2016_maximum_difference_between_increasing_elements.md

Maximum Difference Between Increasing Elements

Problem Number: 2016 Difficulty: Easy Category: Array, Greedy LeetCode Link: https://leetcode.com/problems/maximum-difference-between-increasing-elements/

Problem 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 = 3, the difference nums[j] - nums[i] = 4 - 1 = 3, but it is not the maximum difference.

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] <= 10^9

My Approach

I used a Brute Force approach to find the maximum difference. The key insight is to check all possible pairs (i, j) where i < j and nums[i] < nums[j].

Algorithm:

1. Initialize maximum difference to -1
2. For each index i from 0 to n-2:
   • For each index j from i+1 to n-1:
     • If nums[i] < nums[j], calculate difference
     • Update maximum difference if current difference is larger
3. Return the maximum difference found

Solution

The solution uses brute force to check all valid pairs. See the implementation in the solution file.

Key Points:

• Checks all possible pairs (i, j) where i < j
• Only considers pairs where nums[i] < nums[j]
• Tracks maximum difference found
• Returns -1 if no valid pair exists

Time & Space Complexity

Time Complexity: O(n²)

• Nested loops: outer loop O(n), inner loop O(n)
• Each iteration performs constant time operations
• Total: O(n²)

Space Complexity: O(1)

• Uses only a constant amount of extra space
• No additional data structures needed

Key Insights

1. Brute Force: Checking all possible pairs is straightforward for this problem.

2. Valid Pairs: Only consider pairs where i < j and nums[i] < nums[j].

3. Maximum Tracking: Keep track of the maximum difference found so far.

4. Return Value: Return -1 if no valid pair exists.

5. 0-Indexed: The problem uses 0-indexed arrays.

6. Simple Logic: The solution logic is straightforward and easy to understand.

Mistakes Made

1. Wrong Order: Initially might check pairs where i > j.

2. Missing Condition: Not checking nums[i] < nums[j] condition.

3. Complex Logic: Overcomplicating the pair checking logic.

4. Wrong Return: Not returning -1 when no valid pair exists.

Related Problems

• Best Time to Buy and Sell Stock (Problem 121): Find maximum profit
• Best Time to Buy and Sell Stock II (Problem 122): Multiple transactions
• Container With Most Water (Problem 11): Find maximum area
• Trapping Rain Water (Problem 42): Calculate trapped water

Alternative Approaches

1. Single Pass: Track minimum element and calculate differences - O(n) time
2. Two Pointers: Use two pointers for optimization - O(n) time
3. Dynamic Programming: Use DP to track maximum differences - O(n) time

Common Pitfalls

1. Wrong Order: Checking pairs where i > j.
2. Missing Condition: Not verifying nums[i] < nums[j].
3. Complex Logic: Overcomplicating the pair checking.
4. Wrong Return: Not handling the case where no valid pair exists.
5. Inefficient Approach: Using O(n²) when O(n) is possible.

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Note: This is a simple array problem that demonstrates brute force approach for finding maximum differences.