feat: solve No.2790

Author: BaffinLee

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).