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2016. Maximum Difference Between Increasing Elements

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 the variable $mi$ to represent the minimum value of the elements we have traversed so far, and the variable $ans$ to represent the maximum difference. Initially, $mi$ is set to $+\infty$, and $ans$ is set to $-1$.

We traverse the array. For the current element $x$, if $x > mi$, then we update $ans$ to be $max(ans, x - mi)$, otherwise we update $mi = x$.

After the traversal, we return $ans$.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

  • 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;
    }
}

  • 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;
    }
};

  • 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


  • 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
}

  • function maximumDifference(nums: number[]): number {
    const n = nums.length;
    let min = nums[0];
    let res = -1;
    for (let i = 1; i < n; i++) {
        res = Math.max(res, nums[i] - min);
        min = Math.min(min, nums[i]);
    }
    return res === 0 ? -1 : res;
}


  • /**
 * @param {number[]} nums
 * @return {number}
 */
var maximumDifference = function (nums) {
    let mi = 1 << 30;
    let ans = -1;
    for (const x of nums) {
        if (mi < x) {
            ans = Math.max(ans, x - mi);
        } else {
            mi = x;
        }
    }
    return ans;
};


  • impl Solution {
    pub fn maximum_difference(nums: Vec<i32>) -> i32 {
        let mut min = nums[0];
        let mut res = -1;
        for i in 1..nums.len() {
            res = res.max(nums[i] - min);
            min = min.min(nums[i]);
        }
        match res {
            0 => -1,
            _ => res,
        }
    }
}

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