2016. Maximum Difference Between Increasing Elements
Description
Given a 0-indexed integer array nums
of size n
, find the maximum difference between nums[i]
and nums[j]
(i.e., nums[j] - nums[i]
), such that 0 <= i < j < n
and nums[i] < nums[j]
.
Return the maximum difference. If no such i
and j
exists, return -1
.
Example 1:
Input: nums = [7,1,5,4] Output: 4 Explanation: 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 Explanation: 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 Explanation: 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
Solutions
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 $-1$.
Traverse the array. For the current element $x$, if $x \gt \textit{mi}$, update $\textit{ans}$ to $\max(\textit{ans}, x - \textit{mi})$. Otherwise, update $\textit{mi}$ to $x$.
After the traversal, return $\textit{ans}$.
Time complexity is $O(n)$, where $n$ is the length of the array. Space complexity is $O(1)$.
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