Daily Leads And Partners - Leetcode Solution

Problem Statement:

The problem Daily Leads And Partners on LeetCode asks us to find the daily count of distinct partners who received leads from a company for a given period of time. We are given an array of integers representing the total number of leads generated by a company on each day. The array is of size n. We are also given two integers, partners and days, representing the total number of partners and the number of days for which we need to find the daily count of distinct partners.

Solution:

This problem can be solved using an efficient approach using two pointers. We can maintain a hash table to keep track of the partners who received leads every day. We can use two pointers, left pointer and right pointer, to keep track of the elements in the array. Initially, both the pointers are set to the first element of the array.

We can then iterate through the array using the right pointer. As we iterate through the array, we can update the hash table to include the partner IDs who received the leads on that day. We can also keep track of the number of distinct partners who received the leads on that day using the hash table.

Once we have updated the hash table for the current day, we can check if the number of days for which we need to find the daily count of partners has been reached. If the number of days has been reached, we can add the count of distinct partners who received the leads on the current day to the result array and move the left pointer to the next element in the array.

We can continue iterating through the array using the right pointer and updating the hash table until we have iterated through all the elements in the array. We can return the result array containing the daily count of distinct partners who received the leads.

Implementation:

The following Python code implements the approach described above:

class Solution:

def dailyLeads(self, leads: List[int], partners: int, days: int) -> List[int]:

    if(days == 0):

        return [0] * len(leads)

    

    left = 0

    right = 0

    partnerSet = set()

    dailyPartners = []

    

    while right < len(leads):

        # Update the hash table with the partners who received leads on the current day

        for i in range(leads[right]):

            partnerSet.add(i+1)

            

        # If the number of days has been reached, add the daily count of distinct partners to the result array

        if(right - left + 1 == days):

            dailyPartners.append(len(partnerSet))

            

            # Move the left pointer to the next element

            for i in range(leads[left]):

                if(i+1 in partnerSet):

                    partnerSet.remove(i+1)

            left += 1

        

        right += 1

        

    # If the right pointer has not reached the end of the array, add the count of distinct partners for the remaining days

    while(right - left >= days):

        for i in range(leads[left]):

            if(i+1 in partnerSet):

                partnerSet.remove(i+1)

        dailyPartners.append(len(partnerSet))

        left += 1

        

    return dailyPartners

Time Complexity:

The time complexity of the above solution is O(n * p), where n is the length of the input array and p is the maximum number of partners who can receive leads on a single day. In the worst case, all p partners can receive the leads on each of the n days, resulting in a time complexity of O(n * p). However, in practice, the number of partners who receive leads on a single day is usually limited, resulting in a lower time complexity.

Space Complexity:

The space complexity of the above solution is O(p), where p is the maximum number of partners who can receive leads on a single day. This is because we are using a hash table to keep track of the partners who received leads on each day, and the size of the hash table is limited by the maximum number of partners who can receive leads on a single day.

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