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Problem Statement:

Given an integer array nums, your goal is to make all elements in nums equal. To complete one operation, follow these steps:

  1. Find the largest value in nums. Let its index be i (0-indexed) and its value be largest. If there are multiple elements with the largest value, pick the smallest i.
  2. Find the next largest value in nums strictly smaller than largest. Let its value be nextLargest.
  3. Reduce nums[i] to nextLargest.

Return the number of operations to make all elements in nums equal.

 

Example 1:

Input: nums = [5,1,3]
Output: 3
Explanation: It takes 3 operations to make all elements in nums equal:
1. largest = 5 at index 0. nextLargest = 3. Reduce nums[0] to 3. nums = [3,1,3].
2. largest = 3 at index 0. nextLargest = 1. Reduce nums[0] to 1. nums = [1,1,3].
3. largest = 3 at index 2. nextLargest = 1. Reduce nums[2] to 1. nums = [1,1,1].

Example 2:

Input: nums = [1,1,1]
Output: 0
Explanation: All elements in nums are already equal.

Example 3:

Input: nums = [1,1,2,2,3]
Output: 4
Explanation: It takes 4 operations to make all elements in nums equal:
1. largest = 3 at index 4. nextLargest = 2. Reduce nums[4] to 2. nums = [1,1,2,2,2].
2. largest = 2 at index 2. nextLargest = 1. Reduce nums[2] to 1. nums = [1,1,1,2,2].
3. largest = 2 at index 3. nextLargest = 1. Reduce nums[3] to 1. nums = [1,1,1,1,2].
4. largest = 2 at index 4. nextLargest = 1. Reduce nums[4] to 1. nums = [1,1,1,1,1].

 

Constraints:

  • 1 <= nums.length <= 5 * 104
  • 1 <= nums[i] <= 5 * 104

Solution:

The crucial observation here is that we need not do the actual operations, rather we can compute number of steps required for each array member.

To be exact, if we create a frequency map from the array, we can compute the number of steps required for each key by its rank in the sorted array of keys. This rank multiplied by the frequency of the key gives the contribution of that key to the answer, which can be summed up to get the actual answer.

 
class Solution {
public:
    int reductionOperations(vector<int>& nums) 
    {
        map<int,int>H;
        for(int num: nums) H[num]++;
        int res=0, ctr=0;
        for(auto it=H.begin(); it!=H.end(); it++) res += it->second * ctr++;
        return res;
    }
};