Sweep Line - Question 3

1674. Minimum Moves to Make Array Complementary

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.

Constraints:

n == nums.length

2 <= n <= 10^5

1 <= nums[i] <= limit <= 10^5

n is even.

Analysis:

The brute force method works, for example, the pair sum range is [2, limit*2], so we can check the number of replacements needed for each sum, then pick up the smallest one. But the time complexity is O(N*limit), which could be as high as 10^10!

So we need to find a faster method to pass the OJ.

A faster algorithm is NOT easy to find, if do not have a good understanding about Sweep Line algorithm.

For example, the binary search does not work here, since the optimal pair sum could be more than one. One of the difficulties is the optimal sum is not known.

Some key observations:

1. the pair sum range is [2, limit*2];

2. for each pair a + b = c, the above range can be divided into different intervals base on the number of replacements (to make their sum equals to c):

a. [2, min(a, b)]. In this range, we have to do the replacement for both numbers to make their sum to be c. Remember, for one replacement, we can only replace a number in the range of [1, limit];

b. [min(a, b)+1, a+b-1], needs 1 replacement;

c. [a+b, a+b], needs 0 replacement;

d. [a+b+1, max(a, b) + limit], needs 1 replacement;

e. [max(a, b) + limit + 1, limit*2], needs 2 replacements.

If you have done a similar Sweep Line question previously, then it should be easier to understand the logic here.

One typical feature is that: inside each interval, the number of replacement is the same for one pair sum, which suggests that sweep line algorithm may be a good choice to this question.

The difference here is (compare to a typical Sweep Line question): each pair sum generates 5 intervals (see the above) in the range of [2, limit*2], which makes this question to a hard level.

Two steps:

1. only label the values at the starting and ending for each intervals;

2. after finishing the labeling for all the intervals in step 1, we can do a sweeping scan for all the intervals, to accumulate the replacements needed at each position.

For this question, we need to get the minimum value of it. So we can get the minimum in step 2.

Time complexity: O(max(N, limit))

Space complexity: O(limit)

See the code below:

class Solution {
public:
    int minMoves(vector<int>& nums, int limit) {
        int n = nums.size(), res = n, ct = 0;
        vector<int> data(2*limit+2, 0);
        for(int i=0; i<n/2; ++i) {
            int x = nums[i], y = nums[n-i-1];
            // label the terminals of intervals
            data[2] += 2;
            data[min(x, y) + 1] -= 1;
            data[x+y] -= 1;
            data[x+y+1] += 1;
            data[max(x, y) + limit + 1] += 1;
        }
        for(int i=2; i<2*limit+2; ++i) {
            ct += data[i];
            res = min(res, ct);
        }
        return res;
    }
};

Bonus questions: 

1. can you figure out one sum that gives the optimal (or minimum) replacements?

2. how many sums that can give the optimal replacements?

Upper Layer