2016. Maximum Difference Between Increasing Elements

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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 = 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

Solutions

Solution 1: Maintaining Prefix Minimum

We use a variable $\textit{mi}$ to represent the minimum value among the elements currently being traversed, and a variable $\textit{ans}$ to represent the maximum difference. Initially, $\textit{mi}$ is set to $+\infty$, and $\textit{ans}$ is set to $-1$.

Traverse the array. For the current element $x$, if $x \gt \textit{mi}$, update $\textit{ans}$ to $\max(\textit{ans}, x - \textit{mi})$. Otherwise, update $\textit{mi}$ to $x$.

After the traversal, return $\textit{ans}$.

Time complexity is $O(n)$, where $n$ is the length of the array. Space complexity is $O(1)$.

Python3

class Solution:
    def maximumDifference(self, nums: List[int]) -> int:
        mi = inf
        ans = -1
        for x in nums:
            if x > mi:
                ans = max(ans, x - mi)
            else:
                mi = x
        return ans

Java

class Solution {
    public int maximumDifference(int[] nums) {
        int mi = 1 << 30;
        int ans = -1;
        for (int x : nums) {
            if (x > mi) {
                ans = Math.max(ans, x - mi);
            } else {
                mi = x;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int maximumDifference(vector<int>& nums) {
        int mi = 1 << 30;
        int ans = -1;
        for (int& x : nums) {
            if (x > mi) {
                ans = max(ans, x - mi);
            } else {
                mi = x;
            }
        }
        return ans;
    }
};

Go

func maximumDifference(nums []int) int {
	mi := 1 << 30
	ans := -1
	for _, x := range nums {
		if mi < x {
			ans = max(ans, x-mi)
		} else {
			mi = x
		}
	}
	return ans
}

TypeScript

function maximumDifference(nums: number[]): number {
    let [ans, mi] = [-1, Infinity];
    for (const x of nums) {
        if (x > mi) {
            ans = Math.max(ans, x - mi);
        } else {
            mi = x;
        }
    }
    return ans;
}

Rust

impl Solution {
    pub fn maximum_difference(nums: Vec<i32>) -> i32 {
        let mut mi = i32::MAX;
        let mut ans = -1;

        for &x in &nums {
            if x > mi {
                ans = ans.max(x - mi);
            } else {
                mi = x;
            }
        }

        ans
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
var maximumDifference = function (nums) {
    let [ans, mi] = [-1, Infinity];
    for (const x of nums) {
        if (x > mi) {
            ans = Math.max(ans, x - mi);
        } else {
            mi = x;
        }
    }
    return ans;
};