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Formatted question description: https://leetcode.ca/all/1855.html

1855. Maximum Distance Between a Pair of Values

Level

Medium

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 <= 10^5
  • 1 <= nums2.length <= 10^5
  • 1 <= nums1[i], nums2[j] <= 10^5
  • Both nums1 and nums2 are non-increasing.

Solution

Use two pointers. Loop over index2 from nums2.length - 1 to 0. For each index2, find the smallest index1 such that nums1[index1] <= nums2[index2] and calculate index2 - index1 as the distance. Since the distance is at least 0, if index1 > index2, the distance is negative and will never exceed 0. Only the non-negative instances are stored to maintain the maximum distance.

  • class Solution {
    public int maxDistance(int[] nums1, int[] nums2) {
        int maxDistance = 0;
        int length1 = nums1.length, length2 = nums2.length;
        int index1 = length1 - 1, index2 = length2 - 1;
        while (index1 >= 0 && index2 >= 0) {
            while (index1 >= 0 && nums1[index1] <= nums2[index2]) {
                maxDistance = Math.max(maxDistance, index2 - index1);
                index1--;
            }
            index2--;
        }
        return maxDistance;
    }
}

############

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

  • // OJ: https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int maxDistance(vector<int>& A, vector<int>& B) {
        int ans = 0;
        for (int i = 0; i < A.size(); ++i) {
            int j = upper_bound(begin(B), end(B), A[i], greater()) - begin(B);
            ans = max(ans, max(0, j - 1 - i));
        }
        return ans;
    }
};

  • class Solution:
    def maxDistance(self, nums1: List[int], nums2: List[int]) -> int:
        m, n = len(nums1), len(nums2)
        ans = i = j = 0
        while i < m:
            while j < n and nums1[i] <= nums2[j]:
                j += 1
            ans = max(ans, j - i - 1)
            i += 1
        return ans

############

# 1855. Maximum Distance Between a Pair of Values
# https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/

class Solution:
    def maxDistance(self, nums1: List[int], nums2: List[int]) -> int:
    
        n1, n2 = len(nums1), len(nums2)
        res = 0
        i = j = 0
        
        while i < n1 or j < n2:
            if nums1[i] <= nums2[j]:
                res = max(res, j - i)
                if j < n2 - 1:
                    j += 1
                elif i < n1 - 1:
                    i += 1
                else:
                    break
            else:
                if i < n1 - 1:
                    i += 1
                elif j < n2 - 1:
                    j += 1
                else:
                    break
        
        return res


  • func maxDistance(nums1 []int, nums2 []int) int {
	m, n := len(nums1), len(nums2)
	ans := 0
	for i, j := 0, 0; i < m; i++ {
		for j < n && nums1[i] <= nums2[j] {
			j++
		}
		if ans < j-i-1 {
			ans = j - i - 1
		}
	}
	return ans
}

  • function maxDistance(nums1: number[], nums2: number[]): number {
    let ans = 0;
    const m = nums1.length;
    const n = nums2.length;
    for (let i = 0, j = 0; i < m; ++i) {
        while (j < n && nums1[i] <= nums2[j]) {
            j++;
        }
        ans = Math.max(ans, j - i - 1);
    }
    return ans;
}


  • /**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */
var maxDistance = function (nums1, nums2) {
    let ans = 0;
    const m = nums1.length;
    const n = nums2.length;
    for (let i = 0, j = 0; i < m; ++i) {
        while (j < n && nums1[i] <= nums2[j]) {
            j++;
        }
        ans = Math.max(ans, j - i - 1);
    }
    return ans;
};


  • impl Solution {
    pub fn max_distance(nums1: Vec<i32>, nums2: Vec<i32>) -> i32 {
        let m = nums1.len();
        let n = nums2.len();
        let mut res = 0;
        let mut j = 0;
        for i in 0..m {
            while j < n && nums1[i] <= nums2[j] {
                j += 1
            }
            res = res.max((j - i - 1) as i32)
        }
        res
    }
}

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