Similar to Parity Sorting, the Index Parity Sorting can be applied to an array of integers that have equal number of even numbers and odd numbers. You are to re-arrange the integers so that A[i] is even when i is even and A[j] is odd when j is odd.
It is obvious that you can index parity sort the array more than one way – you may return any of them that satisfy the requirements.
Even and Odd Index Pointers
We can define two index pointers starting at 0 and 1 respectively. Iterating the array and place the numbers at the corresponding index – then increment the index by two.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | class Solution { public: vector<int> sortArrayByParityII(vector<int>& A) { vector<int> r(A.size()); int even = 0, odd = 1; for (int i = 0; i < A.size(); ++ i) { if (A[i] % 2 == 0) { r[even] = A[i]; even += 2; } else { r[odd] = A[i]; odd += 2; } } return r; } }; |
class Solution {
public:
vector<int> sortArrayByParityII(vector<int>& A) {
vector<int> r(A.size());
int even = 0, odd = 1;
for (int i = 0; i < A.size(); ++ i) {
if (A[i] % 2 == 0) {
r[even] = A[i];
even += 2;
} else {
r[odd] = A[i];
odd += 2;
}
}
return r;
}
};The complexity is O(N) and the space complexity is O(N) – the extra array to return the result. We may also transform above idea into a more concise implementation using an indices array.
1 2 3 4 5 6 7 8 9 10 11 | class Solution { public: vector<int> sortArrayByParityII(vector<int>& A) { int indices[2] = { -2, -1 }; vector<int> ret(A.size()); for (int i = 0; i < A.size(); ++ i) { ret[indices[A[i] % 2] += 2] = A[i]; } return ret; } }; |
class Solution {
public:
vector<int> sortArrayByParityII(vector<int>& A) {
int indices[2] = { -2, -1 };
vector<int> ret(A.size());
for (int i = 0; i < A.size(); ++ i) {
ret[indices[A[i] % 2] += 2] = A[i];
}
return ret;
}
};In-Place
If we do it in-place without allocating a new array, the complexity is still O(N^2) but the space complexity is O(1).
1 2 3 4 5 6 7 8 9 10 11 12 13 | class Solution { public: vector<int> sortArrayByParityII(vector<int>& A) { int indexes[2] = { -2, -1 }; for (int i = 0; i < A.size(); ++i) { while (A[i] % 2 != i % 2) { int idx = (indexes[A[i] % 2] += 2); swap(A[i], A[idx]); } } return A; } }; |
class Solution {
public:
vector<int> sortArrayByParityII(vector<int>& A) {
int indexes[2] = { -2, -1 };
for (int i = 0; i < A.size(); ++i) {
while (A[i] % 2 != i % 2) {
int idx = (indexes[A[i] % 2] += 2);
swap(A[i], A[idx]);
}
}
return A;
}
};–EOF (The Ultimate Computing & Technology Blog) —
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