Minimum Increment operations to make Array unique

Given an array A[] of integers. In one move you can choose any element A[i], and increment it by 1. The task is to return the minimum number of moves needed to make every value in the array A[] unique.

Examples:

Input: A[] = [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.

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



A simple solution to make each duplicate values unique is to keep incrementing it repeatedly until it is not unique. However, we might do a lot of extra work, if we have an array of all ones.

So, what we can do instead is to evaluate what our increments should be. If for example we have [1, 1, 1, 3, 5], we don’t need to process all the increments of duplicated 1’s. We could take two ones (taken = [1, 1]) and continue processing. Whenever we find an empty(unused value) place like 2 or 4 we can then recover that our increment will be 2-1, 4-1 respectively.

Thus, we first count the values and for each possible value X in array:

  • If there are 2 or more values X in A, save the extra duplicated values to increment later.
  • If there are 0 values X in A, then a saved value gets incremented to X.

Below is the implementation of above approach:

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# Python Implementation of above approach
  
import collections
  
# function to find minimum increment required
def minIncrementForUnique(A):
  
    # collect frequency of each element
    count = collections.Counter(A)
  
    # array of unique values taken
    taken = []
  
    ans = 0
  
    for x in range(100000):
        if count[x] >= 2:
            taken.extend([x] * (count[x] - 1))
        elif taken and count[x] == 0:
            ans += x - taken.pop()
  
    # return answer
    return ans
  
# Driver code
A = [3, 2, 1, 2, 1, 7]
print(minIncrementForUnique(A))

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Output:

6

Time Complexity: O(N)



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