Skip to content

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 that it is impossible for the array to have all unique values with 5 or less moves.

 

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 $\textit{nums}$.

1
2
3
4
5
6
7
8
class Solution:
    def minIncrementForUnique(self, nums: List[int]) -> int:
        nums.sort()
        ans, y = 0, -1
        for x in nums:
            y = max(y + 1, x)
            ans += y - x
        return ans

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
class Solution {
    public int minIncrementForUnique(int[] nums) {
        Arrays.sort(nums);
        int ans = 0, y = -1;
        for (int x : nums) {
            y = Math.max(y + 1, x);
            ans += y - x;
        }
        return ans;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
class Solution {
public:
    int minIncrementForUnique(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int ans = 0, y = -1;
        for (int x : nums) {
            y = max(y + 1, x);
            ans += y - x;
        }
        return ans;
    }
};

1
2
3
4
5
6
7
8
9
func minIncrementForUnique(nums []int) (ans int) {
    sort.Ints(nums)
    y := -1
    for _, x := range nums {
        y = max(y+1, x)
        ans += y - x
    }
    return
}

1
2
3
4
5
6
7
8
9
function minIncrementForUnique(nums: number[]): number {
    nums.sort((a, b) => a - b);
    let [ans, y] = [0, -1];
    for (const x of nums) {
        y = Math.max(y + 1, x);
        ans += y - x;
    }
    return ans;
}

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.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
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;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
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;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
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
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
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;
}

Comments