Formatted question description: https://leetcode.ca/all/870.html

870. Advantage Shuffle (Medium)

Given two arrays A and B of equal size, the advantage of A with respect to B is the number of indices i for which A[i] > B[i].

Return any permutation of A that maximizes its advantage with respect to B.

 

Example 1:

Input: A = [2,7,11,15], B = [1,10,4,11]
Output: [2,11,7,15]

Example 2:

Input: A = [12,24,8,32], B = [13,25,32,11]
Output: [24,32,8,12]

 

Note:

  1. 1 <= A.length = B.length <= 10000
  2. 0 <= A[i] <= 10^9
  3. 0 <= B[i] <= 10^9

Solution 1.

// OJ: https://leetcode.com/problems/advantage-shuffle/

// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    vector<int> advantageCount(vector<int>& A, vector<int>& B) {
        sort(A.begin(), A.end());
        vector<int> copy(B.begin(), B.end());
        sort(B.begin(), B.end());
        unordered_map<int, queue<int>> m;
        queue<int> leftover;
        int i = 0;
        for (int a : A) {
            if (a > B[i]) m[B[i++]].push(a);
            else leftover.push(a);
        }
        vector<int> ans;
        for (int b : copy) {
            if (m[b].size()) {
                ans.push_back(m[b].front());
                m[b].pop();
            } else {
                ans.push_back(leftover.front());
                leftover.pop();
            }
        }
        return ans;
    }
};

Java

  • class Solution {
        public int[] advantageCount(int[] A, int[] B) {
            int length = A.length;
            int[][] arrayA2D = new int[length][2];
            int[][] arrayB2D = new int[length][2];
            for (int i = 0; i < length; i++) {
                arrayB2D[i][0] = B[i];
                arrayB2D[i][1] = i;
            }
            Arrays.sort(arrayB2D, new Comparator<int[]>() {
                public int compare(int[] array1, int[] array2) {
                    return array1[0] - array2[0];
                }
            });
            for (int i = 0; i < length; i++)
                arrayA2D[i][1] = arrayB2D[i][1];
            Arrays.sort(A);
            int index = 0;
            for (int i = 0; i < length; i++) {
                int num = A[i];
                if (num > arrayB2D[index][0]) {
                    arrayA2D[index][0] = num;
                    index++;
                    A[i] = -1;
                }
            }
            for (int i = 0; i < length; i++) {
                int num = A[i];
                if (num >= 0) {
                    arrayA2D[index][0] = num;
                    index++;
                }
            }
            Arrays.sort(arrayA2D, new Comparator<int[]>() {
                public int compare(int[] array1, int[] array2) {
                    return array1[1] - array2[1];
                }
            });
            int[] permutation = new int[length];
            for (int i = 0; i < length; i++)
                permutation[i] = arrayA2D[i][0];
            return permutation;
        }
    }
    
  • // OJ: https://leetcode.com/problems/advantage-shuffle/
    // Time: O(NlogN)
    // Space: O(N)
    class Solution {
    public:
        vector<int> advantageCount(vector<int>& A, vector<int>& B) {
            int N = A.size(), i = 0;
            vector<int> ia(N), ib(N), ans(N, -1);
            iota(begin(ia), end(ia), 0);
            iota(begin(ib), end(ib), 0);
            sort(begin(ia), end(ia), [&](int a, int b) { return A[a] > A[b]; });
            sort(begin(ib), end(ib), [&](int a, int b) { return B[a] > B[b]; });
            for (int j = 0; i < N && j < N; ++i) {
                while (j < N && B[ib[j]] >= A[ia[i]]) ++j;
                if (j < N) {
                    ans[ib[j++]] = A[ia[i]];
                } else break;
            }
            for (int j = 0; i < N && j < N; ++i) {
                while (j < N && ans[j] != -1) ++j;
                if (j < N) {
                    ans[j++] = A[ia[i]];
                }
            }
            return ans;
        }
    };
    
  • class Solution(object):
        def advantageCount(self, A, B):
            """
            :type A: List[int]
            :type B: List[int]
            :rtype: List[int]
            """
            res = [-1] * len(A)
            A = collections.deque(sorted(A))
            B = collections.deque(sorted((b, i) for i, b in enumerate(B)))
            for i in range(len(A)):
                a = A.popleft()
                b = B[0]
                if a > b[0]:
                    B.popleft()
                else:
                    b = B.pop()
                res[b[1]] = a
            return res
    

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