Nov 19, 2019

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

vector<int> advantageCount1(vector<int>& A, vector<int>& B) {
multiset<int> S(A.begin(), A.end());
for(int i = 0;i < B.size(); i++) {
auto it = S.upper_bound(B[i]);
if (it == S.end()) {
it = S.begin();
}
A[i] = *it;
S.erase(it);
}
return A;
}


vector<int> advantageCount(vector<int>& A, vector<int>& B) {
vector<int> res(A.size());
vector<pair<int, int>> val2index(A.size());

for(int i = 0;i < val2index.size(); i++) val2index[i] = make_pair(B[i], i);
sort(A.begin(), A.end());
sort(val2index.begin(), val2index.end());

int first = 0, last = A.size() - 1;
for(int i = A.size() - 1;i >= 0;i--) {
if (val2index[i].first >= A[last]) {
res[val2index[i].second] = A[first++];
} else {
res[val2index[i].second] = A[last--];
}
}

return res;
}

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Triangle

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