Given an input array of integers and a positive integer k, find out the maximum sum of a contiguous subarray of length exactly k
Example
Input: nums = [5, 6, 1, 2, 6, 6, 4, 3], k = 3
Output: 4 6
Explanation: The subarray nums[4...6] has the maximum sum of 16
The core challenge of this problem is to find the maximum sum of any contiguous subarray of length exactly k. This problem is significant in various applications such as financial analysis, where one might want to find the period with the highest profit.
Potential pitfalls include misunderstanding the requirement for the subarray to be contiguous and not considering edge cases where the array length is less than k.
To solve this problem, we can use a sliding window approach. The naive solution would involve calculating the sum of every possible subarray of length k, which would be inefficient with a time complexity of O(n*k). Instead, we can optimize this using a sliding window technique, which allows us to achieve a time complexity of O(n).
The naive approach involves iterating through the array and calculating the sum of every possible subarray of length k. This is not optimal because it involves redundant calculations.
The sliding window approach involves maintaining a window of size k and sliding it across the array. We start by calculating the sum of the first k elements. Then, for each subsequent element, we slide the window one position to the right by subtracting the element that is left behind and adding the new element that comes into the window.
1. Initialize the sum of the first k elements.
2. Iterate through the array starting from the k-th element.
3. For each new element, update the sum by subtracting the element that is left behind and adding the new element.
4. Keep track of the maximum sum encountered during this process.
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int maxSumSubarray(vector<int>& nums, int k) {
int n = nums.size();
if (n < k) {
// If the array length is less than k, return 0 or an appropriate value
return 0;
}
// Calculate the sum of the first k elements
int max_sum = 0;
for (int i = 0; i < k; i++) {
max_sum += nums[i];
}
int window_sum = max_sum;
// Slide the window from start to end of the array
for (int i = k; i < n; i++) {
window_sum += nums[i] - nums[i - k];
max_sum = max(max_sum, window_sum);
}
return max_sum;
}
int main() {
vector<int> nums = {5, 6, 1, 2, 6, 6, 4, 3};
int k = 3;
cout << "Maximum sum of subarray of length " << k << " is " << maxSumSubarray(nums, k) << endl;
return 0;
}
The time complexity of the sliding window approach is O(n) because we only pass through the array once. The space complexity is O(1) as we are using a constant amount of extra space.
1. If the array length is less than k, the function should handle this gracefully, possibly by returning 0 or an appropriate error value.
2. If all elements are negative, the function should still return the maximum sum, which might be less than 0.
To test the solution comprehensively, consider the following test cases:
When approaching such problems, it is crucial to understand the requirements and constraints fully. Start with a brute-force solution to understand the problem better, then look for patterns or optimizations. Practice similar problems to improve your problem-solving skills.
In this blog post, we discussed how to find the maximum sum of a contiguous subarray of length k using a sliding window approach. This method is efficient with a time complexity of O(n) and is suitable for large arrays. Understanding and implementing such algorithms is essential for solving a wide range of problems in computer science.