Skip to content

Latest commit

 

History

History
230 lines (182 loc) · 5.41 KB

README_EN.md

File metadata and controls

230 lines (182 loc) · 5.41 KB
comments difficulty edit_url tags
true
Medium
Greedy
Array

624. Maximum Distance in Arrays

中文文档

Description

You are given m arrays, where each array is sorted in ascending order.

You can pick up two integers from two different arrays (each array picks one) and calculate the distance. We define the distance between two integers a and b to be their absolute difference |a - b|.

Return the maximum distance.

 

Example 1:

Input: arrays = [[1,2,3],[4,5],[1,2,3]]
Output: 4
Explanation: One way to reach the maximum distance 4 is to pick 1 in the first or third array and pick 5 in the second array.

Example 2:

Input: arrays = [[1],[1]]
Output: 0

 

Constraints:

  • m == arrays.length
  • 2 <= m <= 105
  • 1 <= arrays[i].length <= 500
  • -104 <= arrays[i][j] <= 104
  • arrays[i] is sorted in ascending order.
  • There will be at most 105 integers in all the arrays.

Solutions

Solution 1

Python3

class Solution:
    def maxDistance(self, arrays: List[List[int]]) -> int:
        ans = 0
        mi, mx = arrays[0][0], arrays[0][-1]
        for arr in arrays[1:]:
            a, b = abs(arr[0] - mx), abs(arr[-1] - mi)
            ans = max(ans, a, b)
            mi = min(mi, arr[0])
            mx = max(mx, arr[-1])
        return ans

Java

class Solution {
    public int maxDistance(List<List<Integer>> arrays) {
        int ans = 0;
        int mi = arrays.get(0).get(0);
        int mx = arrays.get(0).get(arrays.get(0).size() - 1);
        for (int i = 1; i < arrays.size(); ++i) {
            var arr = arrays.get(i);
            int a = Math.abs(arr.get(0) - mx);
            int b = Math.abs(arr.get(arr.size() - 1) - mi);
            ans = Math.max(ans, Math.max(a, b));
            mi = Math.min(mi, arr.get(0));
            mx = Math.max(mx, arr.get(arr.size() - 1));
        }
        return ans;
    }
}

C++

class Solution {
public:
    int maxDistance(vector<vector<int>>& arrays) {
        int ans = 0;
        int mi = arrays[0][0], mx = arrays[0][arrays[0].size() - 1];
        for (int i = 1; i < arrays.size(); ++i) {
            auto& arr = arrays[i];
            int a = abs(arr[0] - mx), b = abs(arr[arr.size() - 1] - mi);
            ans = max({ans, a, b});
            mi = min(mi, arr[0]);
            mx = max(mx, arr[arr.size() - 1]);
        }
        return ans;
    }
};

Go

func maxDistance(arrays [][]int) (ans int) {
	mi, mx := arrays[0][0], arrays[0][len(arrays[0])-1]
	for _, arr := range arrays[1:] {
		a, b := abs(arr[0]-mx), abs(arr[len(arr)-1]-mi)
		ans = max(ans, max(a, b))
		mi = min(mi, arr[0])
		mx = max(mx, arr[len(arr)-1])
	}
	return ans
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

TypeScript

function maxDistance(arrays: number[][]): number {
    const n = arrays.length;
    let res = 0;
    let [min, max] = [Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY];

    for (let i = 0; i < n; i++) {
        const a = arrays[i];
        res = Math.max(Math.max(a.at(-1)! - min, max - a[0]), res);
        min = Math.min(min, a[0]);
        max = Math.max(max, a.at(-1)!);
    }

    return res;
}

JavaScript

/**
 * @param {number[][]} arrays
 * @return {number}
 */
var maxDistance = function (arrays) {
    const n = arrays.length;
    let res = 0;
    let [min, max] = [Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY];

    for (let i = 0; i < n; i++) {
        const a = arrays[i];
        res = Math.max(Math.max(a.at(-1) - min, max - a[0]), res);
        min = Math.min(min, a[0]);
        max = Math.max(max, a.at(-1));
    }

    return res;
};

Solution 2: One-line solution

TypeScript

const maxDistance = (arrays: number[][]): number =>
    arrays.reduce(
        ([res, min, max], a) => [
            Math.max(Math.max(a.at(-1)! - min, max - a[0]), res),
            Math.min(min, a[0]),
            Math.max(max, a.at(-1)!),
        ],
        [0, Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY],
    )[0];

JavaScript

/**
 * @param {number[][]} arrays
 * @return {number}
 */
var maxDistance = arrays =>
    arrays.reduce(
        ([res, min, max], a) => [
            Math.max(Math.max(a.at(-1) - min, max - a[0]), res),
            Math.min(min, a[0]),
            Math.max(max, a.at(-1)),
        ],
        [0, Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY],
    )[0];