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1854. Maximum Population Year

Description

You are given a 2D integer array logs where each logs[i] = [birthi, deathi] indicates the birth and death years of the ith person.

The population of some year x is the number of people alive during that year. The ith person is counted in year x's population if x is in the inclusive range [birthi, deathi - 1]. Note that the person is not counted in the year that they die.

Return the earliest year with the maximum population.

 

Example 1:

Input: logs = [[1993,1999],[2000,2010]]
Output: 1993
Explanation: The maximum population is 1, and 1993 is the earliest year with this population.

Example 2:

Input: logs = [[1950,1961],[1960,1971],[1970,1981]]
Output: 1960
Explanation: 
The maximum population is 2, and it had happened in years 1960 and 1970.
The earlier year between them is 1960.

 

Constraints:

  • 1 <= logs.length <= 100
  • 1950 <= birthi < deathi <= 2050

Solutions

Solution 1: Difference Array

We notice that the range of years is $[1950,..2050]$. Therefore, we can map these years to an array $d$ of length $101$, where the index of the array represents the value of the year minus $1950$.

Next, we traverse $logs$. For each person, we increment $d[birth_i - 1950]$ by $1$ and decrement $d[death_i - 1950]$ by $1$. Finally, we traverse the array $d$, find the maximum value of the prefix sum, which is the year with the most population, and add $1950$ to get the answer.

The time complexity is $O(n)$, and the space complexity is $O(C)$. Where $n$ is the length of the array $logs$, and $C$ is the range size of the years, i.e., $2050 - 1950 + 1 = 101$.

  • class Solution {
    public int maximumPopulation(int[][] logs) {
        int[] d = new int[101];
        final int offset = 1950;
        for (var log : logs) {
            int a = log[0] - offset;
            int b = log[1] - offset;
            ++d[a];
            --d[b];
        }
        int s = 0, mx = 0;
        int j = 0;
        for (int i = 0; i < d.length; ++i) {
            s += d[i];
            if (mx < s) {
                mx = s;
                j = i;
            }
        }
        return j + offset;
    }
}

  • class Solution {
public:
    int maximumPopulation(vector<vector<int>>& logs) {
        int d[101]{};
        const int offset = 1950;
        for (auto& log : logs) {
            int a = log[0] - offset;
            int b = log[1] - offset;
            ++d[a];
            --d[b];
        }
        int s = 0, mx = 0;
        int j = 0;
        for (int i = 0; i < 101; ++i) {
            s += d[i];
            if (mx < s) {
                mx = s;
                j = i;
            }
        }
        return j + offset;
    }
};

  • class Solution:
    def maximumPopulation(self, logs: List[List[int]]) -> int:
        d = [0] * 101
        offset = 1950
        for a, b in logs:
            a, b = a - offset, b - offset
            d[a] += 1
            d[b] -= 1
        s = mx = j = 0
        for i, x in enumerate(d):
            s += x
            if mx < s:
                mx, j = s, i
        return j + offset


  • func maximumPopulation(logs [][]int) int {
	d := [101]int{}
	offset := 1950
	for _, log := range logs {
		a, b := log[0]-offset, log[1]-offset
		d[a]++
		d[b]--
	}
	var s, mx, j int
	for i, x := range d {
		s += x
		if mx < s {
			mx = s
			j = i
		}
	}
	return j + offset
}

  • function maximumPopulation(logs: number[][]): number {
    const d: number[] = new Array(101).fill(0);
    const offset = 1950;
    for (const [birth, death] of logs) {
        d[birth - offset]++;
        d[death - offset]--;
    }
    let j = 0;
    for (let i = 0, s = 0, mx = 0; i < d.length; ++i) {
        s += d[i];
        if (mx < s) {
            mx = s;
            j = i;
        }
    }
    return j + offset;
}


  • /**
 * @param {number[][]} logs
 * @return {number}
 */
var maximumPopulation = function (logs) {
    const d = new Array(101).fill(0);
    const offset = 1950;
    for (let [a, b] of logs) {
        a -= offset;
        b -= offset;
        d[a]++;
        d[b]--;
    }
    let j = 0;
    for (let i = 0, s = 0, mx = 0; i < 101; ++i) {
        s += d[i];
        if (mx < s) {
            mx = s;
            j = i;
        }
    }
    return j + offset;
};

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