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Brute Force - Question 1

Recover the Original Array

Alice had a 0-indexed array arr consisting of n positive integers. She chose an arbitrary positive integer k and created two new 0-indexed integer arrays lower and higher in the following manner:

lower[i] = arr[i] - k, for every index i where 0 <= i < n

higher[i] = arr[i] + k, for every index i where 0 <= i < n

Unfortunately, Alice lost all three arrays. However, she remembers the integers that were present in the arrays lower and higher, but not the array each integer belonged to. Help Alice and recover the original array.

Given an array nums consisting of 2n integers, where exactly n of the integers were present in lower and the remaining in higher, return the original array arr. In case the answer is not unique, return any valid array.

Note: The test cases are generated such that there exists at least one valid array arr.


Constraints:


2 * n == nums.length

1 <= n <= 1000

1 <= nums[i] <= 10^9

The test cases are generated such that there exists at least one valid array arr.


Analysis:


This question seems to be very interesting but actually need a brute-force way to solve.

Some observations:

1. the minimum of the array must be from arr[i] - K; similarly, the maximum from arr[i] + K;

2. after sorting the array with size of 2n, the possible ks' can only be the (arr[i] - arr[0])/2, where i can be from 1 to n. Or i cannot be larger than n. 

This is easy to prove: if i can be any value from n+1 to 2*n-1, then we cannot find enough pairs which needs to be n pairs exactly, since the k value becomes too large to make such n pairs.

Or in details: if arr[0] and arr[n+1] is one pair, (one pair here means arr[n+1] - arr[0] == 2*k), then there are not enough arrs on the right-side of arr[n+1] to pair the ones from arr[1] to arr[n].


With the above observations, we can gradually try every possible k.


See the code below:


class Solution {
public:
    vector<int> recoverArray(vector<int>& nums) {
        int n = nums.size();
        sort(nums.begin(), nums.end());
        vector<int> res;
        for(int i=1; i<=n/2; ++i) {
            int k = (nums[i] - nums[0])/2;
            if(k==0) continue;
            vector<int> used(n, 0);
            int j = 0;
            bool find = true;
            while(res.size()<n/2 && j<n) {
                while(j<n&& used[j]) ++j;
                if(j<n) {
                    res.push_back(nums[j]+k);
                    used[j] = 1;
		    // find the first un-used valid element
                    int id = lower_bound(nums.begin(), nums.end(), nums[j]+2*k) - nums.begin();
                    while(id<n&& used[id]) ++id;
		    // if there is no valid element, break and try another k
                    if(id==nums.size() || nums[id] != nums[j]+2*k) {
                        find = false;
                        res.clear();
                        break;
                    }
                    used[id] = 1;
                }
            }
	    // when find one, done
            if(find) break;
        }
        return res;
    }
};



Upper Layer


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