README_EN.md

945. Minimum Increment to Make Array Unique

中文文档

Description

You are given an integer array nums. In one move, you can pick an index i where 0 <= i < nums.length and increment nums[i] by 1.

Return the minimum number of moves to make every value in nums unique.

The test cases are generated so that the answer fits in a 32-bit integer.

 

Example 1:

Input: nums = [1,2,2]
Output: 1
Explanation: After 1 move, the array could be [1, 2, 3].

Example 2:

Input: nums = [3,2,1,2,1,7]
Output: 6
Explanation: After 6 moves, the array could be [3, 4, 1, 2, 5, 7].
It can be shown with 5 or less moves that it is impossible for the array to have all unique values.

 

Constraints:

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

Solutions

Python3

class Solution:
    def minIncrementForUnique(self, nums: List[int]) -> int:
        nums.sort()
        ans = 0
        for i in range(1, len(nums)):
            if nums[i] <= nums[i - 1]:
                d = nums[i - 1] - nums[i] + 1
                nums[i] += d
                ans += d
        return ans

Java

class Solution {
    public int minIncrementForUnique(int[] nums) {
        Arrays.sort(nums);
        int ans = 0;
        for (int i = 1; i < nums.length; ++i) {
            if (nums[i] <= nums[i - 1]) {
                int d = nums[i - 1] - nums[i] + 1;
                nums[i] += d;
                ans += d;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int minIncrementForUnique(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int ans = 0;
        for (int i = 1; i < nums.size(); ++i) {
            if (nums[i] <= nums[i - 1]) {
                int d = nums[i - 1] - nums[i] + 1;
                nums[i] += d;
                ans += d;
            }
        }
        return ans;
    }
};

Go

func minIncrementForUnique(nums []int) int {
	sort.Ints(nums)
	ans := 0
	for i := 1; i < len(nums); i++ {
		if nums[i] <= nums[i-1] {
			d := nums[i-1] - nums[i] + 1
			nums[i] += d
			ans += d
		}
	}
	return ans
}

...