3254. Find the Power of K-Size Subarrays I - Detailed Explanation
Problem Statement
You are given an array of positive integers nums and an integer k. A K-size subarray is any contiguous segment of the array that has exactly ( k ) elements. The power of a subarray is defined as the sum of its elements. Your task is to return the maximum power (i.e. the maximum subarray sum) among all contiguous subarrays of size ( k ).
Examples
-
Example 1:
- Input:
nums = [1, 4, 2, 5, 3],k = 3 - Subarrays of size 3:
- ([1, 4, 2]) with sum = (1 + 4 + 2 = 7)
- ([4, 2, 5]) with sum = (4 + 2 + 5 = 11)
- ([2, 5, 3]) with sum = (2 + 5 + 3 = 10)
- Output:
11
Explanation: The subarray ([4, 2, 5]) has the maximum power (11).
- Input:
-
Example 2:
- Input:
nums = [2, 3, 4],k = 2 - Subarrays of size 2:
- ([2, 3]) with sum = (5)
- ([3, 4]) with sum = (7)
- Output:
7
- Input:
-
Example 3:
- Input:
nums = [5, 1, 3, 2, 4],k = 1 - Output:
5
Explanation: When ( k = 1 ), each subarray is a single element. The maximum power is simply the maximum element.
- Input:
Constraints
- (1 \leq \text{nums.length} \leq 10^5)
- (1 \leq \text{nums}[i] \leq 10^4)
- (1 \leq k \leq \text{nums.length})
Hints
-
Sliding Window:
Because the subarrays are contiguous and of fixed size ( k ), a sliding window approach is ideal. -
Efficient Update:
Instead of computing the sum from scratch for every window, use the sum of the previous window—subtract the element that’s leaving and add the new element coming in. -
Initialization:
First compute the sum for the first ( k ) elements, then slide the window across the array while updating the maximum sum found.
Approaches
Approach 1: Sliding Window (Optimal)
Idea:
Use a sliding window of size ( k ) to compute the sum of each subarray in one pass:
-
Compute the sum of the first ( k ) elements.
-
For each subsequent position, subtract the element that is left behind and add the new element that enters the window.
-
Keep track of the maximum sum encountered.
Time Complexity: (O(n))
Space Complexity: (O(1))
Approach 2: Prefix Sum Array
Idea:
Compute a prefix sum array where each index stores the sum of all elements up to that index. Then, for every index ( i ) from ( 0 ) to ( n-k ), the sum of the subarray starting at ( i ) is:
[
\text{sum} = \text{prefix}[i+k] - \text{prefix}[i]
]
and update the maximum.
Time Complexity: (O(n))
Space Complexity: (O(n))
Note: This approach uses extra space compared to the sliding window but is equally efficient.
Code Implementations
Python Implementation
Java Implementation
Step-by-Step Walkthrough (Sliding Window Approach)
Let’s walk through the example:
Input: nums = [1, 4, 2, 5, 3], ( k = 3 )
-
Initialization:
- Compute the sum of the first ( k ) elements: [ \text{window\_sum} = 1 + 4 + 2 = 7 ]
- Set
max_sum = 7.
-
Slide the Window:
- First Slide (Window: indices 1 to 3):
- Remove
nums[0](value 1) and addnums[3](value 5): [ \text{window\_sum} = 7 - 1 + 5 = 11 ] - Update
max_sum = \max(7, 11) = 11.
- Remove
- Second Slide (Window: indices 2 to 4):
- Remove
nums[1](value 4) and addnums[4](value 3): [ \text{window\_sum} = 11 - 4 + 3 = 10 ] - Update
max_sum = \max(11, 10) = 11.
- Remove
- First Slide (Window: indices 1 to 3):
-
Completion:
- No more windows (since there are 5 elements and window size is 3).
- The maximum power among all ( 3 )-size subarrays is 11.
Complexity Analysis
-
Time Complexity:
(O(n)) — Each element is added and subtracted exactly once when the window slides. -
Space Complexity:
(O(1)) — Only a few extra variables are used.
Common Mistakes
-
Off-by-One Errors:
Make sure the window covers exactly ( k ) elements and that the loop iterates correctly from index ( k ) to ( n-1 ). -
Not Updating Maximum After the Loop:
Sometimes the last window’s sum isn’t compared with the current maximum; be sure to update the maximum each time. -
Recomputing the Sum from Scratch:
Avoid a nested loop that computes the sum for every subarray independently (which would lead to (O(n \times k)) time).
Edge Cases
- When ( k = 1 ):
The answer is simply the maximum element of the array. - When ( k ) equals the array length:
The answer is the sum of the entire array. - Single Element Array:
For an array with one element and ( k = 1 ), the result is that element.
Alternative Variations & Related Problems
-
Maximum Sum Subarray of Variable Length (Kadane’s Algorithm):
Instead of a fixed size ( k ), find the contiguous subarray with the maximum sum. -
Minimum Sum Subarray of Size ( k ):
A similar sliding window problem where you find the subarray with the minimum sum.
-
Average of Subarrays:
Instead of the sum, you might be asked to compute the average of all ( k )-size subarrays.
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