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:
CPP
// C++ Implementation of above approach #include <bits/stdc++.h> using namespace std; // function to find minimum increment required int minIncrementForUnique(vector<int> A) { // collect frequency of each element unordered_map<int, int> mpp; for(int i:A) mpp[i]++; // array of unique values taken vector<int> taken; int ans = 0; for (int x = 0; x < 100000; x++) { if (mpp[x] >= 2) taken.push_back(x * (mpp[x] - 1)); else if(taken.size() > 0 and mpp[x] == 0) { ans += x - taken.back(); taken.pop_back(); } } // return answer return ans; } // Driver code int main() { vector<int> A = {3, 2, 1, 2, 1, 7}; cout << minIncrementForUnique(A); return 0; } // This code is contributed by mohit kumar 29 |
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Python
# 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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Improved By : mohit kumar 29
