Continuous Subarray Sum Leetcode Problem 523 [Python Solution]
In this blog post, we will tackle the LeetCode problem titled Continuous Subarray Sum This problem falls under the "Arrays & Hashing" category and is categorized as having a "Medium" difficulty level.
We will explore the problem statement, constraints, and then dive into an efficient Python solution.
But before we get started, let's take a look at the problem overview.
Problem Overview
Problem Statement:
Given an integer array nums and an integer k, your task is to determine whether nums contains a contiguous subarray of at least two elements whose sum is a multiple of k.
If such a subarray exists, return true; otherwise, return false.
Definition of a Good Subarray:
A good subarray is a contiguous subarray that meets the following conditions:
1. It has at least two elements.
2. The sum of the elements in the subarray is a multiple of k.
Note:
– A subarray is a part of the array that consists of consecutive elements.
– An integer x is considered a multiple of k if there exists an integer n such that x = n * k.
It's worth noting that 0 is always considered a multiple of k.
Example 1:
“`
Input: nums = [23, 2, 4, 6, 7], k = 6
Output: true
Explanation: The subarray [2, 4] is a continuous subarray of size 2 whose elements sum up to 6, which is a multiple of k.
Example 2:
Input: nums = [23, 2, 6, 4, 7], k = 6
Output: true
Explanation: The entire array [23, 2, 6, 4, 7] is a continuous subarray of size 5, and its elements sum up to 42. Since 42 is a multiple of 6 (7 * 6), the answer is true.
Example 3:
Input: nums = [23, 2, 6, 4, 7], k = 13
Output: false
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Constraints:
– 1 <= nums.length <= 105
– 0 <= nums[i] <= 109
– 0 <= sum(nums[i]) <= 231 - 1
– 1 <= k <= 231 - 1
Now that we have a clear understanding of the problem, let's explore an efficient Python solution.
Efficient Python Code Solution
We will use a hashmap to efficiently solve this problem.
The key insight is that we can use the remainder of the running sum modulo k to keep track of subarrays with a sum that is a multiple of k.
The basic idea is to maintain a hashmap where the keys are the remainders of the running sum, and the values are the indices at which that remainder occurs in the nums array.
Here's the Python code that implements this efficient solution:
def checkSubarraySum(nums, k):
# Initialize a hashmap with one entry for the remainder 0 at index -1.
# This handles the case where the subarray starts at index 0. hashmap = {0: -1}
summ = 0
for i, num in enumerate(nums):
summ += num
# Calculate the remainder of the current sum modulo k.
current_remainder = summ % k
# If the current remainder exists in the hashmap and the length of the subarray is at least 2, return True.
if current_remainder in hashmap and i - hashmap[current_remainder] >= 2:
return True
# If the current remainder is not in the hashmap, add it with the current index.
if current_remainder not in hashmap:
hashmap[current_remainder] = i
# If no valid subarray is found, return False.
return False
This code efficiently processes the input array, keeping track of the remainders and their corresponding indices.
If it finds a subarray that satisfies the conditions of having at least two elements and a sum that is a multiple of k, it returns True.
If no such subarray exists, it returns False.
Time and Space Complexity
Now, let's analyze the time and space complexity of our solution:
Time Complexity: The code iterates through the nums array once, which takes O(n) time, where n is the length of the nums array.
Space Complexity: We use a hashmap to store the remainders, which can have a maximum of n entries.
Therefore, the space complexity is O(n).
Our solution is both time and space efficient, making it a suitable approach for solving the Continuous Subarray Sum problem.
Reasoning Behind Our Approach
The reasoning behind this approach lies in the observation that if we encounter the same remainder more than once during the iteration, it means we've found a subarray whose sum is a multiple of k.
By keeping track of the remainders and their corresponding indices, we can efficiently determine if such a subarray exists without the need for a brute-force search.
This approach leverages the properties of remainders and running sums to optimize the solution.
It's a clever way to address the problem's requirements while keeping the time complexity linear, which is especially important for large input arrays.
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Conclusion
In this blog post, we explored the LeetCode problem Continuous Subarray Sum We discussed the problem statement, constraints, and presented an efficient Python solution using a hashmap to keep track of remainders and their indices.
The provided code allows us to determine whether a given array contains a subarray with the desired properties efficiently.
Remember that efficient algorithms are often based on clever insights and observations, and this problem is a great example of how thinking outside the box can lead to an optimal solution.
If you have any questions or suggestions, please feel free to comment, ask questions, and share this content.
If you'd like to further explore this problem on LeetCode, you can find it here.
Happy coding!
