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624. Maximum Distance in Arrays

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

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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/**
 * @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

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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];

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/**
 * @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];

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