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Leetcode 2016. Maximum Difference Between Increasing Elements

Problem Statement:

Given a 0-indexed integer array nums of size n, you need to 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:

Input: nums = [7,1,5,4]
Output: 4
Explanation:
The maximum difference occurs with i = 1 and j = 2 (nums[2] - nums[1] = 5 - 1 = 4).

Clarifying Questions:

  1. What is the range of the size of the array nums?
    • The size n will be within the range from 1 to 10^3.
  2. What is the range of the elements in the array nums?
    • The elements in the array will be within the range from 1 to 10^9.
  3. Are there any constraints regarding possible duplicates in the array?
    • No, there are no special constraints mentioned, duplicates are allowed but may affect the difference.

Strategy:

A straightforward approach to solve this problem is to use a single pass scan while keeping track of the minimum element seen so far. At every position j, the difference nums[j] - min_value is computed, and the maximum difference is kept track of. Here is a more detailed algorithm:

  1. Initialize min_value with the first element of the array.
  2. Initialize max_diff as -1 (since we need to return -1 if no such i and j exists).
  3. Iterate through each element of the array starting from the second element.
    • For each element, calculate the difference with min_value.
    • Update max_diff if the difference is larger than the current max_diff.
    • Update min_value if the current element is smaller than min_value.
  4. Return max_diff.

Code in C++:

#include <vector>
#include <algorithm>
#include <iostream>

class Solution {
public:
    int maximumDifference(std::vector<int>& nums) {
        int min_value = nums[0];
        int max_diff = -1;

        for (int j = 1; j < nums.size(); ++j) {
            if (nums[j] > min_value) {
                max_diff = std::max(max_diff, nums[j] - min_value);
            } else {
                min_value = std::min(min_value, nums[j]);
            }
        }
        
        return max_diff;
    }
};

int main() {
    Solution sol;
    std::vector<int> nums = {7, 1, 5, 4};
    std::cout << "Maximum Difference: " << sol.maximumDifference(nums) << std::endl;
    return 0;
}

Time Complexity:

The time complexity of this solution is O(n), where n is the length of the input array. This is because we traverse the array only once.

The space complexity is O(1) since we use a fixed amount of extra space to store min_value and max_diff.

By maintaining these values during a single pass over the array, we ensure a performant solution that meets the problem constraints efficiently.

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