|
| 1 | +class Solution { |
| 2 | +public: |
| 3 | + int maxFrequency(vector<int>& nums, int k, int numOperations) { |
| 4 | + const int MAXN = 200005; |
| 5 | + static vector<int> freq(MAXN), prefixSum(MAXN); |
| 6 | + |
| 7 | + int maxValue = (*max_element(nums.begin(), nums.end())); |
| 8 | + int limit = (maxValue + k + 2); |
| 9 | + |
| 10 | + // Reset frequency array |
| 11 | + fill(freq.begin(), freq.begin() + limit, 0); |
| 12 | + |
| 13 | + // Count occurrences of each number |
| 14 | + for (int num : nums) { |
| 15 | + freq[num]++; |
| 16 | + } |
| 17 | + |
| 18 | + // If no operations are allowed, return the highest existing frequency |
| 19 | + if (numOperations == 0) { |
| 20 | + return (*max_element(freq.begin(), freq.begin() + limit)); |
| 21 | + } else { |
| 22 | + // Build prefix sum array |
| 23 | + prefixSum[0] = freq[0]; |
| 24 | + for (int i = 1; i < limit; i++) { |
| 25 | + prefixSum[i] = (prefixSum[i - 1] + freq[i]); |
| 26 | + } |
| 27 | + |
| 28 | + int best = 0; |
| 29 | + |
| 30 | + // Check each possible target value |
| 31 | + for (int target = 0; target <= maxValue; target++) { |
| 32 | + int left; |
| 33 | + if (target > k) { |
| 34 | + left = (target - k); |
| 35 | + } else { |
| 36 | + left = 0; |
| 37 | + } |
| 38 | + |
| 39 | + int right; |
| 40 | + if ((target + k) < limit) { |
| 41 | + right = (target + k); |
| 42 | + } else { |
| 43 | + right = (limit - 1); |
| 44 | + } |
| 45 | + |
| 46 | + int total = (prefixSum[right] - ((left > 0) ? prefixSum[left - 1] : 0)); |
| 47 | + int changeable = (total - freq[target]); |
| 48 | + |
| 49 | + int possible; |
| 50 | + if (numOperations < changeable) { |
| 51 | + possible = (freq[target] + numOperations); |
| 52 | + } else { |
| 53 | + possible = (freq[target] + changeable); |
| 54 | + } |
| 55 | + |
| 56 | + if (possible > best) { |
| 57 | + best = possible; |
| 58 | + } |
| 59 | + } |
| 60 | + |
| 61 | + return best; |
| 62 | + } |
| 63 | + } |
| 64 | +}; |
0 commit comments