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Find the prefix common array of two arrays leetcode 2657

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okay, let's break down the leetcode problem "2657. find the prefix common array of two arrays" with a detailed explanation, approach, code, and considerations.
**problem statement:**
you are given two integer arrays `a` and `b` of length `n`. you need to find an array `c` of the same length such that `c[i]` is equal to the number of common elements between the prefixes `a[0...i]` and `b[0...i]`. in other words, `c[i]` represents the number of elements that appear in both `a[0...i]` and `b[0...i]`.
**example:**
* **c[0]:** a[0...0] = [1], b[0...0] = [3]. no common elements. c[0] = 0
* **c[1]:** a[0...1] = [1, 3], b[0...1] = [3, 1]. common elements are 1 and 3. c[1] = 2
* **c[2]:** a[0...2] = [1, 3, 2], b[0...2] = [3, 1, 2]. common elements are 1, 2, and 3. c[2] = 3
* **c[3]:** a[0...3] = [1, 3, 2, 4], b[0...3] = [3, 1, 2, 4]. common elements are 1, 2, 3, and 4. c[3] = 4
**approach:**
the core idea is to iterate through the arrays `a` and `b` up to index `i`. for each index `i`, we determine the prefixes `a[0...i]` and `b[0...i]`. then, we count how many elements are present in both prefixes.
we can use a set (or hash table) to efficiently keep track of the elements encountered in the prefixes. here's the breakdown of the common algorithm:
1. **initialization:**
* create an empty array `c` of size `n` to store the results.
* create two sets (or hash tables): `seta` and `setb`. these will store elements of the prefixes of a and b, respectively.
* initialize `commoncount = 0`. this variable tracks the number of elements common to both prefix sets.
2. **iteration:**
* iterate from `i = 0` to `n - 1`:
* add `a[i]` to `seta`.
* add `b[i]` to `setb`.
* now, iterate through `seta`. for each element in `seta`, check if it's present in `setb`.
* if an element from `seta` is also in `setb`, increment `commoncount`.
* set `c[i] = commoncount`.
* reset `commoncount = ...
#LeetCode #CodingChallenge #ArrayAlgorithms
prefix common array
LeetCode 2657
two arrays
common elements
array manipulation
prefix sum
element comparison
algorithm challenge
coding interview
Python solution
array intersection
problem-solving
data structures
performance optimization
competitive programming
**problem statement:**
you are given two integer arrays `a` and `b` of length `n`. you need to find an array `c` of the same length such that `c[i]` is equal to the number of common elements between the prefixes `a[0...i]` and `b[0...i]`. in other words, `c[i]` represents the number of elements that appear in both `a[0...i]` and `b[0...i]`.
**example:**
* **c[0]:** a[0...0] = [1], b[0...0] = [3]. no common elements. c[0] = 0
* **c[1]:** a[0...1] = [1, 3], b[0...1] = [3, 1]. common elements are 1 and 3. c[1] = 2
* **c[2]:** a[0...2] = [1, 3, 2], b[0...2] = [3, 1, 2]. common elements are 1, 2, and 3. c[2] = 3
* **c[3]:** a[0...3] = [1, 3, 2, 4], b[0...3] = [3, 1, 2, 4]. common elements are 1, 2, 3, and 4. c[3] = 4
**approach:**
the core idea is to iterate through the arrays `a` and `b` up to index `i`. for each index `i`, we determine the prefixes `a[0...i]` and `b[0...i]`. then, we count how many elements are present in both prefixes.
we can use a set (or hash table) to efficiently keep track of the elements encountered in the prefixes. here's the breakdown of the common algorithm:
1. **initialization:**
* create an empty array `c` of size `n` to store the results.
* create two sets (or hash tables): `seta` and `setb`. these will store elements of the prefixes of a and b, respectively.
* initialize `commoncount = 0`. this variable tracks the number of elements common to both prefix sets.
2. **iteration:**
* iterate from `i = 0` to `n - 1`:
* add `a[i]` to `seta`.
* add `b[i]` to `setb`.
* now, iterate through `seta`. for each element in `seta`, check if it's present in `setb`.
* if an element from `seta` is also in `setb`, increment `commoncount`.
* set `c[i] = commoncount`.
* reset `commoncount = ...
#LeetCode #CodingChallenge #ArrayAlgorithms
prefix common array
LeetCode 2657
two arrays
common elements
array manipulation
prefix sum
element comparison
algorithm challenge
coding interview
Python solution
array intersection
problem-solving
data structures
performance optimization
competitive programming