feat: add solutions to lc problems: No.1600+ (doocs#2119)

Author: yanglbme

solution/1600-1699/1674.Minimum Moves to Make Array Complementary/README.md

@@ -80,7 +80,7 @@ nums[3] + nums[0] = 3 + 1 = 4.
 
 可以发现，这实际上是在对一个连续区间内的元素进行加减操作，因此我们可以使用差分数组来实现。
 
-时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 `nums` 的长度。
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $nums$ 的长度。
 
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solution/1600-1699/1674.Minimum Moves to Make Array Complementary/README_EN.md

@@ -52,6 +52,30 @@ Therefore, nums[i] + nums[n-1-i] = 4 for every i, so nums is complementary.
 
 ## Solutions
 
+**Solution 1: Difference Array**
+
+Let's denote $a$ as the smaller value between $nums[i]$ and $nums[n-i-1]$, and $b$ as the larger value between $nums[i]$ and $nums[n-i-1]$.
+
+Suppose that after replacement, the sum of the two numbers is $x$. From the problem, we know that the minimum value of $x$ is $2$, which means both numbers are replaced by $1$; the maximum value is $2 \times limit$, which means both numbers are replaced by $limit$. Therefore, the range of $x$ is $[2,... 2 \times limit]$.
+
+How to find the minimum number of replacements for different $x$?
+
+We analyze and find:
+
+-   If $x = a + b$, then the number of replacements we need is $0$, which means the current pair of numbers already meets the complement requirement;
+-   Otherwise, if $1 + a \le x \le limit + b $, then the number of replacements we need is $1$, which means we can replace one of the numbers;
+-   Otherwise, if $2 \le x \le 2 \times limit$, then the number of replacements we need is $2$, which means we need to replace both numbers.
+
+Therefore, we can iterate over each pair of numbers and perform the following operations:
+
+1. First, add $2$ to the number of operations required in the range $[2,... 2 \times limit]$.
+1. Then, subtract $1$ from the number of operations required in the range $[1 + a,... limit + b]$.
+1. Finally, subtract $1$ from the number of operations required in the range $[a + b,... a + b]$.
+
+We can see that this is actually adding and subtracting elements in a continuous interval, so we can use a difference array to implement it.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.
+
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 ### **Python3**