Formatted question description: https://leetcode.ca/all/945.html

945. Minimum Increment to Make Array Unique (Medium)

Given an array of integers A, a move consists of choosing any A[i], and incrementing it by 1.

Return the least number of moves to make every value in A unique.

 

Example 1:

Input: [1,2,2]
Output: 1
Explanation:  After 1 move, the array could be [1, 2, 3].

Example 2:

Input: [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.

 

Note:

  1. 0 <= A.length <= 40000
  2. 0 <= A[i] < 40000
 

Related Topics:
Array

Solution 1.

// OJ: https://leetcode.com/problems/minimum-increment-to-make-array-unique/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int minIncrementForUnique(vector<int>& A) {
        sort(A.begin(), A.end());
        int ans = 0;
        for (int i = 1; i < A.size(); ++i) {
            if (A[i] > A[i - 1]) continue;
            ans += A[i - 1] + 1 - A[i];
            A[i] = A[i - 1] + 1;
        }
        return ans;
    }
};

Java

  • class Solution {
        public int minIncrementForUnique(int[] A) {
            if (A == null || A.length <= 1)
                return 0;
            int increments = 0;
            Arrays.sort(A);
            int curValue = A[0];
            int length = A.length;
            for (int i = 1; i < length; i++) {
                curValue = Math.max(curValue + 1, A[i]);
                increments += curValue - A[i];
            }
            return increments;
        }
    }
    
  • // OJ: https://leetcode.com/problems/minimum-increment-to-make-array-unique/
    // Time: O(N)
    // Space: O(1)
    class Solution {
    public:
        int minIncrementForUnique(vector<int>& A) {
            sort(A.begin(), A.end());
            int ans = 0;
            for (int i = 1; i < A.size(); ++i) {
                if (A[i] > A[i - 1]) continue;
                ans += A[i - 1] + 1 - A[i];
                A[i] = A[i - 1] + 1;
            }
            return ans;
        }
    };
    
  • class Solution(object):
        def minIncrementForUnique(self, A):
            """
            :type A: List[int]
            :rtype: int
            """
            N = len(A)
            seats = [0] * 80010
            res = 0
            for a in A:
                if not seats[a]:
                    seats[a] = 1
                else:
                    pos = a
                    while pos < 80010 and seats[pos] == 1:
                        pos += 1
                    seats[pos] = 1
                    res += pos - a
            return res
    

All Problems

All Solutions