LeetCode #2610 — MEDIUM

Convert an Array Into a 2D Array With Conditions

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

Solve on LeetCode

The Problem

Problem Statement

You are given an integer array nums. You need to create a 2D array from nums satisfying the following conditions:

The 2D array should contain only the elements of the array nums.
Each row in the 2D array contains distinct integers.
The number of rows in the 2D array should be minimal.

Return the resulting array. If there are multiple answers, return any of them.

Note that the 2D array can have a different number of elements on each row.

Example 1:

Input: nums = [1,3,4,1,2,3,1]
Output: [[1,3,4,2],[1,3],[1]]
Explanation: We can create a 2D array that contains the following rows:
- 1,3,4,2
- 1,3
- 1
All elements of nums were used, and each row of the 2D array contains distinct integers, so it is a valid answer.
It can be shown that we cannot have less than 3 rows in a valid array.

Example 2:

Input: nums = [1,2,3,4]
Output: [[4,3,2,1]]
Explanation: All elements of the array are distinct, so we can keep all of them in the first row of the 2D array.

Constraints:

1 <= nums.length <= 200
1 <= nums[i] <= nums.length

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums. You need to create a 2D array from nums satisfying the following conditions: The 2D array should contain only the elements of the array nums. Each row in the 2D array contains distinct integers. The number of rows in the 2D array should be minimal. Return the resulting array. If there are multiple answers, return any of them. Note that the 2D array can have a different number of elements on each row.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

[1,3,4,1,2,3,1]

Example 2

[2,1,1]

Step 02

Core Insight

What unlocks the optimal approach

Process the elements in the array one by one in any order and only create a new row in the matrix when we cannot put it into the existing rows
We can simply iterate over the existing rows of the matrix to see if we can place each element.

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 #2610: Convert an Array Into a 2D Array With Conditions
class Solution {
    public List<List<Integer>> findMatrix(int[] nums) {
        List<List<Integer>> ans = new ArrayList<>();
        int n = nums.length;
        int[] cnt = new int[n + 1];
        for (int x : nums) {
            ++cnt[x];
        }
        for (int x = 1; x <= n; ++x) {
            int v = cnt[x];
            for (int j = 0; j < v; ++j) {
                if (ans.size() <= j) {
                    ans.add(new ArrayList<>());
                }
                ans.get(j).add(x);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2610: Convert an Array Into a 2D Array With Conditions
func findMatrix(nums []int) (ans [][]int) {
	n := len(nums)
	cnt := make([]int, n+1)
	for _, x := range nums {
		cnt[x]++
	}
	for x, v := range cnt {
		for j := 0; j < v; j++ {
			if len(ans) <= j {
				ans = append(ans, []int{})
			}
			ans[j] = append(ans[j], x)
		}
	}
	return
}

# Accepted solution for LeetCode #2610: Convert an Array Into a 2D Array With Conditions
class Solution:
    def findMatrix(self, nums: List[int]) -> List[List[int]]:
        cnt = Counter(nums)
        ans = []
        for x, v in cnt.items():
            for i in range(v):
                if len(ans) <= i:
                    ans.append([])
                ans[i].append(x)
        return ans

// Accepted solution for LeetCode #2610: Convert an Array Into a 2D Array With Conditions
impl Solution {
    pub fn find_matrix(nums: Vec<i32>) -> Vec<Vec<i32>> {
        let n = nums.len();
        let mut cnt = vec![0; n + 1];
        let mut ans: Vec<Vec<i32>> = Vec::new();

        for &x in &nums {
            cnt[x as usize] += 1;
        }

        for x in 1..=n as i32 {
            for j in 0..cnt[x as usize] {
                if ans.len() <= j {
                    ans.push(Vec::new());
                }
                ans[j].push(x);
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #2610: Convert an Array Into a 2D Array With Conditions
function findMatrix(nums: number[]): number[][] {
    const ans: number[][] = [];
    const n = nums.length;
    const cnt: number[] = Array(n + 1).fill(0);
    for (const x of nums) {
        ++cnt[x];
    }
    for (let x = 1; x <= n; ++x) {
        for (let j = 0; j < cnt[x]; ++j) {
            if (ans.length <= j) {
                ans.push([]);
            }
            ans[j].push(x);
        }
    }
    return ans;
}

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

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.

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.