Leetcode Problem 2170. Minimum Operations to Make the Array Alternating
Leetcode Solutions
Frequency Count and Greedy Selection
- Initialize two frequency arrays (or hashmaps) for even and odd indices.
- Traverse the input array, incrementing the frequency of the current element in the corresponding frequency array.
- Find the max and second max frequency elements for both even and odd indices.
- If the max elements for even and odd indices are different, the minimum operations required are
n - freqMaxEven - freqMaxOdd. - If the max elements for even and odd indices are the same, calculate the minimum operations required by considering the second max elements:
min(n - freqMaxEven - freqSecondMaxOdd, n - freqSecondMaxEven - freqMaxOdd). - Return the calculated minimum number of operations.
Greedy Approach with Hash Maps
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