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Problem Statement
Given an array of integers nums
, and an integer k
, find the count of "good" subarrays within nums
.
A subarray is considered "good" if it contains at least k
pairs of elements (i, j)
where i < j
and nums[i] == nums[j]
.
A subarray is a contiguous sequence
of elements in the original array.
Examples
-
Example 1:
- Input:
nums = [2, 2, 2, 3, 3], k = 3
- Expected Output:
3
- Justification: There are 3 good subarrays that meet the criteria:
[2, 2, 2, 3]
,[2, 2, 2, 3, 3]
, and[2, 2, 2]
. Each of these subarrays has at least 3 pairs of equal elements.
- Input:
-
Example 2:
- Input:
nums = [4, 4, 4, 4], k = 6
- Expected Output:
1
- Justification: The only good subarray that meets the criteria is the entire array itself
[4, 4, 4, 4]
, which contains 6 pairs of equal elements.
- Input:
-
Example 3:
- Input:
nums = [1, 2, 3, 1, 2], k = 2
- Expected Output:
2
- Justification: There is only 1 good subarray, which is
[1, 2, 3, 1, 2]
. It contains at least 2 pairs of equal elements.
- Input:
Solution
To solve this problem, an efficient strategy utilizing a sliding window technique alongside a frequency map is employed to meticulously count the subarrays fulfilling the criterion of containing at least k
pairs of identical elements. This method hinges on dynamically adjusting the boundaries of the window to encompass various segments of the array, methodically tracking the occurrence of each element to ascertain the formation of pairs. As the window expands, it meticulously updates the pair count, ensuring that each segment's potential to meet the specified condition is fully explored.
The effectiveness of this approach is underscored by its capacity to seamlessly navigate through the array, leveraging the frequency map for swift updates and checks on pair formation. This optimization minimizes redundant computations, thereby enhancing efficiency. By judiciously adjusting the window's size based on the evolving pair count, the algorithm ensures a comprehensive evaluation of all possible subarrays, accurately identifying those that qualify as "good" by the problem's standards.
Step-by-Step Algorithm
-
Initialize Variables: Start with two pointers,
i = 0
andj = 0
, a long counterans = 0
for the answer, a long countercount = 0
for the number of pairs, andn = nums.length
for the size of the input array. Also, initialize a frequency mapm
. -
Iterate Through the Array: Loop through the array with
j
from 0 ton - 1
. For each elementnums[j]
, increment its frequency in the mapm
. -
Update Pair Count: If the frequency of
nums[j]
after incrementing is more than 1, it forms pairs. Incrementcount
bym[nums[j]] - 1
, which represents the new pairs formed by adding the current element. -
Check and Adjust Window: While
count >= k
(the subarray has enough pairs) andi <= j
, calculate the number of new good subarrays that can be formed with the currentj
by addingn - j
toans
. Then, decrement the frequency ofnums[i]
asi
moves forward. Ifnums[i]
's frequency is still at least 1, decrementcount
bym[nums[i]]
because movingi
means losing some pairs. Incrementi
to shrink the window from the start. -
Return Answer: Once the loop completes, return
ans
as the total count of good subarrays found.
Algorithm Walkthrough
Let's consider the Input [2, 2, 2, 3, 3], k = 3
-
Initialization:
- Set
i = 0
to mark the start of the sliding window. - Initialize
ans = 0
to keep track of the total count of good subarrays. - Initialize
count = 0
to track the current number of pairs within the window. - Determine
n = 5
as the size of the input array. - Create a frequency map
m
to track the occurrences of each number within the current window.
- Set
-
First Iteration (j = 0):
- Process the first element,
2
. The frequency map is updated to{2: 1}
. No pairs are formed yet, socount
remains at 0.
- Process the first element,
-
Second Iteration (j = 1):
- Process the second element,
2
. The frequency map updates to{2: 2}
. We now have one pair of2
s, incrementingcount
by 1 (count = count + 2 - 1 = 0 + 2 - 1 = 1
).
- Process the second element,
-
Third Iteration (
j = 2
):- Add another
2
tom
.m = {2: 3}
,count = count + 3 - 1 = 1 + 3 - 1 = 1
. - Now,
count
is equal tok(3)
andi<=j
. So, execute the inner loop.ans = ans + n - j = 0 + 5 - 2 = 3
- Update the map to
{2: 2}
. - Update the count to
count(3) - (m[2](3) - 1) = 3 - 2 = 1
. - increment
i
to1
.
- Add another
-
Fourth Iteration (
j = 3
):- Add
3
tom
.m = {2: 2, 3: 1}
,count
remains1
.
- Add
-
Fifth Iteration (
j = 4
):- Add another
3
tom
.m = {2: 3, 3: 2}
,count
increases to2
.
- Add another
-
Final Answer:
ans = 3
is the total count of good subarrays found.
Code
Complexity Analysis
-
Time Complexity: The overall time complexity of the solution is O(n^2). This is because the solution uses two nested loops to explore all possible subarrays of the input array, where
n
is the length of the input array. Within the inner loop, operations such as updating the frequency map and calculating the number of pairs have a time complexity of O(1) per iteration, but the repeated calculation for each subarray leads to the quadratic time complexity. -
Space Complexity: The space complexity is O(n) for all solutions. This is due to the use of a hash map (or dictionary in Python) to count the frequency of elements within each subarray. In the worst case, if all elements of the array are distinct, the space required for the frequency map would be proportional to the size of the subarray being considered, which can be up to
n
in size.
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