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 the given array that sums up to the 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.
The optimized solution involves using a hash map to store the cumulative sum up to each index. By doing this, we can quickly check if there is a subarray that sums to the target by looking for the difference between the current cumulative sum and the target in the hash map.
1. Initialize a hash map to store the cumulative sum and its corresponding index.
2. Iterate through the array, updating the cumulative sum.
3. For each element, check if the cumulative sum minus the target exists in the hash map.
4. If it exists, return the indices. Otherwise, store the cumulative sum and its index in the hash map.
#include <iostream>
#include <unordered_map>
#include <vector>
std::vector<int> subarraySum(std::vector<int> &nums, int target) {
std::unordered_map<int, int> sumMap; // To store cumulative sum and its index
int cumulativeSum = 0;
sumMap[0] = -1; // To handle the case when subarray starts from index 0
for (int i = 0; i < nums.size(); ++i) {
cumulativeSum += nums[i];
// Check if (cumulativeSum - target) exists in the map
if (sumMap.find(cumulativeSum - target) != sumMap.end()) {
return {sumMap[cumulativeSum - target] + 1, i};
}
// Store the cumulative sum with its index
sumMap[cumulativeSum] = i;
}
return {}; // Return an empty vector if no subarray is found
}
int main() {
std::vector<int> nums = {1, 1, 5, 2, 1, 3, 10, 2, 1};
int target = 21;
std::vector<int> result = subarraySum(nums, target);
if (!result.empty()) {
std::cout << "Subarray found from index " << result[0] << " to " << result[1] << std::endl;
} else {
std::cout << "No subarray found" << std::endl;
}
return 0;
}
The time complexity of this approach is O(n) because we traverse the array once. The space complexity is also O(n) due to the hash map storing 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:
When approaching such problems, consider using hash maps to store intermediate results for quick lookups. Practice similar problems to improve your problem-solving skills.
Understanding and solving this problem helps in developing efficient algorithms for finding subarrays with specific properties. Practice and explore further to enhance your skills.