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| 1 | +class Solution { |
| 2 | +public: |
| 3 | + // Helper function to solve the subproblem of picking non-adjacent slices |
| 4 | + int solve(int index, int endIndex, vector<int>& slices, int n) { |
| 5 | + // Base case: If we've chosen n slices or we exceed the valid range, return 0 (no sum) |
| 6 | + if (n == 0 || index > endIndex) return 0; |
| 7 | + |
| 8 | + // Option 1: Take the slice at the current index and move two steps forward |
| 9 | + int takeSlice = slices[index] + solve(index + 2, endIndex, slices, n - 1); |
| 10 | + |
| 11 | + // Option 2: Skip the slice at the current index and move one step forward |
| 12 | + int noTakeSlice = 0 + solve(index + 1, endIndex, slices, n); |
| 13 | + |
| 14 | + // Return the maximum of taking or skipping the current slice |
| 15 | + return max(takeSlice, noTakeSlice); |
| 16 | + } |
| 17 | + |
| 18 | + // Main function to find the maximum sum of slices |
| 19 | + int maxSizeSlices(vector<int>& slices) { |
| 20 | + int k = slices.size(); // Total number of slices |
| 21 | + // Case 1: Exclude the last slice (pick slices from index 0 to k-2) |
| 22 | + int case1 = solve(0, k - 2, slices, k / 3); |
| 23 | + |
| 24 | + // Case 2: Exclude the first slice (pick slices from index 1 to k-1) |
| 25 | + int case2 = solve(1, k - 1, slices, k / 3); |
| 26 | + |
| 27 | + // Return the maximum of the two cases |
| 28 | + return max(case1, case2); |
| 29 | + } |
| 30 | +}; |
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