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Given a 0-indexed integer array nums
, determine whether there exist two subarrays of length 2
with equal sum. Note that the two subarrays must begin at different indices.
Return true
if these subarrays exist, and false
otherwise.
A subarray is a contiguous non-empty sequence of elements within an array.
Example 1:
Input: nums = [4,2,4]
Output: true
Explanation: The subarrays with elements [4,2] and [2,4] have the same sum of 6.
Example 2:
Input: nums = [1,2,3,4,5]
Output: false
Explanation: No two subarrays of size 2 have the same sum.
Example 3:
Input: nums = [0,0,0]
Output: true
Explanation: The subarrays [nums[0],nums[1]] and [nums[1],nums[2]] have the same sum of 0.
Note that even though the subarrays have the same content, the two subarrays are considered different because they are in different positions in the original array.
Constraints:
2 <= nums.length <= 1000
-109 <= nums[i] <= 109
class Solution {
public:
bool findSubarrays(vector<int>& nums) {
int n=nums.size();
int count=0;
for(int i=0;i<n-1;i++){
for(int j=i+1;j<n-1;j++){
if(nums[j]+nums[j+1]==nums[i]+nums[i+1]) count++;
}
}
return count>=1;
}
};
class Solution:
def findSubarrays(self, nums: List[int]) -> bool:
sums = set()
for i in range(len(nums)-1):
t = nums[i]+nums[i+1]
if t in sums:
return True
else:
sums.add(t)
return False
class Solution {
public boolean findSubarrays(int[] nums) {
if(nums.length<2)
return false;
HashSet<Integer> set=new HashSet<>(); // To Track previous sum calculated
for(int i=0;i<nums.length-1;i++){
if(!set.add(nums[i]+nums[i+1])) // Check if sum already exist
return true;
}
return false;
}
}
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