1855. Maximum Distance Between a Pair of Values

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).

 

Constraints:

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

Solutions

Solution 1: Binary Search

Assume the lengths of $nums1$ and $nums2$ are $m$ and $n$ respectively.

Traverse array $nums1$, for each number $nums1[i]$, perform a binary search for numbers in $nums2$ in the range $[i,n)$, find the last position $j$ that is greater than or equal to $nums1[i]$, calculate the distance between this position and $i$, and update the maximum distance value $ans$.

The time complexity is $O(m \times \log n)$, where $m$ and $n$ are the lengths of $nums1$ and $nums2$ respectively. The space complexity is $O(1)$.

Python3JavaC++GoTypeScriptRustJavaScript

class Solution:
    def maxDistance(self, nums1: List[int], nums2: List[int]) -> int:
        ans = 0
        nums2 = nums2[::-1]
        for i, v in enumerate(nums1):
            j = len(nums2) - bisect_left(nums2, v) - 1
            ans = max(ans, j - i)
        return ans

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

class Solution {
public:
    int maxDistance(vector<int>& nums1, vector<int>& nums2) {
        int ans = 0;
        reverse(nums2.begin(), nums2.end());
        for (int i = 0; i < nums1.size(); ++i) {
            int j = nums2.size() - (lower_bound(nums2.begin(), nums2.end(), nums1[i]) - nums2.begin()) - 1;
            ans = max(ans, j - i);
        }
        return ans;
    }
};

func maxDistance(nums1 []int, nums2 []int) int {
    ans, n := 0, len(nums2)
    for i, num := range nums1 {
        left, right := i, n-1
        for left < right {
            mid := (left + right + 1) >> 1
            if nums2[mid] >= num {
                left = mid
            } else {
                right = mid - 1
            }
        }
        if ans < left-i {
            ans = left - i
        }
    }
    return ans
}

function maxDistance(nums1: number