README_EN.md

2592. Maximize Greatness of an Array

中文文档

Description

You are given a 0-indexed integer array nums. You are allowed to permute nums into a new array perm of your choosing.

We define the greatness of nums be the number of indices 0 <= i < nums.length for which perm[i] > nums[i].

Return the maximum possible greatness you can achieve after permuting nums.

 

Example 1:

Input: nums = [1,3,5,2,1,3,1]
Output: 4
Explanation: One of the optimal rearrangements is perm = [2,5,1,3,3,1,1].
At indices = 0, 1, 3, and 4, perm[i] > nums[i]. Hence, we return 4.

Example 2:

Input: nums = [1,2,3,4]
Output: 3
Explanation: We can prove the optimal perm is [2,3,4,1].
At indices = 0, 1, and 2, perm[i] > nums[i]. Hence, we return 3.

 

Constraints:

• 1 <= nums.length <= 105
• 0 <= nums[i] <= 109

Solutions

Solution 1: Greedy

We can sort the array $nums$ first.

Then we define a pointer $i$ pointing to the first element of the array $nums$. We traverse the array $nums$, and for each element $x$ we encounter, if $x$ is greater than $nums[i]$, then we move the pointer $i$ to the right.

Finally, we return the value of the pointer $i$.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$, where $n$ is the length of the array $nums$.

Python3

class Solution:
    def maximizeGreatness(self, nums: List[int]) -> int:
        nums.sort()
        i = 0
        for x in nums:
            i += x > nums[i]
        return i

Java

class Solution {
    public int maximizeGreatness(int[] nums) {
        Arrays.sort(nums);
        int i = 0;
        for (int x : nums) {
            if (x > nums[i]) {
                ++i;
            }
        }
        return i;
    }
}

C++

class Solution {
public:
    int maximizeGreatness(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int i = 0;
        for (int x : nums) {
            i += x > nums[i];
        }
        return i;
    }
};

Go

func maximizeGreatness(nums []int) int {
	sort.Ints(nums)
	i := 0
	for _, x := range nums {
		if x > nums[i] {
			i++
		}
	}
	return i
}

TypeScript

function maximizeGreatness(nums: number[]): number {
    nums.sort((a, b) => a - b);
    let i = 0;
    for (const x of nums) {
        if (x > nums[i]) {
            i += 1;
        }
    }
    return i;
}

...