# 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 the variable $mi$ to represent the minimum value of the elements we have traversed so far, and the variable $ans$ to represent the maximum difference. Initially, $mi$ is set to $+\infty$, and $ans$ is set to $-1$.

We traverse the array. For the current element $x$, if $x > mi$, then we update $ans$ to be $max(ans, x - mi)$, otherwise we update $mi = x$.

After the traversal, we return $ans$.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

• class Solution {
public int maximumDifference(int[] nums) {
int mi = 1 << 30;
int ans = -1;
for (int x : nums) {
if (x > mi) {
ans = Math.max(ans, x - mi);
} else {
mi = x;
}
}
return ans;
}
}

• class Solution {
public:
int maximumDifference(vector<int>& nums) {
int mi = 1 << 30;
int ans = -1;
for (int& x : nums) {
if (x > mi) {
ans = max(ans, x - mi);
} else {
mi = x;
}
}
return ans;
}
};

• 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


• func maximumDifference(nums []int) int {
mi := 1 << 30
ans := -1
for _, x := range nums {
if mi < x {
ans = max(ans, x-mi)
} else {
mi = x
}
}
return ans
}

• function maximumDifference(nums: number[]): number {
const n = nums.length;
let min = nums[0];
let res = -1;
for (let i = 1; i < n; i++) {
res = Math.max(res, nums[i] - min);
min = Math.min(min, nums[i]);
}
return res === 0 ? -1 : res;
}


• /**
* @param {number[]} nums
* @return {number}
*/
var maximumDifference = function (nums) {
let mi = 1 << 30;
let ans = -1;
for (const x of nums) {
if (mi < x) {
ans = Math.max(ans, x - mi);
} else {
mi = x;
}
}
return ans;
};


• impl Solution {
pub fn maximum_difference(nums: Vec<i32>) -> i32 {
let mut min = nums[0];
let mut res = -1;
for i in 1..nums.len() {
res = res.max(nums[i] - min);
min = min.min(nums[i]);
}
match res {
0 => -1,
_ => res,
}
}
}