Is Delayed Count of Equal Elements easy or hard?

The problem is typically classified as Medium difficulty. The brute force idea is simple, but recognizing that you can track future occurrences with a hash table to achieve O(n) time requires familiarity with frequency counting patterns.

Delayed Count of Equal Elements Python/Java solution

In Python, use a dictionary to track element frequencies while iterating from the end of the array. In Java, use a HashMap for the same purpose. For each element, add the current frequency from the map to the answer, then update the map with the new count.

How to solve Delayed Count of Equal Elements in O(n)?

Iterate through the array from right to left and maintain a hash map storing how many times each value has appeared so far. When processing nums[i], check the map for its frequency and add that count to the result. Then increment the frequency of nums[i] in the map. Each lookup and update takes constant time, giving an overall O(n) solution.

What is the best approach for Delayed Count of Equal Elements?

The most efficient approach uses a hash table combined with reverse enumeration. Traverse the array from right to left while storing frequencies of elements already seen. For each element, the frequency in the map directly tells how many equal elements appear later. This reduces the complexity to O(n) time with O(n) extra space.

Is Delayed Count of Equal Elements asked at Google/Amazon/Meta?

Pair counting and frequency tracking problems are common interview patterns at companies like Google, Amazon, and Meta. While the exact problem title may vary, the technique of using a hash table to count future or past occurrences frequently appears in coding interviews.

What data structure is used in Delayed Count of Equal Elements?

A hash table (hash map) is the primary data structure used. It stores the frequency of elements encountered during the reverse traversal. This allows constant-time lookups and updates when counting equal elements appearing later in the array.

What is the time complexity of Delayed Count of Equal Elements?

The optimal solution runs in O(n) time because each element is processed exactly once with constant-time hash map operations. A naive brute force approach checks all pairs and requires O(n^2) time. Space complexity for the optimal method is O(n) due to the frequency map.

3837. Delayed Count of Equal Elements - Practice Online

You are given an integer array nums of length n and an integer k.

For each index i, define the delayed count as the number of indices j such that:

i + k < j <= n - 1, and
nums[j] == nums[i]

Return an array ans where ans[i] is the delayed count of index i.

Example 1:

Input: nums = [1,2,1,1], k = 1

Output: [2,0,0,0]

Explanation:

`i`	`nums[i]`	possible `j`	`nums[j]`	satisfying `nums[j] == nums[i]`	`ans[i]`
0	1	[2, 3]	[1, 1]	[2, 3]	2
1	2	[3]	[1]	[]	0
2	1	[]	[]	[]	0
3	1	[]	[]	[]	0

Thus, ans = [2, 0, 0, 0].

Example 2:

Input: nums = [3,1,3,1], k = 0

Output: [1,1,0,0]

Explanation:

`i`	`nums[i]`	possible `j`	`nums[j]`	satisfying `nums[j] == nums[i]`	`ans[i]`
0	3	[1, 2, 3]	[1, 3, 1]	[2]	1
1	1	[2, 3]	[3, 1]	[3]	1
2	3	[3]	[1]	[]	0
3	1	[]	[]	[]	0

Thus, ans = [1, 1, 0, 0].

Constraints:

1 <= n == nums.length <= 10⁵
1 <= nums[i] <= 10⁵
0 <= k <= n - 1