Minimum Product Complete Solution

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Solution Approach

We are only concerned of frequency of four smallest numbers of the array. There are multiple cases: Let's discuss them one by one -

Case 1 - If frequency of smallest element is greater than 3. We have to select all four indices of this number. So, if frequency of smallest number is x. Answer would be xC4
Case 2 - If sum of frequency of smallest two numbers is greater than 3. Let frequency of smallest element is y and the frequency of second smallest element is x. Case a: y = 3 We have to select only 1 index of all second smallest element index. So, answer would be x. Case b: y = 2. We have to select 2 indices of all second smallest element index. So, answer would be xC2. Case c: y = 1. We have to select 3 indices of all second smallest element. So, answer would be xC3.
Case 3 - If sum of frequency of smallest three numbers is greater than 3. Let the frequency of third smallest element is x. Case a: If the frequency of smallest or second smallest element is 2. In this case frequency sum of smallest two elements is 3. So, we have to select only 1 index of all third smallest element index. So, answer would be x. Case b: If the frequency of smallest and second smallest element is 1 In this case we have to select 2 indices of all third smallest element index. So, answer for this case would be xC2.
Case 4 - If the frequency of smallest three elements is 1. Let the frequency of fourth smallest element is x. So, we have to select only 1 index of all fourth smallest element index. So, answer for this case would be x.
So, we can sort the array A and store the frequency of all elements in HashMap.
And, then find the answer according to the case satisfied.