#2016

Easy

auto_awesomeArray-driven solution strategy

LeetCode Problem Workspace

Maximum Difference Between Increasing Elements

Find the maximum difference between two increasing elements in an array using a linear scan to track the minimum value efficiently.

array

Problem Statement

Given a 0-indexed integer array nums, identify two indices i and j where 0 <= i < j < nums.length such that nums[i] < nums[j], and find the maximum value of nums[j] - nums[i].

Return the maximum difference found. If no pair exists where a later element exceeds an earlier element, return -1. For example, nums = [7,1,5,4] yields 4, while nums = [9,4,3,2] yields -1.

Examples

Example 1

Input: nums = [7,1,5,4]

Output: 4

The maximum difference occurs with i = 1 and j = 2, nums[j] - nums[i] = 5 - 1 = 4. Note that with i = 1 and j = 0, the difference nums[j] - nums[i] = 7 - 1 = 6, but i > j, so it is not valid.

Example 2

Input: nums = [9,4,3,2]

Output: -1

There is no i and j such that i < j and nums[i] < nums[j].

Example 3

Input: nums = [1,5,2,10]

Output: 9

The maximum difference occurs with i = 0 and j = 3, nums[j] - nums[i] = 10 - 1 = 9.

Constraints

n == nums.length
2 <= n <= 1000
1 <= nums[i] <= 109

Solution Approach

Single-pass minimum tracking

Traverse the array from start to finish, maintaining the smallest value encountered. At each element, calculate the difference with this minimum and update the maximum difference accordingly. This ensures O(n) time complexity without extra space.

Handle edge cases

If the array is strictly decreasing, there will be no valid increasing pair. Track whether any valid difference was found; if none, return -1 to correctly handle these scenarios.

Optimal space usage

Only store the current minimum and maximum difference, avoiding additional arrays or data structures. This approach preserves O(1) space complexity while adhering to the array-driven pattern.

Complexity Analysis

Metric	Value
Time	O(n)
Space	O(1)

The algorithm performs a single traversal of the array, making the time complexity O(n). Only two variables are tracked throughout, resulting in O(1) space usage.

What Interviewers Usually Probe

Looks for a linear scan approach rather than nested loops.
Checks if candidate tracks the minimum element efficiently.
Tests understanding of handling arrays with no increasing pairs.

Common Pitfalls or Variants

Common pitfalls

Using nested loops which increases time complexity to O(n^2).
Failing to return -1 when no valid increasing pair exists.
Incorrectly updating minimum after computing difference, missing optimal solution.

Follow-up variants

Find maximum ratio nums[j] / nums[i] with i < j and nums[i] < nums[j].
Determine maximum difference where elements must be k indices apart.
Compute maximum difference in a circular array considering wrap-around.

FAQ

What is the core strategy to solve Maximum Difference Between Increasing Elements?

Track the minimum element while iterating and compute the difference with each current element, updating the maximum difference as you go.

What should I return if the array has no increasing elements?

Return -1, as there is no valid pair satisfying nums[i] < nums[j].

Can this problem be solved with nested loops?

Yes, but nested loops increase time complexity to O(n^2) and are inefficient for larger arrays.

What is the time complexity using the optimal approach?

O(n), because the array is scanned once while maintaining the minimum element and maximum difference.

Does the array-driven solution pattern apply to other difference problems?

Yes, tracking running minima or maxima while iterating is a common pattern for array difference calculations.

terminal

Solution

Solution 1: Maintaining Prefix Minimum

We use a variable $\textit{mi}$ to represent the minimum value among the elements currently being traversed, and a variable $\textit{ans}$ to represent the maximum difference. Initially, $\textit{mi}$ is set to $+\infty$, and $\textit{ans}$ is set to $-1$.

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class Solution:
    def maximumDifference(self, nums: List[int]) -> int:
        mi = inf
        ans = -1
        for x in nums:
            if x > mi:
                ans = max(ans, x - mi)
            else:
                mi = x
        return ans

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