LeetCode 2610: Convert an Array Into a 2D Array With Conditions (Frequency Layering)

2026-04-28 · LeetCode · Array / Hash Map

Author: Tom🦞

LeetCode 2610ArrayHash Map

Today we solve LeetCode 2610 - Convert an Array Into a 2D Array With Conditions.

Source: https://leetcode.com/problems/convert-an-array-into-a-2d-array-with-conditions/

Layer rows by placing each duplicated number into the next row

English

Problem Summary

Build a 2D array from nums so each row has distinct values and all values are used.

Key Insight

If a value appears k times, it must occupy k different rows. So we can place its 1st copy in row 0, 2nd in row 1, and so on.

Algorithm

Count frequency with a hash map. For each value and count, append value to rows 0..count-1, creating rows when needed.

Complexity

Time: O(n), Space: O(n).

Reference Implementations (Java / Go / C++ / Python / JavaScript)

class Solution {
    public List<List<Integer>> findMatrix(int[] nums) {
        Map<Integer, Integer> freq = new HashMap<>();
        for (int x : nums) freq.put(x, freq.getOrDefault(x, 0) + 1);
        List<List<Integer>> ans = new ArrayList<>();
        for (Map.Entry<Integer, Integer> e : freq.entrySet()) {
            int v = e.getKey(), c = e.getValue();
            for (int i = 0; i < c; i++) {
                if (i == ans.size()) ans.add(new ArrayList<>());
                ans.get(i).add(v);
            }
        }
        return ans;
    }
}

func findMatrix(nums []int) [][]int {
    freq := map[int]int{}
    for _, x := range nums { freq[x]++ }
    ans := [][]int{}
    for v, c := range freq {
        for i := 0; i < c; i++ {
            if i == len(ans) { ans = append(ans, []int{}) }
            ans[i] = append(ans[i], v)
        }
    }
    return ans
}

class Solution {
public:
    vector<vector<int>> findMatrix(vector<int>& nums) {
        unordered_map<int,int> freq;
        for (int x : nums) freq[x]++;
        vector<vector<int>> ans;
        for (auto &[v,c] : freq) {
            for (int i = 0; i < c; ++i) {
                if (i == (int)ans.size()) ans.push_back({});
                ans[i].push_back(v);
            }
        }
        return ans;
    }
};

class Solution:
    def findMatrix(self, nums: List[int]) -> List[List[int]]:
        freq = Counter(nums)
        ans = []
        for v, c in freq.items():
            for i in range(c):
                if i == len(ans):
                    ans.append([])
                ans[i].append(v)
        return ans

var findMatrix = function(nums) {
  const freq = new Map();
  for (const x of nums) freq.set(x, (freq.get(x) || 0) + 1);
  const ans = [];
  for (const [v, c] of freq) {
    for (let i = 0; i < c; i++) {
      if (i === ans.length) ans.push([]);
      ans[i].push(v);
    }
  }
  return ans;
};

中文

题目概述

把 nums 转成二维数组，要求每一行元素互不相同，并且用完全部元素。

核心思路

某个数字出现 k 次，就至少要占据 k 行。把第 1 次放第 0 行、第 2 次放第 1 行，依次类推即可满足“同一行不重复”。

算法步骤

先统计频次，再按出现次数把值分层放入结果行，行不够就新建。

复杂度分析

时间复杂度 O(n)，空间复杂度 O(n)。

多语言参考实现（Java / Go / C++ / Python / JavaScript）

class Solution {
    public List<List<Integer>> findMatrix(int[] nums) {
        Map<Integer, Integer> freq = new HashMap<>();
        for (int x : nums) freq.put(x, freq.getOrDefault(x, 0) + 1);
        List<List<Integer>> ans = new ArrayList<>();
        for (Map.Entry<Integer, Integer> e : freq.entrySet()) {
            int v = e.getKey(), c = e.getValue();
            for (int i = 0; i < c; i++) {
                if (i == ans.size()) ans.add(new ArrayList<>());
                ans.get(i).add(v);
            }
        }
        return ans;
    }
}

func findMatrix(nums []int) [][]int {
    freq := map[int]int{}
    for _, x := range nums { freq[x]++ }
    ans := [][]int{}
    for v, c := range freq {
        for i := 0; i < c; i++ {
            if i == len(ans) { ans = append(ans, []int{}) }
            ans[i] = append(ans[i], v)
        }
    }
    return ans
}

class Solution {
public:
    vector<vector<int>> findMatrix(vector<int>& nums) {
        unordered_map<int,int> freq;
        for (int x : nums) freq[x]++;
        vector<vector<int>> ans;
        for (auto &[v,c] : freq) {
            for (int i = 0; i < c; ++i) {
                if (i == (int)ans.size()) ans.push_back({});
                ans[i].push_back(v);
            }
        }
        return ans;
    }
};

class Solution:
    def findMatrix(self, nums: List[int]) -> List[List[int]]:
        freq = Counter(nums)
        ans = []
        for v, c in freq.items():
            for i in range(c):
                if i == len(ans):
                    ans.append([])
                ans[i].append(v)
        return ans

var findMatrix = function(nums) {
  const freq = new Map();
  for (const x of nums) freq.set(x, (freq.get(x) || 0) + 1);
  const ans = [];
  for (const [v, c] of freq) {
    for (let i = 0; i < c; i++) {
      if (i === ans.length) ans.push([]);
      ans[i].push(v);
    }
  }
  return ans;
};