945. Minimum Increment to Make Array Unique
Problem 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.
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 <= 10^50 <= nums[i] <= 10^5
Solution
Python Solution
class Solution:
def minIncrementForUnique(self, nums: List[int]) -> int:
nums.sort()
moves = 0
for i in range(1, len(nums)):
if nums[i] <= nums[i-1]:
moves += nums[i-1] - nums[i] + 1
nums[i] = nums[i-1] + 1
return moves
Time Complexity: O(n log n)
Where n is the length of the array. The sorting operation takes O(n log n) time.
Space Complexity: O(1)
We modify the input array in-place.
Java Solution
class Solution {
public int minIncrementForUnique(int[] nums) {
Arrays.sort(nums);
int moves = 0;
for (int i = 1; i < nums.length; i++) {
if (nums[i] <= nums[i-1]) {
moves += nums[i-1] - nums[i] + 1;
nums[i] = nums[i-1] + 1;
}
}
return moves;
}
}
Time Complexity: O(n log n)
Where n is the length of the array. The sorting operation takes O(n log n) time.
Space Complexity: O(1)
We modify the input array in-place.
C++ Solution
class Solution {
public:
int minIncrementForUnique(vector& nums) {
sort(nums.begin(), nums.end());
int moves = 0;
for (int i = 1; i < nums.size(); i++) {
if (nums[i] <= nums[i-1]) {
moves += nums[i-1] - nums[i] + 1;
nums[i] = nums[i-1] + 1;
}
}
return moves;
}
};
Time Complexity: O(n log n)
Where n is the length of the array. The sorting operation takes O(n log n) time.
Space Complexity: O(1)
We modify the input array in-place.
JavaScript Solution
/**
* @param {number[]} nums
* @return {number}
*/
var minIncrementForUnique = function(nums) {
nums.sort((a, b) => a - b);
let moves = 0;
for (let i = 1; i < nums.length; i++) {
if (nums[i] <= nums[i-1]) {
moves += nums[i-1] - nums[i] + 1;
nums[i] = nums[i-1] + 1;
}
}
return moves;
};
Time Complexity: O(n log n)
Where n is the length of the array. The sorting operation takes O(n log n) time.
Space Complexity: O(1)
We modify the input array in-place.
C# Solution
public class Solution {
public int MinIncrementForUnique(int[] nums) {
Array.Sort(nums);
int moves = 0;
for (int i = 1; i < nums.Length; i++) {
if (nums[i] <= nums[i-1]) {
moves += nums[i-1] - nums[i] + 1;
nums[i] = nums[i-1] + 1;
}
}
return moves;
}
}
Time Complexity: O(n log n)
Where n is the length of the array. The sorting operation takes O(n log n) time.
Space Complexity: O(1)
We modify the input array in-place.