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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: Maintain Maximum and Minimum Values

We notice that the maximum distance must be the distance between the maximum value in one array and the minimum value in another array. Therefore, we can maintain two variables $\textit{mi}$ and $\textit{mx}$, representing the minimum and maximum values of the arrays we have traversed. Initially, $\textit{mi}$ and $\textit{mx}$ are the first and last elements of the first array, respectively.

Next, we traverse from the second array. For each array, we first calculate the distance between the first element of the current array and $\textit{mx}$, and the distance between the last element of the current array and $\textit{mi}$. Then, we update the maximum distance. At the same time, we update $\textit{mi} = \min(\textit{mi}, \textit{arr}[0])$ and $\textit{mx} = \max(\textit{mx}, \textit{arr}[\textit{size} - 1])$.

After traversing all arrays, we get the maximum distance.

The time complexity is $O(m)$, where $m$ is the number of arrays. The space complexity is $O(1)$.

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 {
    let ans = 0;
    let [mi, mx] = [arrays[0][0], arrays[0].at(-1)!];
    for (let i = 1; i < arrays.length; ++i) {
        const arr = arrays[i];
        const a = Math.abs(arr[0] - mx);
        const b = Math.abs(arr.at(-1)! - mi);
        ans = Math.max(ans, a, b);
        mi = Math.min(mi, arr[0]);
        mx = Math.max(mx, arr.at(-1)!);
    }
    return ans;
}

Rust

impl Solution {
    pub fn max_distance(arrays: Vec<Vec<i32>>) -> i32 {
        let mut ans = 0;
        let mut mi = arrays[0][0];
        let mut mx = arrays[0][arrays[0].len() - 1];

        for i in 1..arrays.len() {
            let arr = &arrays[i];
            let a = (arr[0] - mx).abs();
            let b = (arr[arr.len() - 1] - mi).abs();
            ans = ans.max(a).max(b);

            mi = mi.min(arr[0]);
            mx = mx.max(arr[arr.len() - 1]);
        }

        ans
    }
}

JavaScript

/**
 * @param {number[][]} arrays
 * @return {number}
 */
var maxDistance = function (arrays) {
    let ans = 0;
    let [mi, mx] = [arrays[0][0], arrays[0].at(-1)];
    for (let i = 1; i < arrays.length; ++i) {
        const arr = arrays[i];
        const a = Math.abs(arr[0] - mx);
        const b = Math.abs(arr.at(-1) - mi);
        ans = Math.max(ans, a, b);
        mi = Math.min(mi, arr[0]);
        mx = Math.max(mx, arr.at(-1));
    }
    return ans;
};