LeetCode #1855 — MEDIUM

Maximum Distance Between a Pair of Values

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Two Pointers · Binary Search

Example 1

[55,30,5,4,2]
[100,20,10,10,5]

Example 2

[2,2,2]
[10,10,1]

Example 3

[30,29,19,5]
[25,25,25,25,25]

Core Insight

What unlocks the optimal approach

Since both arrays are sorted in a non-increasing way this means that for each value in the first array. We can find the farthest value smaller than it using binary search.
There is another solution using a two pointers approach since the first array is non-increasing the farthest j such that nums2[j] ≥ nums1[i] is at least as far as the farthest j such that nums2[j] ≥ nums1[i-1]

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1855: Maximum Distance Between a Pair of Values
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;
    }
}

// Accepted solution for LeetCode #1855: Maximum Distance Between a Pair of Values
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
}

# Accepted solution for LeetCode #1855: Maximum Distance Between a Pair of Values
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

// Accepted solution for LeetCode #1855: Maximum Distance Between a Pair of Values
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
    }
}

// Accepted solution for LeetCode #1855: Maximum Distance Between a Pair of Values
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.