notes/OJ notes/pages/Leetcode Maximum-Difference-Between-Increasing-Elements.md

Leetcode Maximum-Difference-Between-Increasing-Elements

2022-06-27 11:09

Algorithms:

#algorithm #Kadane_s_algorithm

Difficulty:

#coding_problem #difficulty-easy

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

tag:#Kadane_s_algorithm 

• Kadane's Algorithm
• Leetcode Best-Time-To-Buy-And-Sell-Stock
• Leetcode Maxinum-subarray

Links:

• Link to problem

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Problem

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.

Examples

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

Thoughts

Since 0 <= i < j <=n, this can be completed using kadane's algo in one pass.

[!summary] This is a #Kadane_s_algorithm

Solution

class Solution {
public:
  int maximumDifference(vector<int>& nums) {
    // Kadane's algorithm, since 0 <= i < j < n, and it can be done using one loop.
    int minNum = nums[0];
    int maxNum = 0;
    
    for (int i : nums) {
      minNum = min(minNum, i);
      maxNum = max(maxNum, i - minNum);
    }
    
    if (maxNum == 0) {
      return -1;
    } else {
      return maxNum;
    }
  }
};