feat: add No.1464

Author: BaffinLee

1401-1500/1464. Maximum Product of Two Elements in an Array.md

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+# 1464. Maximum Product of Two Elements in an Array
+
+- Difficulty: Easy.
+- Related Topics: Array, Sorting, Heap (Priority Queue).
+- Similar Questions: .
+
+## Problem
+
+Given the array of integers `nums`, you will choose two different indices `i` and `j` of that array. **Return the maximum value of** `(nums[i]-1)*(nums[j]-1)`.
+ 
+Example 1:
+
+```
+Input: nums = [3,4,5,2]
+Output: 12 
+Explanation: If you choose the indices i=1 and j=2 (indexed from 0), you will get the maximum value, that is, (nums[1]-1)*(nums[2]-1) = (4-1)*(5-1) = 3*4 = 12. 
+```
+
+Example 2:
+
+```
+Input: nums = [1,5,4,5]
+Output: 16
+Explanation: Choosing the indices i=1 and j=3 (indexed from 0), you will get the maximum value of (5-1)*(5-1) = 16.
+```
+
+Example 3:
+
+```
+Input: nums = [3,7]
+Output: 12
+```
+
+ 
+**Constraints:**
+
+
+	
+- `2 <= nums.length <= 500`
+	
+- `1 <= nums[i] <= 10^3`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var maxProduct = function(nums) {
+    if (nums.length === 2) {
+        return (nums[0] - 1) * (nums[1] - 1);
+    }
+    var minNegativeNum = 0;
+    var secondMinNegativeNum = 0;
+    var maxPositiveNum = 0;
+    var secondMaxPositiveNum = 0;
+    for (var i = 0; i < nums.length; i++) {
+        var num = nums[i] - 1;
+        if (num < minNegativeNum) {
+            secondMinNegativeNum = minNegativeNum;
+            minNegativeNum = num;
+        } else if (num < secondMinNegativeNum) {
+            secondMinNegativeNum = num;
+        } else if (num > maxPositiveNum) {
+            secondMaxPositiveNum = maxPositiveNum;
+            maxPositiveNum = num;
+        } else if (num > secondMaxPositiveNum) {
+            secondMaxPositiveNum = num;
+        }
+    }
+    return Math.max(
+        minNegativeNum * secondMinNegativeNum,
+        maxPositiveNum * secondMaxPositiveNum,
+    );
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(1).