vault backup: 2022-06-27 11:19:16
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@ -17,6 +17,7 @@ tag:#coding_problem tag:#Kadane_s_algorithm -tag:#template_remove_me
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- [[Leetcode Best-Time-To-Buy-And-Sell-Stock]]
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- [[Leetcode Best-Time-To-Buy-And-Sell-Stock]]
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- [[Leetcode Maximum-Difference-Between-Increasing-Elements]]
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- [[Leetcode Maxinum-subarray]]
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- [[Leetcode Maxinum-subarray]]
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@ -0,0 +1,96 @@
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# Leetcode Maximum-Difference-Between-Increasing-Elements
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#### 2022-06-27 11:09
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> ##### Algorithms:
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> #algorithm #Kadane_s_algorithm
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> ##### Difficulty:
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> #coding_problem #difficulty-easy
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> ##### Additional tags:
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> #leetcode
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> ##### Revisions:
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> Initial encounter: 2022-06-27
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##### Related topics:
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```expander
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tag:#Kadane_s_algorithm
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```
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- [[Kadane's Algorithm]]
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- [[Leetcode Best-Time-To-Buy-And-Sell-Stock]]
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- [[Leetcode Maxinum-subarray]]
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##### Links:
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- [Link to problem](https://leetcode.com/problems/maximum-difference-between-increasing-elements/)
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___
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### Problem
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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]`.
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Return _the **maximum difference**._ If no such `i` and `j` exists, return `-1`.
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#### Examples
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Example 1:
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```
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Input: nums = [7,1,5,4]
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Output: 4
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Explanation:
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The maximum difference occurs with i = 1 and j = 2, nums[j] - nums[i] = 5 - 1 = 4.
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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.
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```
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Example 2:
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```
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Input: nums = [9,4,3,2]
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Output: -1
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Explanation:
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There is no i and j such that i < j and nums[i] < nums[j].
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```
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Example 3:
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```
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Input: nums = [1,5,2,10]
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Output: 9
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Explanation:
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The maximum difference occurs with i = 0 and j = 3, nums[j] - nums[i] = 10 - 1 = 9.
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```
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#### Constraints
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- n == nums.length
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- 2 <= n <= 1000
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- 1 <= nums[i] <= 109
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### Thoughts
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Since 0 <= i < j <=n, this can be completed using kadane's algo in one pass.
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> [!summary]
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> This is a #Kadane_s_algorithm
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### Solution
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```cpp
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class Solution {
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public:
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int maximumDifference(vector<int>& nums) {
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// Kadane's algorithm, since 0 <= i < j < n, and it can be done using one loop.
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int minNum = nums[0];
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int maxNum = 0;
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for (int i : nums) {
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minNum = min(minNum, i);
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maxNum = max(maxNum, i - minNum);
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}
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if (maxNum == 0) {
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return -1;
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} else {
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return maxNum;
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}
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}
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};
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```
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