LeetCode in Kotlin
2016. Maximum Difference Between Increasing Elements
Easy
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.length2 <= n <= 10001 <= nums[i] <= 109
Solution
class Solution {
fun maximumDifference(nums: IntArray): Int {
var mini = nums[0]
var ans = -1
for (i in 0 until nums.size - 1) {
if (nums[i] < mini) {
mini = nums[i]
}
if (nums[i + 1] - mini > ans) {
ans = nums[i + 1] - mini
}
}
return if (ans <= 0) -1 else ans
}
}