Question

945. Minimum Increment to Make Array Unique

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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

Solution 1

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

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;
  }
}

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;
  }
};

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
}