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945. Minimum Increment to Make Array Unique

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: Sorting + Greedy

First, we sort the array $\textit{nums}$, and use a variable $\textit{y}$ to record the current maximum value, initially $\textit{y} = -1$.

Then, we iterate through the array $\textit{nums}$. For each element $x$, we update $y$ to $\max(y + 1, x)$, and accumulate the operation count $y - x$ into the result.

After completing the iteration, we return the result.

The time complexity is $O(n \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array $\textit{nums}$.

Solution 2: Counting + Greedy

According to the problem description, the maximum value of the result array $m = \max(\textit{nums}) + \textit{len}(\textit{nums})$. We can use a counting array $\textit{cnt}$ to record the occurrence count of each element.

Then, we iterate from $0$ to $m - 1$. For each element $i$, if its occurrence count $\textit{cnt}[i]$ is greater than $1$, then we add $\textit{cnt}[i] - 1$ elements to $i + 1$, and accumulate the operation count into the result.

After completing the iteration, we return the result.

The time complexity is $O(m)$, and the space complexity is $O(m)$. Here, $m$ is the length of the array $\textit{nums}$ plus the maximum value in the array.

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  }
  
  
  // Solution 2
  class Solution {
      public int minIncrementForUnique(int[] nums) {
          int m = Arrays.stream(nums).max().getAsInt() + nums.length;
          int[] cnt = new int[m];
          for (int x : nums) {
              ++cnt[x];
          }
          int ans = 0;
          for (int i = 0; i < m - 1; ++i) {
              int diff = cnt[i] - 1;
              if (diff > 0) {
                  cnt[i + 1] += diff;
                  ans += diff;
              }
          }
          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;
      }
  };
  
  
  // Solution 2
  class Solution {
  public:
      int minIncrementForUnique(vector<int>& nums) {
          int m = *max_element(nums.begin(), nums.end()) + nums.size();
          int cnt[m];
          memset(cnt, 0, sizeof(cnt));
          for (int x : nums) {
              ++cnt[x];
          }
          int ans = 0;
          for (int i = 0; i < m - 1; ++i) {
              int diff = cnt[i] - 1;
              if (diff > 0) {
                  cnt[i + 1] += diff;
                  ans += diff;
              }
          }
          return ans;
      }
  };
  
  

• 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
  
  
  # Solution 2
  class Solution:
      def minIncrementForUnique(self, nums: List[int]) -> int:
          m = max(nums) + len(nums)
          cnt = Counter(nums)
          ans = 0
          for i in range(m - 1):
              if (diff := cnt[i] - 1) > 0:
                  cnt[i + 1] += diff
                  ans += diff
          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
  }
  
  
  // Solution 2
  func minIncrementForUnique(nums []int) (ans int) {
  	m := slices.Max(nums) + len(nums)
  	cnt := make([]int, m)
  	for _, x := range nums {
  		cnt[x]++
  	}
  	for i := 0; i < m-1; i++ {
  		if diff := cnt[i] - 1; diff > 0 {
  			cnt[i+1] += diff
  			ans += diff
  		}
  	}
  	return ans
  }
  
  

• function minIncrementForUnique(nums: number[]): number {
      nums.sort((a, b) => a - b);
      let ans = 0;
      for (let i = 1; i < nums.length; ++i) {
          if (nums[i] <= nums[i - 1]) {
              ans += nums[i - 1] - nums[i] + 1;
              nums[i] = nums[i - 1] + 1;
          }
      }
      return ans;
  }
  
  
  // Solution 2
  function minIncrementForUnique(nums: number[]): number {
      const m = Math.max(...nums) + nums.length;
      const cnt: number[] = Array(m).fill(0);
      for (const x of nums) {
          cnt[x]++;
      }
      let ans = 0;
      for (let i = 0; i < m - 1; ++i) {
          const diff = cnt[i] - 1;
          if (diff > 0) {
              cnt[i + 1] += diff;
              ans += diff;
          }
      }
      return ans;
  }
  
  

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