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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
-
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; } }
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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; } };
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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
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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 }
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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; }