Given an array of integers and a number target, find a continous subarray whose sum is equal to target.
Return the start and end indices denoting this subarray. If there are multiple solutions, you can return any of them.
Example:
Input: nums = [1, 1, 5, 2, 1, 3, 10, 2, 1], k = 21
Output: [2, 6]
Explanation: 5 + 2 + 1 + 3 + 10 = 21
1. The array can consist of positive and negative numbers.
2. Your algorithm should run in O(n) time and use O(n) extra space.
The core challenge of this problem is to find a continuous subarray within an array of integers that sums up to a given target value. This problem is significant in various applications such as financial analysis, where one might need to find a period with a specific sum of transactions.
Potential pitfalls include handling negative numbers and ensuring the solution runs efficiently within the given constraints.
To solve this problem, we can use a sliding window approach combined with a hash map to keep track of the cumulative sums. This allows us to find the subarray in O(n) time complexity.
A naive solution would involve checking all possible subarrays and their sums, which would result in O(n^2) time complexity. This is not optimal for large arrays.
We can optimize the solution using a hash map to store the cumulative sums and their corresponding indices. By iterating through the array and maintaining a running sum, we can check if the difference between the current sum and the target has been seen before. If it has, we have found our subarray.
1. Initialize a hash map to store cumulative sums and their indices.
2. Initialize a variable to keep track of the cumulative sum.
3. Iterate through the array, updating the cumulative sum.
4. For each element, check if the cumulative sum minus the target exists in the hash map.
5. If it exists, return the indices. Otherwise, store the current cumulative sum and its index in the hash map.
def subarray_sum(nums, target):
# Dictionary to store the cumulative sum and its index
sum_map = {}
# Initialize the cumulative sum
current_sum = 0
# Iterate through the array
for i, num in enumerate(nums):
# Update the cumulative sum
current_sum += num
# Check if the current sum is equal to the target
if current_sum == target:
return [0, i]
# Check if (current_sum - target) exists in the map
if (current_sum - target) in sum_map:
return [sum_map[current_sum - target] + 1, i]
# Store the current sum and its index in the map
sum_map[current_sum] = i
# If no subarray is found, return an empty list
return []
# Example usage
nums = [1, 1, 5, 2, 1, 3, 10, 2, 1]
target = 21
print(subarray_sum(nums, target)) # Output: [2, 6]
The time complexity of this approach is O(n) because we iterate through the array once. The space complexity is also O(n) due to the hash map storing the cumulative sums.
1. The array contains only one element equal to the target.
2. The array contains negative numbers.
3. The target is zero.
4. No subarray sums to the target.
To test the solution comprehensively, consider the following test cases:
# Test case 1: Simple case assert subarray_sum([1, 2, 3, 4, 5], 9) == [1, 3] # Test case 2: Negative numbers assert subarray_sum([-1, -2, 3, 4, -5], 2) == [1, 3] # Test case 3: Single element equal to target assert subarray_sum([5], 5) == [0, 0] # Test case 4: No subarray sums to target assert subarray_sum([1, 2, 3], 7) == []
1. Break down the problem into smaller parts and understand the requirements.
2. Consider edge cases and how your solution handles them.
3. Practice similar problems to improve your problem-solving skills.
Understanding and solving the subarray sum problem is crucial for various applications. By using an optimized approach with a hash map, we can achieve an efficient solution with O(n) time complexity.
1. LeetCode: Subarray Sum Equals K