Leetcode 2395: Find Subarrays With Equal Sum
You are given a 0-indexed integer array
nums. Your task is to determine if there are two subarrays of length 2 that have the same sum, but these subarrays must begin at different indices in the array. A subarray is defined as a contiguous sequence of elements within the array.Problem
Approach
Steps
Complexity
Input: The input consists of an integer array `nums` of length n, where 2 <= n <= 1000 and -10^9 <= nums[i] <= 10^9.
Example: nums = [3,1,2,3,5]
Constraints:
• 2 <= nums.length <= 1000
• -10^9 <= nums[i] <= 10^9
Output: Return `true` if there are two distinct subarrays of length 2 with the same sum, and `false` otherwise.
Example: Output: true
Constraints:
Goal: The goal is to find if there are two distinct subarrays of length 2 with the same sum by checking each pair of consecutive elements in the array.
Steps:
• 1. Traverse the array and for each pair of consecutive elements, calculate their sum.
• 2. Store the sum of each subarray of length 2 in a set.
• 3. If any sum is encountered more than once, return true.
• 4. If no sums are repeated, return false.
Goal: The solution must handle the array size efficiently and account for large values of array elements.
Steps:
• Ensure the solution works within time limits for an array of size up to 1000.
Assumptions:
• There will always be at least one pair of subarrays to check.
• Input: nums = [4,2,4]
• Explanation: The subarrays [4,2] and [2,4] have the same sum of 6. Since they are at different indices, the answer is true.
• Input: nums = [1,2,3,4,5]
• Explanation: No two subarrays of length 2 have the same sum. Hence, the answer is false.
Approach: The approach involves calculating the sum of all possible subarrays of length 2 and checking for duplicates using a set.
Observations:
• The problem requires checking pairs of elements for matching sums.
• We can use a set to track sums and easily check for duplicates.
Steps:
• 1. Initialize an empty set to store subarray sums.
• 2. Iterate over the `nums` array, calculate the sum of each pair of consecutive elements.
• 3. For each sum, check if it already exists in the set.
• 4. If a duplicate sum is found, return true.
• 5. If the loop finishes without finding duplicates, return false.
Empty Inputs:
• If the array contains less than 2 elements, no subarray of length 2 can exist.
Large Inputs:
• Handle cases where the array contains the maximum allowed number of elements (1000).
Special Values:
• Consider cases where all elements are the same.
Constraints:
• Ensure the solution works within time limits for large inputs and handles extreme values of array elements.
bool findSubarrays(vector<int>& nums) {
unordered_set<int> s;
for (int i = 1; i < nums.size(); ++i)
if(!s.insert(nums[i - 1] + nums[i]).second)
return true;
return false;
}
1 : Function Declaration
bool findSubarrays(vector<int>& nums) {
The function `findSubarrays` is declared, which takes a reference to a vector of integers `nums` as input and returns a boolean value.
2 : Set Initialization
unordered_set<int> s;
An unordered set `s` is initialized to store the sums of consecutive pairs of elements in the `nums` array. This ensures that only unique sums are stored.
3 : Loop Initialization
for (int i = 1; i < nums.size(); ++i)
A loop starts from the second element of the `nums` vector (index 1) and iterates through the rest of the vector.
4 : Checking for Duplicates
if(!s.insert(nums[i - 1] + nums[i]).second)
For each consecutive pair of elements, their sum is calculated and inserted into the set `s`. If the insertion is unsuccessful (i.e., the sum is already in the set), it indicates that a duplicate sum exists.
5 : Return True
return true;
If a duplicate sum is found, the function immediately returns `true`, indicating that there are two consecutive subarrays with the same sum.
6 : Return False
return false;
If the loop completes without finding a duplicate sum, the function returns `false`, indicating no such subarrays exist.
Best Case: O(n), where n is the length of the input array.
Average Case: O(n), where n is the length of the input array.
Worst Case: O(n), where n is the length of the input array.
Description: The algorithm processes each pair of consecutive elements once, making the time complexity O(n).
Best Case: O(n), where n is the length of the input array.
Worst Case: O(n), where n is the length of the input array, due to the space used by the set to store sums.
Description: The space complexity is O(n) as we need a set to store the subarray sums.
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