Maximum Population Year

The task is to determine the year with the highest population. Here’s a brief explanation:

  1. We have data about the birth and death years of different people, represented as 2D integer array logs. Each entry in logs represents one person, with the birth year and death year as the two elements in the array.

  2. The population of a particular year is the count of people alive during that year. A person is considered alive in a particular year if that year is between their birth year and one year less than their death year, inclusive.

  3. We need to identify the year with the maximum population. If there are multiple years with the same maximum population, we should return the earliest year.

To put it in a simple way, imagine you have a timeline from 1950 to 2050. For each person, mark the years they were alive. Now, find the year with the most marks. If there’s a tie for the most marks, pick the earliest year.

This problem can be solved by a simple frequency count.

Python solution:

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class Solution:
    def maximumPopulation(self, logs: List[List[int]]) -> int:
        population = [0] * 101  # because the year range is from 1950 to 2050, 101 years in total
        for birth, death in logs:
            for i in range(birth - 1950, death - 1950):  # translate the year to the index of the list
                population[i] += 1
        max_population_year = population.index(max(population)) + 1950  # find the year with maximum population
        return max_population_year

This code first initializes a list with 101 zeros, where each index i represents the year 1950 + i. It then counts the population of each year and stores it in the list. Finally, it finds the index of the maximum population in the list, which represents the earliest year with maximum population, and returns this year.

  • population = [0] * 101: This line creates a list population of size 101 (representing years from 1950 to 2050) and initializes all the values to 0.

  • The nested for loop goes over every year in the lifespan of each person (from their birth year to their death year). For each of these years, it increments the corresponding value in the population list by 1. This effectively counts the number of people alive in each year.

  • max_population_year = population.index(max(population)) + 1950: This line first finds the maximum value in the population list (which represents the maximum population) and then finds the earliest year in which this maximum population occurred.

So, this code essentially creates a histogram of the population for each year and then finds the year with the maximum population. There isn’t an accumulation step like in a prefix sum. The list population stores the total population for each year directly, not cumulative totals. I hope this clarifies things. My apologies for the previous confusion.

The problem does have similarities with the “Meeting Rooms” type problems, specifically “Meeting Rooms III” where you need to find the maximum number of concurrent meetings.

Here, you’re finding the year with the maximum population, which is analogous to finding the time slot with the maximum number of overlapping meetings. The birth year can be thought of as the start of a meeting, and the death year as the end. The concept of tracking the start and end of intervals is a common approach in both problem types.

The line population[i] += 1 is counting the frequency, where ‘frequency’ refers to the number of people born in a particular year. This will help to build the ‘population’ list where each index represents a year (from 1950 to 2050), and the value at each index represents the number of people born in that year. This method essentially generates a population distribution across the years.

Range Addition

Range addition is a technique used when you have a range (start, end) and you want to increment or decrement all the values in that range by a certain amount in an array.

Let’s use a simple example to illustrate this. Suppose you have an array of length 5, initialized with zeros.

array = [0, 0, 0, 0, 0]

Now, let’s say you want to increment all values in the range [1, 3] (0-indexed) by 2. Using range addition, you would add 2 at the start index and subtract 2 at the end index + 1:

array = [0, 2, 0, 0, -2]

At this point, the array doesn’t reflect the correct state, because we want array[1], array[2], and array[3] to each be incremented by 2. To fix this, we perform a prefix sum operation, where each element is the sum of itself and the element before it:

array = [0, 2, 2, 2, 0]

Now the array correctly represents that the elements at indices 1, 2, and 3 have been incremented by 2.

This is a simplified version of the range addition technique. In the problem about finding the year with the maximum population, we similarly increment the start year (birth year) and decrement the end year (death year) for each person’s lifespan, then use prefix sum to compute the population for each year.

Here is a way to solve the problem using the concept of Range Addition:

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class Solution:
    def maximumPopulation(self, logs: List[List[int]]) -> int:
        delta = [0] * 101  # because the year range is from 1950 to 2050, 101 years in total
        for birth, death in logs:
            delta[birth - 1950] += 1  # increment the birth year
            delta[death - 1950] -= 1  # decrement the death year
        max_population = 0
        max_year = 0
        current_population = 0
        for year, change in enumerate(delta):
            current_population += change
            if current_population > max_population:
                max_population = current_population
                max_year = year
        return max_year + 1950  # convert the year index back to actual year

This approach increments the population at the start of each person’s life and decrements it at the end. This results in the delta list representing the change in population each year. We then iterate over delta, keeping track of the current population. When we find a year where the population is higher than any previous year, we update max_population and max_year.

This is effectively a variant of the Range Addition concept, as we increment the start of the range and decrement at the end of the range for each log of birth and death years. This allows us to track the changes in population over the years.

The optimized solution to this problem uses a prefix sum. After applying the range addition technique to increment the population count for the start year and decrement it for the end year, the array contains the changes in population for each year relative to the previous year.

In order to convert these relative changes into absolute population counts for each year, a prefix sum operation is performed. In this operation, each element in the array is replaced by the sum of all the elements before it and itself. The prefix sum operation essentially accumulates the population changes, yielding the total population for each year.

The prefix sum operation is performed with this line of code:

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for i in range(1, 101):
    population[i] += population[i - 1]

Here, each element of the population array is updated to be the sum of itself and the previous element, which effectively calculates the prefix sum.

The concept of “Range Addition” is applied when we have a range [start, end] and we need to increment or decrement the values in this range by a certain amount. In this problem, for each person’s lifespan, we increment the birth year and decrement the death year in the population array:

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population[birth - 1950] += 1
population[death - 1950] -= 1

This operation effectively marks the changes in population at the beginning and end of each person’s life. Then, when we perform the Prefix Sum, it accumulates these changes over all years to yield the total population for each year.

These two techniques, Range Addition and Prefix Sum, work together to solve this problem efficiently.

Line Sweep

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class Solution:
    def maximumPopulation(self, logs: List[List[int]]) -> int:
        pop = [0] * 2051
        res = 0
        for l in logs:
            pop[l[0]] += 1
            pop[l[1]] -= 1

        for i in range(1950, 2050):
            pop[i] += pop[i-1]
            if pop[i] > pop[res]:
                res = i

        return res

The function maximumPopulation is designed to find the year with the maximum population given a list of birth and death years. The logs list contains multiple lists where each sublist has two elements: the birth year and the death year.

Here’s how the function works:

  1. The function starts by initializing a list pop of size 2051 with all elements set to 0. This list will be used to store the population count for each year.

  2. Then, it iterates over all the logs. For each log, it increments the population count for the birth year and decrements the population count for the death year in the pop list. This approach is known as a prefix sum or cumulative sum approach, where the value at any index i in the pop list represents the net change in population in year i.

  3. After this, it iterates over the years from 1950 to 2049. For each year i, it updates the population count for that year by adding the population count for the previous year. This effectively calculates the cumulative population count for each year.

  4. During this process, it also keeps track of the year with the maximum population count. If the population count for year i is greater than the population count for the current maximum year, it updates the maximum year to i.

  5. Finally, it returns the year with the maximum population count.

This function effectively uses the prefix sum approach to calculate the cumulative population count for each year in a single pass, resulting in a time complexity of O(n), where n is the number of logs. The space complexity is O(1) as the size of the pop list is fixed.

Q&A

Do we have any similar line sweep questions on leetcode ?

Yes, there are several problems on LeetCode that utilize the Line Sweep algorithm. Here are a few examples:

  1. “Meeting Rooms II” (LeetCode 253) - Given an array of meeting time intervals, find the minimum number of conference rooms required.

  2. “Merge Intervals” (LeetCode 56) - Given an array of intervals, merge all overlapping intervals.

  3. “Car Pooling” (LeetCode 1094) - You are driving a vehicle that has capacity empty seats initially available for passengers. The vehicle only drives east (i.e., it cannot turn around and drive west). Given a list of trips, trip[i] = [num_passengers, start_location, end_location] contains information about the i-th trip: the number of passengers that must be picked up, and the locations to pick them up and drop them off. The locations are given as the number of kilometers due east from your vehicle’s initial location.

  4. “Skyline Problem” (LeetCode 218) - A city’s skyline is the outer contour of the silhouette formed by all the buildings in that city when viewed from a distance. Given the locations and heights of all the buildings, return the skyline formed by these buildings collectively.

Line Sweep is not a solution but a strategy used in solving these problems. It involves sorting events that have a start and end, then ‘sweeping’ over them from left to right to find useful information.