Description
You are given an integer array nums of even length n and an integer limit. In one move, you can replace any integer from nums with another integer between 1 and limit, inclusive.
The array nums is complementary if for all indices i (0-indexed), nums[i] + nums[n - 1 - i] equals the same number. For example, the array [1,2,3,4] is complementary because for all indices i, nums[i] + nums[n - 1 - i] = 5.
Return the minimum number of moves required to make nums complementary.
Example 1:
Input: nums = [1,2,4,3], limit = 4 Output: 1 Explanation: In 1 move, you can change nums to [1,2,2,3] (underlined elements are changed). nums[0] + nums[3] = 1 + 3 = 4. nums[1] + nums[2] = 2 + 2 = 4. nums[2] + nums[1] = 2 + 2 = 4. nums[3] + nums[0] = 3 + 1 = 4. Therefore, nums[i] + nums[n-1-i] = 4 for every i, so nums is complementary.
Example 2:
Input: nums = [1,2,2,1], limit = 2 Output: 2 Explanation: In 2 moves, you can change nums to [2,2,2,2]. You cannot change any number to 3 since 3 > limit.
Example 3:
Input: nums = [1,2,1,2], limit = 2 Output: 0 Explanation: nums is already complementary.
Constraints:
n == nums.length2 <= n <= 1051 <= nums[i] <= limit <= 105nis even.
Solutions
Nearly identical to Solution 5 but with clearer variable naming (low and high instead of conditional assignments), using the same difference array technique to mark one-move ranges and zero-move points. It accumulates the delta array to efficiently compute the cost of changes needed for each target sum, tracking the minimum as it iterates through all possible targets.
/**
* @param {number[]} nums
* @param {number} limit
* @return {number}
*/
var minMoves = function(nums, limit) {
const n = nums.length;
const halfLen = n / 2;
const dp = Array(limit * 2 + 2).fill(0);
let res = n;
let cur = n;
for (let i = 0; i < halfLen; i++) {
const sum = nums[i] + nums[n - 1 - i];
let low;
let high;
if (nums[i] < nums[n - 1 - i]) {
low = nums[i];
high = nums[n - 1 - i];
} else {
low = nums[n - 1 - i];
high = nums[i];
}
dp[low + 1]--;
dp[sum]--;
dp[sum + 1]++;
dp[high + limit + 1]++;
}
for (const val of dp) {
cur += val;
res = Math.min(res, cur);
}
return res;
};