【每日算法】从「枚举」到「排序 + 前缀和 + 二分 + 滑动窗口」｜Python 主题月

输入：nums = [1,2,4], k = 5
输出：3
解释：对第一个元素执行 3 次递增操作，对第二个元素执 2 次递增操作，此时 nums = [4,4,4] 。
4 是数组中最高频元素，频数是 3 。
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输入：nums = [1,4,8,13], k = 5
输出：2
解释：存在多种最优解决方案：
- 对第一个元素执行 3 次递增操作，此时 nums = [4,4,8,13] 。4 是数组中最高频元素，频数是 2 。
- 对第二个元素执行 4 次递增操作，此时 nums = [1,8,8,13] 。8 是数组中最高频元素，频数是 2 。
- 对第三个元素执行 5 次递增操作，此时 nums = [1,4,13,13] 。13 是数组中最高频元素，频数是 2 。
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输入：nums = [3,9,6], k = 2
输出：1
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class Solution {
    public int maxFrequency(int[] nums, int k) {
        int n = nums.length;
        Map<Integer, Integer> map = new HashMap<>();
        for (int i : nums) map.put(i, map.getOrDefault(i, 0) + 1);
        List<Integer> list = new ArrayList<>(map.keySet());
        Collections.sort(list);
        int ans = 1;
        for (int i = 0; i < list.size(); i++) {
            int x = list.get(i), cnt = map.get(x);
            if (i > 0) {
                int p = k;
                for (int j = i - 1; j >= 0; j--) {
                    int y = list.get(j);
                    int diff = x - y;
                    if (p >= diff) {
                        int add = p / diff;
                        int min = Math.min(map.get(y), add);
                        p -= min * diff;
                        cnt += min;
                    } else {
                        break;
                    }
                }
            }
            ans = Math.max(ans, cnt);
        }
        return ans;
    }
}
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class Solution:
    def maxFrequency(self, nums: List[int], k: int) -> int:
        n = len(nums)
        hashmap = defaultdict(int)
        for i in nums:
            hashmap[i] += 1
        ans = 1
        lt = sorted(hashmap.keys())
        for i in range(len(lt)):
            x = lt[i]
            cnt = hashmap[lt[i]]
            if i > 0:
                p = k
                for j in range(i - 1, -1, -1):
                    y = lt[j]
                    diff = x - y
                    if p >= diff:
                        add = p // diff
                        m = min(hashmap[y], add)
                        p -= m * diff
                        cnt += m
                    else:
                        break
            ans = max(ans, cnt)
        return ans
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class Solution {
    int[] nums, sum;
    int n, k;
    public int maxFrequency(int[] _nums, int _k) {
        nums = _nums;
        k = _k;
        n = nums.length;
        Arrays.sort(nums);
        sum = new int[n + 1];
        for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + nums[i - 1];
        int l = 0, r = n;
        while (l < r) {
            int mid = l + r + 1 >> 1;
            if (check(mid)) l = mid;
            else r = mid - 1;
        }
        return r;
    }
    boolean check(int len) {
        for (int l = 0; l + len - 1 < n; l++) {
            int r = l + len - 1;
            int cur = sum[r + 1] - sum[l];
            int t = nums[r] * len;
            if (t - cur <= k) return true;
        }
        return false;
    }
}
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class Solution:
    def maxFrequency(self, nums: List[int], k: int) -> int:
        def check(length):
            for i in range(n+1-length):
                j = i + length - 1
                cur = presum[j + 1] - presum[i]
                t = nums[j] * length
                if t - cur <= k:
                    return True
            return False
        n = len(nums)
        nums.sort()
        presum = [0] + list(accumulate(nums))
        l, r = 0, n
        while l < r:
            mid = l + r + 1 >> 1
            if check(mid):
                l = mid
            else:
                r = mid - 1
        return r
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