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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 $O(1)$.

• Java
• C++
• Python
• Go
• TypeScript
• Javascript
• RenderScript

• 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;
      }
  };
  

• 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
  
  

• 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[], nums2: number[]): number {
      let ans = 0;
      let m = nums1.length;
      let n = nums2.length;
      for (let i = 0; i < m; ++i) {
          let left = i;
          let right = n - 1;
          while (left < right) {
              const mid = (left + right + 1) >> 1;
              if (nums2[mid] >= nums1[i]) {
                  left = mid;
              } else {
                  right = mid - 1;
              }
          }
          ans = Math.max(ans, left - i);
      }
      return ans;
  }
  
  

• /**
   * @param {number[]} nums1
   * @param {number[]} nums2
   * @return {number}
   */
  var maxDistance = function (nums1, nums2) {
      let ans = 0;
      let m = nums1.length;
      let n = nums2.length;
      for (let i = 0; i < m; ++i) {
          let left = i;
          let right = n - 1;
          while (left < right) {
              const mid = (left + right + 1) >> 1;
              if (nums2[mid] >= nums1[i]) {
                  left = mid;
              } else {
                  right = mid - 1;
              }
          }
          ans = Math.max(ans, left - i);
      }
      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;
          for i in 0..m {
              let mut left = i;
              let mut right = n;
              while left < right {
                  let mid = left + (right - left) / 2;
                  if nums2[mid] >= nums1[i] {
                      left = mid + 1;
                  } else {
                      right = mid;
                  }
              }
              res = res.max((left - i - 1) as i32);
          }
          res
      }
  }
  
  

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