Merge branch 'master' of github.com:BaffinLee/leetcode-javascript

Author: BaffinLee

1201-1300/1282. Group the People Given the Group Size They Belong To.md

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+# 1282. Group the People Given the Group Size They Belong To
+
+- Difficulty: Medium.
+- Related Topics: Array, Hash Table.
+- Similar Questions: Maximum Number of Groups With Increasing Length.
+
+## Problem
+
+There are `n` people that are split into some unknown number of groups. Each person is labeled with a **unique ID** from `0` to `n - 1`.
+
+You are given an integer array `groupSizes`, where `groupSizes[i]` is the size of the group that person `i` is in. For example, if `groupSizes[1] = 3`, then person `1` must be in a group of size `3`.
+
+Return **a list of groups such that each person `i` is in a group of size `groupSizes[i]`**.
+
+Each person should appear in **exactly one group**, and every person must be in a group. If there are multiple answers, **return any of them**. It is **guaranteed** that there will be **at least one** valid solution for the given input.
+
+ 
+Example 1:
+
+```
+Input: groupSizes = [3,3,3,3,3,1,3]
+Output: [[5],[0,1,2],[3,4,6]]
+Explanation: 
+The first group is [5]. The size is 1, and groupSizes[5] = 1.
+The second group is [0,1,2]. The size is 3, and groupSizes[0] = groupSizes[1] = groupSizes[2] = 3.
+The third group is [3,4,6]. The size is 3, and groupSizes[3] = groupSizes[4] = groupSizes[6] = 3.
+Other possible solutions are [[2,1,6],[5],[0,4,3]] and [[5],[0,6,2],[4,3,1]].
+```
+
+Example 2:
+
+```
+Input: groupSizes = [2,1,3,3,3,2]
+Output: [[1],[0,5],[2,3,4]]
+```
+
+ 
+**Constraints:**
+
+
+	
+- `groupSizes.length == n`
+	
+- `1 <= n <= 500`
+	
+- `1 <= groupSizes[i] <= n`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number[]} groupSizes
+ * @return {number[][]}
+ */
+var groupThePeople = function(groupSizes) {
+    var map = Array(groupSizes.length + 1).fill(0).map(() => []);
+    var res = [];
+    for (var i = 0; i < groupSizes.length; i++) {
+        var size = groupSizes[i];
+        map[size].push(i);
+        if (map[size].length === size) {
+            res.push(map[size]);
+            map[size] = [];
+        }
+    }
+    return res;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(n).

2701-2800/2790. Maximum Number of Groups With Increasing Length.md

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+# 2790. Maximum Number of Groups With Increasing Length
+
+- Difficulty: Hard.
+- Related Topics: Array, Math, Binary Search, Greedy, Sorting.
+- Similar Questions: Group the People Given the Group Size They Belong To.
+
+## Problem
+
+You are given a **0-indexed** array `usageLimits` of length `n`.
+
+Your task is to create **groups** using numbers from `0` to `n - 1`, ensuring that each number, `i`, is used no more than `usageLimits[i]` times in total **across all groups**. You must also satisfy the following conditions:
+
+
+	
+- Each group must consist of **distinct **numbers, meaning that no duplicate numbers are allowed within a single group.
+	
+- Each group (except the first one) must have a length **strictly greater** than the previous group.
+
+
+Return **an integer denoting the **maximum** number of groups you can create while satisfying these conditions.**
+
+ 
+Example 1:
+
+```
+Input: usageLimits = [1,2,5]
+Output: 3
+Explanation: In this example, we can use 0 at most once, 1 at most twice, and 2 at most five times.
+One way of creating the maximum number of groups while satisfying the conditions is: 
+Group 1 contains the number [2].
+Group 2 contains the numbers [1,2].
+Group 3 contains the numbers [0,1,2]. 
+It can be shown that the maximum number of groups is 3. 
+So, the output is 3. 
+```
+
+Example 2:
+
+```
+Input: usageLimits = [2,1,2]
+Output: 2
+Explanation: In this example, we can use 0 at most twice, 1 at most once, and 2 at most twice.
+One way of creating the maximum number of groups while satisfying the conditions is:
+Group 1 contains the number [0].
+Group 2 contains the numbers [1,2].
+It can be shown that the maximum number of groups is 2.
+So, the output is 2. 
+```
+
+Example 3:
+
+```
+Input: usageLimits = [1,1]
+Output: 1
+Explanation: In this example, we can use both 0 and 1 at most once.
+One way of creating the maximum number of groups while satisfying the conditions is:
+Group 1 contains the number [0].
+It can be shown that the maximum number of groups is 1.
+So, the output is 1. 
+```
+
+ 
+**Constraints:**
+
+
+	
+- `1 <= usageLimits.length <= 105`
+	
+- `1 <= usageLimits[i] <= 109`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number[]} usageLimits
+ * @return {number}
+ */
+var maxIncreasingGroups = function(usageLimits) {
+    var count = 0;
+    var num = 0;
+    var minCount = 1;
+    usageLimits.sort((a, b) => a - b);
+    for (var i = 0; i < usageLimits.length; i++) {
+        count += usageLimits[i];
+        if (count >= minCount) {
+            num++;
+            minCount += num + 1;
+        }
+    }
+    return num;
+};
+```
+
+**Explain:**
+
+Math.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(1).