
Problem Statement
In this problem, we have to work with two arrays, A
and B
, each containing a permutation of integers from 1
to n
. Our goal is to construct a new array C
such that each element C[i]
captures the count of numbers that appear in both A
and B
up to the index i
. This concept is described as the prefix common array for A
and B
.
The challenge involves efficiently comparing elements up to index i
in both permutations and counting common numbers to form each corresponding index in array C
. We are essentially tracking the accumulation of commonalities between the two arrays as we progress from the start to the end of the arrays.
Examples
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 <= 50
1 <= A[i], B[i] <= n
It is guaranteed that A and B are both a permutation of n integers.
Approach and Intuition
Given the constraints and the nature of the problem, a direct approach can be practical:
Initialize an empty array
C
which will store the count of common numbers for each index. Also, set up a Set data structure,seen
, to keep track of numbers encountered inB
as we iterate.For each index
i
from 0 ton-1
:- Add the number at the current index of
B
to theseen
set. This step continually updates which numbers we've encountered inB
up to the current index. - Initialize a variable,
common_count
, to zero. This variable will hold the count of common elements for the current indexi
. - Iterate over array
A
from index 0 toi
.- For each number in this slice of
A
, check if it also exists inseen
(indicates the current number has been seen inB
up to indexi
). - If a number is found in
seen
, increment thecommon_count
.
- For each number in this slice of
- After finishing this check for all indices up to
i
, appendcommon_count
to arrayC
.
- Add the number at the current index of
This approach takes advantage of the fact that by using a set, checking for existence is average-case O(1), making it efficient to determine if elements of A
up to index i
have appeared in B
. We deliberately accept a probably higher complexity for clarity and directness due to manageable constraint limits (n <= 50). This method ensures that we correctly build the prefix common array by leveraging the properties of sets for quick lookups and aggregating counts as we progress through the arrays.
Solutions
- C++
- Java
- Python
class Solution {
public:
vector<int> calculateCommonPrefix(vector<int>& array1, vector<int>& array2) {
int length = array1.size();
vector<int> result(length), elementCounts(length + 1, 0);
int countShared = 0;
for (int i = 0; i < length; ++i) {
if (++elementCounts[array1[i]] == 2) ++countShared;
if (++elementCounts[array2[i]] == 2) ++countShared;
result[i] = countShared;
}
return result;
}
};
The provided C++ code defines a function calculateCommonPrefix
within a class named Solution
. This function aims to find the common prefix array of two integer arrays based on the following principles:
- The function takes two vectors
array1
andarray2
as input. - It initializes a vector
result
to store the cumulative number of common prefix elements between the two arrays up to each index. - Another vector
elementCounts
is used to track the occurrences of each element across both arrays. - A variable
countShared
keeps track of the count of shared elements between the two input vectors at each point.
The key steps in the function are:
- Iterate through the elements of the input arrays using a for-loop, simultaneously processing elements from both arrays.
- Increment the count of each element in
elementCounts
as it appears in either array. If after incrementing, an element's count reaches two, this indicates that the element now exists in botharray1
andarray2
. - For every element found in both arrays, increment
countShared
. - Store the current value of
countShared
in theresult
vector at the current index, which represents the total number of shared elements up to that point between the two arrays.
Finally, the vector result
is returned, which contains the count of common elements up to each index of the input arrays. This function efficiently computes the number of elements common between any two positions of the two arrays using a single loop and provides a direct count by leveraging the indexing property of vectors.
class Solution {
public int[] computeIntersectionCounts(int[] arr1, int[] arr2) {
int len = arr1.length;
int[] resultCounts = new int[len];
int[] elemCount = new int[len + 1];
int intersectCount = 0;
// Processing each element for intersections
for (int idx = 0; idx < len; ++idx) {
// Increment element counts for current elements from both arrays
elemCount[arr1[idx]] += 1;
if (elemCount[arr1[idx]] == 2) ++intersectCount;
elemCount[arr2[idx]] += 1;
if (elemCount[arr2[idx]] == 2) ++intersectCount;
// Accumulate intersect counts in result array
resultCounts[idx] = intersectCount;
}
// Return the array of intersect counts
return resultCounts;
}
}
In the provided Java solution, focus on determining the intersection counts between elements of two arrays at each index. The method computeIntersectionCounts
in the Solution
class achieves this by:
- Initializing
len
to the length of the first array (arr1
). - Setting up
resultCounts
, an array to store the result, andelemCount
to keep track of the occurrences of each element across both arrays. - A loop iterates through each index of the arrays:
- For each index, the method increments the count of the current elements from both arrays in
elemCount
. - If the count of an element reaches 2 (indicating that element is found in both arrays up to this index), the
intersectCount
is incremented. - The total
intersectCount
up to the current index is stored inresultCounts
.
- For each index, the method increments the count of the current elements from both arrays in
- The method returns
resultCounts
, which provides the cumulative intersection count at each index.
This solution ensures you can track how many times elements of two arrays have matched up to each index, providing a detailed intersection profile across the arrays.
class Solution:
def computeCommonPrefixes(self, numbersA: List[int], numbersB: List[int]) -> List[int]:
size = len(numbersA)
result_arr = [0] * size
occurrences = [0] * (size + 1)
common_element_count = 0
for idx in range(size):
occurrences[numbersA[idx]] += 1
if occurrences[numbersA[idx]] == 2:
common_element_count += 1
occurrences[numbersB[idx]] += 1
if occurrences[numbersB[idx]] == 2:
common_element_count += 1
result_arr[idx] = common_element_count
return result_arr
The given Python code defines a method to compute the prefix common array between two integer arrays, numbersA
and numbersB
. The function returns an array wherein each index captures the count of common elements encountered up to that point in the two input arrays. Here's how the computation process works:
- Determine the size of the input arrays, assuming both are of equal length, and initialize
result_arr
with zero values of that size. - Create an
occurrences
array, sized one larger than the input arrays, initialized with zeros to keep track of the occurrence count of numbers from both arrays. - Initialize
common_element_count
to zero to count the number of common elements. - Traverse through the input arrays:
- For each number in
numbersA
, increase its count inoccurrences
. If the count reaches two, incrementcommon_element_count
. - Similarly, update counts for
numbersB
and adjustcommon_element_count
accordingly. - Update
result_arr
at the current index with the value ofcommon_element_count
.
- For each number in
- Return
result_arr
, which holds the cumulative count of common elements at each index.
This operation helps in understanding how many common elements are found progressively when moving through both arrays simultaneously.
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