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Commit 38c51ac

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feat: solve No.2610,478
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# 2610. Convert an Array Into a 2D Array With Conditions
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- Difficulty: Medium.
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- Related Topics: Array, Hash Table.
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- Similar Questions: .
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## Problem
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You are given an integer array `nums`. You need to create a 2D array from `nums` satisfying the following conditions:
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- The 2D array should contain **only** the elements of the array `nums`.
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- Each row in the 2D array contains **distinct** integers.
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- The number of rows in the 2D array should be **minimal**.
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Return **the resulting array**. If there are multiple answers, return any of them.
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**Note** that the 2D array can have a different number of elements on each row.
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Example 1:
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```
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Input: nums = [1,3,4,1,2,3,1]
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Output: [[1,3,4,2],[1,3],[1]]
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Explanation: We can create a 2D array that contains the following rows:
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- 1,3,4,2
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- 1,3
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- 1
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All elements of nums were used, and each row of the 2D array contains distinct integers, so it is a valid answer.
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It can be shown that we cannot have less than 3 rows in a valid array.
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```
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Example 2:
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```
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Input: nums = [1,2,3,4]
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Output: [[4,3,2,1]]
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Explanation: All elements of the array are distinct, so we can keep all of them in the first row of the 2D array.
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```
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**Constraints:**
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- `1 <= nums.length <= 200`
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- `1 <= nums[i] <= nums.length`
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## Solution
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```javascript
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/**
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* @param {number[]} nums
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* @return {number[][]}
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*/
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var findMatrix = function(nums) {
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var numMap = Array(nums.length + 1).fill(0);
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var res = [];
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for (var i = 0; i < nums.length; i++) {
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var num = nums[i];
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var row = numMap[nums[i]];
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if (!res[row]) {
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res.push([]);
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}
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res[row].push(num);
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numMap[num]++;
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}
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return res;
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};
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```
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**Explain:**
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nope.
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**Complexity:**
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* Time complexity : O(n).
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* Space complexity : O(n).
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# 478. Generate Random Point in a Circle
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- Difficulty: Medium.
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- Related Topics: Math, Geometry, Rejection Sampling, Randomized.
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- Similar Questions: Random Point in Non-overlapping Rectangles.
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## Problem
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Given the radius and the position of the center of a circle, implement the function `randPoint` which generates a uniform random point inside the circle.
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Implement the `Solution` class:
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- `Solution(double radius, double x_center, double y_center)` initializes the object with the radius of the circle `radius` and the position of the center `(x_center, y_center)`.
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- `randPoint()` returns a random point inside the circle. A point on the circumference of the circle is considered to be in the circle. The answer is returned as an array `[x, y]`.
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Example 1:
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```
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Input
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["Solution", "randPoint", "randPoint", "randPoint"]
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[[1.0, 0.0, 0.0], [], [], []]
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Output
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[null, [-0.02493, -0.38077], [0.82314, 0.38945], [0.36572, 0.17248]]
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Explanation
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Solution solution = new Solution(1.0, 0.0, 0.0);
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solution.randPoint(); // return [-0.02493, -0.38077]
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solution.randPoint(); // return [0.82314, 0.38945]
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solution.randPoint(); // return [0.36572, 0.17248]
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```
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**Constraints:**
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- `0 < radius <= 108`
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- `-107 <= x_center, y_center <= 107`
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- At most `3 * 104` calls will be made to `randPoint`.
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## Solution
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```javascript
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/**
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* @param {number} radius
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* @param {number} x_center
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* @param {number} y_center
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*/
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var Solution = function(radius, x_center, y_center) {
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this.radius = radius;
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this.centerX = x_center;
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this.centerY = y_center;
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};
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/**
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* @return {number[]}
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*/
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Solution.prototype.randPoint = function() {
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var radius = Math.sqrt(Math.random()) * this.radius;
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var rand = Math.random();
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var degree = Math.PI / 2 * (rand === 1 ? 0 : rand);
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var x = Math.cos(degree) * radius;
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var y = Math.sin(degree) * radius;
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return [
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this.centerX + (Math.random() > 0.5 ? 1 : -1) * x,
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this.centerY + (Math.random() > 0.5 ? 1 : -1) * y,
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];
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};
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/**
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* Your Solution object will be instantiated and called as such:
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* var obj = new Solution(radius, x_center, y_center)
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* var param_1 = obj.randPoint()
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*/
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```
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**Explain:**
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nope.
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**Complexity:**
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* Time complexity : O(n).
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* Space complexity : O(1).

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