feat: solve No.1814

Author: BaffinLee

1801-1900/1814. Count Nice Pairs in an Array.md

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+# 1814. Count Nice Pairs in an Array
+
+- Difficulty: Medium.
+- Related Topics: Array, Hash Table, Math, Counting.
+- Similar Questions: Number of Pairs of Interchangeable Rectangles, Count Number of Bad Pairs, Number of Pairs Satisfying Inequality.
+
+## Problem
+
+You are given an array `nums` that consists of non-negative integers. Let us define `rev(x)` as the reverse of the non-negative integer `x`. For example, `rev(123) = 321`, and `rev(120) = 21`. A pair of indices `(i, j)` is **nice** if it satisfies all of the following conditions:
+
+
+	
+- `0 <= i < j < nums.length`
+	
+- `nums[i] + rev(nums[j]) == nums[j] + rev(nums[i])`
+
+
+Return **the number of nice pairs of indices**. Since that number can be too large, return it **modulo** `109 + 7`.
+
+ 
+Example 1:
+
+```
+Input: nums = [42,11,1,97]
+Output: 2
+Explanation: The two pairs are:
+ - (0,3) : 42 + rev(97) = 42 + 79 = 121, 97 + rev(42) = 97 + 24 = 121.
+ - (1,2) : 11 + rev(1) = 11 + 1 = 12, 1 + rev(11) = 1 + 11 = 12.
+```
+
+Example 2:
+
+```
+Input: nums = [13,10,35,24,76]
+Output: 4
+```
+
+ 
+**Constraints:**
+
+
+	
+- `1 <= nums.length <= 105`
+	
+- `0 <= nums[i] <= 109`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var countNicePairs = function(nums) {
+    var diffArray = nums.map(num => {
+        var rev = Number(String(num).split('').reverse().join(''));
+        return num - rev;
+    }).sort((a, b) => a - b);
+    var tmp = 0;
+    var res = 0;
+    var mod = Math.pow(10, 9) + 7;
+    for (var i = 0; i < diffArray.length; i++) {
+        tmp += 1;
+        if (diffArray[i + 1] !== diffArray[i]) {
+            res += tmp * (tmp - 1) / 2;
+            res %= mod;
+            tmp = 0;
+        }
+    }
+    return res;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n * log(n)).
+* Space complexity : O(n).