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

• Java
• C++
• Python

• 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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