2170. Minimum Operations to Make the Array Alternating
You are given a 0indexed array nums consisting of n positive integers.
The array nums is called alternating if:
 nums[i  2] == nums[i], where 2 <= i <= n  1.
 nums[i  1] != nums[i], where 1 <= i <= n  1.
In one operation, you can choose an index i and change nums[i] into any positive integer.
Return the minimum number of operations required to make the array alternating.
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 class Solution { pair<int,int> getMaxFrequency(unordered_map<int, int>& m) { int num = 1, cnt = 0; for(auto& [k, count]: m) { if(cnt < count) { cnt = count; num = k; } } return {num, cnt}; }
pair<int,int> getSecondMaxFrequency(unordered_map<int, int>& m, int skip) { int num = 1, cnt = 0; for(auto& [k, count]: m) { if(k == skip) continue; if(cnt < count) { cnt = count; num = k; } } return {num, cnt}; } public: int minimumOperations(vector<int>& nums) { if(nums.size() <= 1) return 0; if(nums.size() == 2) return nums[0] == nums[1]; int res = 0, n = nums.size(); unordered_map<int, int> odd, even; for(int i = 0; i < n; i++) { if(i&1) odd[nums[i]]++; else even[nums[i]]++; } auto [oddNum, oddMaxFreq] = getMaxFrequency(odd); auto [evenNum, evenMaxFreq] = getMaxFrequency(even); if(oddNum != evenNum) { return n  oddMaxFreq  evenMaxFreq; } auto [secondOddNum, secondOddFreq] = getSecondMaxFrequency(odd, oddNum); auto [secondEvenNum, secondEvenFreq] = getSecondMaxFrequency(even, evenNum); return n + min(oddMaxFreqsecondEvenFreq, secondOddFreqevenMaxFreq); } };
