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

• 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;
}
};

• 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


• 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
}

• 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;
}