You are given two 0-indexed integerpermutations A and B of length n.
A prefix common array of A and B is an array C such that C[i] is equal to the count of numbers that are present at or before the index i in both A and B.
Return the prefix common array of A and B.
A sequence of n integers is called a permutation if it contains all integers from 1 to n exactly once.
Example 1:
Input: A = [1,3,2,4], B = [3,1,2,4] Output: [0,2,3,4] Explanation: At i = 0: no number is common, so C[0] = 0. At i = 1: 1 and 3 are common in A and B, so C[1] = 2. At i = 2: 1, 2, and 3 are common in A and B, so C[2] = 3. At i = 3: 1, 2, 3, and 4 are common in A and B, so C[3] = 4.
Example 2:
Input: A = [2,3,1], B = [3,1,2] Output: [0,1,3] Explanation: At i = 0: no number is common, so C[0] = 0. At i = 1: only 3 is common in A and B, so C[1] = 1. At i = 2: 1, 2, and 3 are common in A and B, so C[2] = 3.
Constraints:
1 <= A.length == B.length == n <= 501 <= A[i], B[i] <= nIt is guaranteed that A and B are both a permutation of n integers.
Approach 01:
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C++
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Python
#include <vector>
#include <unordered_set>
#include <algorithm> // for set_intersection
using namespace std;
class Solution {
public:
vector<int> findThePrefixCommonArray(vector<int>& A, vector<int>& B) {
int n = A.size();
vector<int> commonPrefixCount;
// Iterate through each index and compute common elements in prefixes
for (int i = 0; i < n; ++i) {
unordered_set<int> prefixA(A.begin(), A.begin() + i + 1);
unordered_set<int> prefixB(B.begin(), B.begin() + i + 1);
// Calculate the intersection of the sets
unordered_set<int> intersection;
for (const auto& elem : prefixA) {
if (prefixB.find(elem) != prefixB.end()) {
intersection.insert(elem);
}
}
// Add the size of the intersection to the result
commonPrefixCount.push_back(intersection.size());
}
return commonPrefixCount;
}
};
Time Complexity:
- Prefix Set Construction:For each index \( i \), we construct two sets (prefixA and prefixB) of size \( O(i) \), which takes \( O(i) \) time. So for all indices, the total time for set construction is \( O(1 + 2 + \ldots + n) = O(n^2) \), where \( n \) is the size of the array.
- Set Intersection:For each index \( i \), we perform an intersection between the two sets, which can take up to \( O(i) \) time. The total time for all intersections is \( O(1 + 2 + \ldots + n) = O(n^2) \).
- Overall Time Complexity:Since both the set construction and intersection take \( O(n^2) \), the overall time complexity is \( O(n^2) \).
Space Complexity:
- Set Space:For each index \( i \), we store two sets of size \( O(i) \), leading to a total space usage of \( O(n) \) for each set. The space complexity is \( O(n) \) for the sets used in each iteration.
- Overall Space Complexity:Since we are using sets and the final result array, the total space complexity is \( O(n) \).
class Solution:
def findThePrefixCommonArray(self, A: List[int], B: List[int]) -> List[int]:
n = len(A)
commonPrefixCount = []
# Iterate through each index and compute common elements in prefixes
for i in range(n):
prefixA = set(A[:i+1])
prefixB = set(B[:i+1])
commonPrefixCount.append(len(prefixA & prefixB))
return commonPrefixCount
Time Complexity:
- Prefix Set Construction:
For each index \( i \), we construct two sets (prefixA and prefixB) of size \( O(i) \), which takes \( O(i) \) time. So for all indices, the total time for set construction is \( O(1 + 2 + \ldots + n) = O(n^2) \), where \( n \) is the size of the array.
- Set Intersection:
For each index \( i \), we perform an intersection between the two sets, which can take up to \( O(i) \) time. The total time for all intersections is \( O(1 + 2 + \ldots + n) = O(n^2) \).
- Overall Time Complexity:
Since both the set construction and intersection take \( O(n^2) \), the overall time complexity is \( O(n^2) \).
Space Complexity:
- Set Space:
For each index \( i \), we store two sets of size \( O(i) \), leading to a total space usage of \( O(n) \) for each set. The space complexity is \( O(n) \) for the sets used in each iteration.
- Overall Space Complexity:
Since we are using sets and the final result array, the total space complexity is \( O(n) \).