feat: solve No.1887

Author: BaffinLee

1801-1900/1887. Reduction Operations to Make the Array Elements Equal.md

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+# 1887. Reduction Operations to Make the Array Elements Equal
+
+- Difficulty: Medium.
+- Related Topics: Array, Sorting.
+- Similar Questions: .
+
+## Problem
+
+Given an integer array `nums`, your goal is to make all elements in `nums` equal. To complete one operation, follow these steps:
+
+
+	
+- Find the **largest** value in `nums`. Let its index be `i` (**0-indexed**) and its value be `largest`. If there are multiple elements with the largest value, pick the smallest `i`.
+	
+- Find the **next largest** value in `nums` **strictly smaller** than `largest`. Let its value be `nextLargest`.
+	
+- Reduce `nums[i]` to `nextLargest`.
+
+
+Return **the number of operations to make all elements in **`nums`** equal**.
+
+ 
+Example 1:
+
+```
+Input: nums = [5,1,3]
+Output: 3
+Explanation: It takes 3 operations to make all elements in nums equal:
+1. largest = 5 at index 0. nextLargest = 3. Reduce nums[0] to 3. nums = [3,1,3].
+2. largest = 3 at index 0. nextLargest = 1. Reduce nums[0] to 1. nums = [1,1,3].
+3. largest = 3 at index 2. nextLargest = 1. Reduce nums[2] to 1. nums = [1,1,1].
+```
+
+Example 2:
+
+```
+Input: nums = [1,1,1]
+Output: 0
+Explanation: All elements in nums are already equal.
+```
+
+Example 3:
+
+```
+Input: nums = [1,1,2,2,3]
+Output: 4
+Explanation: It takes 4 operations to make all elements in nums equal:
+1. largest = 3 at index 4. nextLargest = 2. Reduce nums[4] to 2. nums = [1,1,2,2,2].
+2. largest = 2 at index 2. nextLargest = 1. Reduce nums[2] to 1. nums = [1,1,1,2,2].
+3. largest = 2 at index 3. nextLargest = 1. Reduce nums[3] to 1. nums = [1,1,1,1,2].
+4. largest = 2 at index 4. nextLargest = 1. Reduce nums[4] to 1. nums = [1,1,1,1,1].
+```
+
+ 
+**Constraints:**
+
+
+	
+- `1 <= nums.length <= 5 * 104`
+	
+- `1 <= nums[i] <= 5 * 104`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var reductionOperations = function(nums) {
+    nums.sort((a, b) => a - b);
+    var uniqueNums = [];
+    for (var i = 0; i < nums.length; i++) {
+        if (nums[i] === nums[i - 1]) {
+            uniqueNums[uniqueNums.length - 1]++;
+        } else {
+            uniqueNums.push(1);
+        }
+    }
+    var res = 0;
+    while (uniqueNums.length > 1) {
+        var last = uniqueNums.pop();
+        res += last;
+        uniqueNums[uniqueNums.length - 1] += last;
+    }
+    return res;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n * log(n)).
+* Space complexity : O(n).