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1855. Maximum Distance Between a Pair of Values

中文文档

Description

You are given two non-increasing 0-indexed integer arrays nums1​​​​​​ and nums2​​​​​​.

A pair of indices (i, j), where 0 <= i < nums1.length and 0 <= j < nums2.length, is valid if both i <= j and nums1[i] <= nums2[j]. The distance of the pair is j - i​​​​.

Return the maximum distance of any valid pair (i, j). If there are no valid pairs, return 0.

An array arr is non-increasing if arr[i-1] >= arr[i] for every 1 <= i < arr.length.

 

Example 1:

Input: nums1 = [55,30,5,4,2], nums2 = [100,20,10,10,5]
Output: 2
Explanation: The valid pairs are (0,0), (2,2), (2,3), (2,4), (3,3), (3,4), and (4,4).
The maximum distance is 2 with pair (2,4).

Example 2:

Input: nums1 = [2,2,2], nums2 = [10,10,1]
Output: 1
Explanation: The valid pairs are (0,0), (0,1), and (1,1).
The maximum distance is 1 with pair (0,1).

Example 3:

Input: nums1 = [30,29,19,5], nums2 = [25,25,25,25,25]
Output: 2
Explanation: The valid pairs are (2,2), (2,3), (2,4), (3,3), and (3,4).
The maximum distance is 2 with pair (2,4).

Example 4:

Input: nums1 = [5,4], nums2 = [3,2]
Output: 0
Explanation: There are no valid pairs, so return 0.

 

Constraints:

  • 1 <= nums1.length <= 105
  • 1 <= nums2.length <= 105
  • 1 <= nums1[i], nums2[j] <= 105
  • Both nums1 and nums2 are non-increasing.

Solutions

Python3

class Solution:
    def maxDistance(self, nums1: List[int], nums2: List[int]) -> int:
        res = 0
        for i in range(len(nums1)):
            l, r = i, len(nums2) - 1
            while l <= r:
                mid = (l + r) >> 1
                if nums2[mid] >= nums1[i]:
                    res = max(res, mid - i)
                    l = mid + 1
                else:
                    r = mid - 1
        return res

Java

class Solution {
    public int maxDistance(int[] nums1, int[] nums2) {
        int res = 0;
        for (int i = 0; i < nums1.length; ++i) {
            int l = i, r = nums2.length - 1;
            while (l <= r) {
                int mid = (l + r) >>> 1;
                if (nums2[mid] >= nums1[i]) {
                    res = Math.max(res, mid - i);
                    l = mid + 1;
                } else {
                    r = mid - 1;
                }
            }
        }
        return res;
    }
}

C++

class Solution {
public:
    int maxDistance(vector<int>& nums1, vector<int>& nums2) {
        int res = 0;
        for (int i = 0; i < nums1.size(); ++i) {
            int l = i, r = nums2.size() - 1;
            while (l <= r) {
                int mid = (l + r) >> 1;
                if (nums2[mid] >= nums1[i]) {
                    res = max(res, mid - i);
                    l = mid + 1;
                } else {
                    r = mid - 1;
                }
            }
        }
        return res;
    }
};

JavaScript

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */
var maxDistance = function(nums1, nums2) {
    let res = 0;
    for (let i = 0; i < nums1.length; i++) {
        let left = 0, right = nums2.length - 1;
        while (left <= right) {
            mid = (left + right) >> 1;
            if (nums2[mid] >= nums1[i]) {
                res = Math.max(res, mid - i);
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
    }
    return res;
};

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