Arithmetic Slices – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
An integer array is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same.
- For example,
[1,3,5,7,9],[7,7,7,7], and[3,-1,-5,-9]are arithmetic sequences.
Given an integer array nums, return the number of arithmetic subarrays of nums.
A subarray is a contiguous subsequence of the array.
Example 1:
Input: nums = [1,2,3,4] Output: 3 Explanation: We have 3 arithmetic slices in nums: [1, 2, 3], [2, 3, 4] and [1,2,3,4] itself.
Example 2:
Input: nums = [1] Output: 0
Constraints:
1 <= nums.length <= 5000-1000 <= nums[i] <= 1000
C++ Arithmetic Slices LeetCode Solution
class Solution {
public:
int numberOfArithmeticSlices(vector<int>& nums) {
// if nums size is less than 3 return false
if(nums.size() < 3)
return 0;
int cnt = 0, diff;
for(int i = 0; i<nums.size()-2; ++i)
{
// storing diff of first 2 elements
diff = nums[i+1] - nums[i];
// checking for consecutive elements with same difference.
for(int j = i+2; j<nums.size(); ++j)
{
// if we find the same diff of next 2 elements
// this means we find consecutive elements
// increase the Count
if(nums[j] - nums[j-1] == diff)
++cnt;
else
// break as we need to cnt for consecutive diff elements
break;
}
}
// return cnt
return cnt;
}
};
Java Arithmetic Slices LeetCode Solution
public int numberOfArithmeticSlices(int[] nums) {
var slices = 0;
for (int i = 2, prev = 0; i < nums.length; i++)
slices += (nums[i] - nums[i - 1] == nums[i - 1] - nums[i - 2])
? ++prev
: (prev = 0);
return slices;
}
Python 3 Arithmetic Slices LeetCode Solution
class Solution:
def numberOfArithmeticSlices(self, nums: List[int]) -> int:
n = len(nums)
dp = [0] * n
ans = 0
for i in range(2, n):
if nums[i-1] - nums[i-2] == nums[i] - nums[i-1]:
dp[i] = dp[i-1] + 1
ans += dp[i]
return ans
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