LeetCode 2367: Number of Arithmetic Triplets Solution
Master LeetCode problem 2367 (Number of Arithmetic Triplets), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
2367. Number of Arithmetic Triplets
Problem Explanation
Explanation:
To solve this problem, we can use a hashmap to keep track of the count of potential triplets as we iterate through the array. For each element num in the array, we check if num - diff exists in the map (i.e., a potential j) and num - 2*diff exists in the map (i.e., a potential i). If both exist, we increment the count of triplets by the value at num - diff. This is because we have found a valid triplet (i, j, k) where j is the current element, i is num - diff, and k is num.
Time Complexity:
The time complexity of this approach is O(N), where N is the number of elements in the array.
Space Complexity:
The space complexity is also O(N) due to the hashmap storing potential triplets.
:
Solution Code
class Solution {
public int numberOfArithmeticSlices(int[] nums, int diff) {
int count = 0;
Map<Integer, Integer> map = new HashMap<>();
for (int num : nums) {
int prevCount = map.getOrDefault(num - diff, 0);
count += prevCount;
map.put(num, map.getOrDefault(num, 0) + prevCount);
}
return count;
}
}Try It Yourself
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