Merge pull request #1078 from Anshika14528/2610

Added Leetcode Problem 2610: #1069

Author: ajay-dhangar

dsa-solutions/lc-solutions/2500 - 3000/2610- convert-an-array-into-a-2d-array-with-conditions.md

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+---
+id: convert-an-array-into-a-2d-array-with-conditions
+title: Convert an Array Into a 2D Array With Conditions
+sidebar_label: 2610 Convert an Array Into a 2D Array With Conditions
+tags:
+  - Array
+  - Hash Table
+  - LeetCode
+  - C++
+description: "This is a solution to the Convert an Array Into a 2D Array With Conditions problem on LeetCode."
+---
+
+## Problem Description
+
+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.
+
+### Examples
+
+**Example 1:**
+
+```
+
+Input:  nums = [1,2]
+Output: 2
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,3,4,1,2,3,1]
+Output: [[1,3,4,2],[1,3],[1]]
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,2,3,4]
+Output: [[4,3,2,1]]
+```
+
+### Constraints
+
+- $1 \leq \text{nums.length} \leq 200$.
+- $1 \leq \text{nums[i]} \leq \text{nums.length}$.
+
+### Approach 
+
+To solve this problem(convert an array into a 2d array with conditions)we simply have to use an unordered map which will keep track if we can insert a particular number in the row. we will 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.
+
+#### Code in C++
+
+```cpp
+class Solution {
+public:
+    vector<vector<int>> findMatrix(vector<int>& nums) {
+        vector<vector<int>>a;
+       unordered_map<int,int>b;
+       int r=0;
+       for(int i=0;i<nums.size();i++){
+        b[nums[i]]++;
+        if(b[nums[i]]>r){
+            r++;
+            a.push_back({nums[i]});
+        }
+        else{
+            a[b[nums[i]]-1].push_back(nums[i]);
+        }
+       }
+        return a;
+    }
+};
+```
+
+#### Code in Java
+
+```java
+class Solution {
+
+    public List<List<Integer>> findMatrix(int[] nums) {
+        HashMap<Integer, Integer> map = new HashMap();
+        for (int n : nums) {
+            map.put(n, map.getOrDefault(n, 0) + 1);
+        }
+
+        List<List<Integer>> ll = new ArrayList();
+        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
+            int key = entry.getKey();
+            int val = entry.getValue();
+            for (int i = ll.size(); i < val; i++) ll.add(new ArrayList());
+            int i = 0;
+            while (val-- > 0) {
+                ll.get(i++).add(key);
+            }
+        }
+
+        return ll;
+    }
+}
+
+```
+
+