to its original configuration and returns it.int[] shuffle() Returns a random shuffling of the array.Example 1:
Input
["Solution", "shuffle", "reset", "shuffle"]
[[[1, 2, 3]], [], [], []]
Output
[null, [3, 1, 2], [1, 2, 3], [1, 3, 2]]
Explanation
Solution solution = new Solution([1, 2, 3]);
solution.shuffle(); // Shuffle the array [1,2,3] and return its result.
// Any permutation of [1,2,3] must be equally likely to be returned.
// Example: return [3, 1, 2]
solution.reset(); // Resets the array back to its original configuration [1,2,3]. Return [1, 2, 3]
solution.shuffle(); // Returns the random shuffling of array [1,2,3]. Example: return [1, 3, 2]
Constraints:
1 <= nums.length <= 50-106 <= nums[i] <= 106nums are unique.104 calls in total will be made to reset and shuffle.Problem summary: Given an integer array nums, design an algorithm to randomly shuffle the array. All permutations of the array should be equally likely as a result of the shuffling. Implement the Solution class: Solution(int[] nums) Initializes the object with the integer array nums. int[] reset() Resets the array to its original configuration and returns it. int[] shuffle() Returns a random shuffling of the array.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Math · Design
["Solution","shuffle","reset","shuffle"] [[[1,2,3]],[],[],[]]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #384: Shuffle an Array
class Solution {
private int[] nums;
private int[] original;
private Random rand;
public Solution(int[] nums) {
this.nums = nums;
this.original = Arrays.copyOf(nums, nums.length);
this.rand = new Random();
}
public int[] reset() {
nums = Arrays.copyOf(original, original.length);
return nums;
}
public int[] shuffle() {
for (int i = 0; i < nums.length; ++i) {
swap(i, i + rand.nextInt(nums.length - i));
}
return nums;
}
private void swap(int i, int j) {
int t = nums[i];
nums[i] = nums[j];
nums[j] = t;
}
}
/**
* Your Solution object will be instantiated and called as such:
* Solution obj = new Solution(nums);
* int[] param_1 = obj.reset();
* int[] param_2 = obj.shuffle();
*/
// Accepted solution for LeetCode #384: Shuffle an Array
type Solution struct {
nums, original []int
}
func Constructor(nums []int) Solution {
return Solution{nums, append([]int(nil), nums...)}
}
func (this *Solution) Reset() []int {
copy(this.nums, this.original)
return this.nums
}
func (this *Solution) Shuffle() []int {
n := len(this.nums)
for i := range this.nums {
j := i + rand.Intn(n-i)
this.nums[i], this.nums[j] = this.nums[j], this.nums[i]
}
return this.nums
}
/**
* Your Solution object will be instantiated and called as such:
* obj := Constructor(nums);
* param_1 := obj.Reset();
* param_2 := obj.Shuffle();
*/
# Accepted solution for LeetCode #384: Shuffle an Array
class Solution:
def __init__(self, nums: List[int]):
self.nums = nums
self.original = nums.copy()
def reset(self) -> List[int]:
self.nums = self.original.copy()
return self.nums
def shuffle(self) -> List[int]:
for i in range(len(self.nums)):
j = random.randrange(i, len(self.nums))
self.nums[i], self.nums[j] = self.nums[j], self.nums[i]
return self.nums
# Your Solution object will be instantiated and called as such:
# obj = Solution(nums)
# param_1 = obj.reset()
# param_2 = obj.shuffle()
// Accepted solution for LeetCode #384: Shuffle an Array
use rand::Rng;
struct Solution {
nums: Vec<i32>,
}
/**
* `&self` means the method takes an immutable reference.
* If you need a mutable reference, change it to `&mut self` instead.
*/
impl Solution {
fn new(nums: Vec<i32>) -> Self {
Self { nums }
}
fn reset(&self) -> Vec<i32> {
self.nums.clone()
}
fn shuffle(&mut self) -> Vec<i32> {
let n = self.nums.len();
let mut res = self.nums.clone();
for i in 0..n {
let j = rand::thread_rng().gen_range(0, n);
res.swap(i, j);
}
res
}
}
// Accepted solution for LeetCode #384: Shuffle an Array
class Solution {
private nums: number[];
constructor(nums: number[]) {
this.nums = nums;
}
reset(): number[] {
return this.nums;
}
shuffle(): number[] {
const n = this.nums.length;
const res = [...this.nums];
for (let i = 0; i < n; i++) {
const j = Math.floor(Math.random() * n);
[res[i], res[j]] = [res[j], res[i]];
}
return res;
}
}
/**
* Your Solution object will be instantiated and called as such:
* var obj = new Solution(nums)
* var param_1 = obj.reset()
* var param_2 = obj.shuffle()
*/
Use this to step through a reusable interview workflow for this problem.
Use a simple list or array for storage. Each operation (get, put, remove) requires a linear scan to find the target element — O(n) per operation. Space is O(n) to store the data. The linear search makes this impractical for frequent operations.
Design problems target O(1) amortized per operation by combining data structures (hash map + doubly-linked list for LRU, stack + min-tracking for MinStack). Space is always at least O(n) to store the data. The challenge is achieving constant-time operations through clever structure composition.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.