↓Skip to main content
  1. Posts/

LeetCode-2790 长度递增组的最大数目

·1 min·
Table of Contents

LeetCode-2790 长度递增组的最大数目 #

Solution 1 #

二分搜索答案. 考虑从大到小排序后的数组与目标数组的差异, 注意前面少的可以用后面多的补.

using ll = long long;

class Solution {
public:
    int maxIncreasingGroups(vector<int>& usageLimits) {
        sort(usageLimits.begin(), usageLimits.end(), greater<int>());
        auto check = [&](int m) {
            int res = 0;
            for (int i = 0; i < usageLimits.size(); i++) {
                res = min(res + usageLimits[i] - max(m - i, 0), 0);
            }
            return res >= 0;
        };

        ll left = 2, right = usageLimits.size() + 1;
        while (left < right) {
            ll mid = left + (right - left) / 2;
            if (check(mid)) {
                left = mid + 1;
            }
            else {
                right = mid;
            }
        }
        return left - 1;
    }
};

Solution 2 #

考虑如下的构造方法:

  • 考虑一个形如上三角矩阵的网格, 我们从数量小的数字开始填格子;
  • 前 $k$ 行至多形成 $k$ 组, 需要 $\frac{k(k+1)}{2}$ 个数字;
  • 统计数字数量是否够就可以.

这种构造方法的精妙之处在于, 从数量小的数字开始填, 详细的思路讲解参见0x3f的周赛讲解:证明题？构造题！【力扣周赛 355】

using ll = long long;

class Solution {
public:
    int maxIncreasingGroups(vector<int>& usageLimits) {
        sort(usageLimits.begin(), usageLimits.end());
        ll res = 0, ans = 0;
        for (int v: usageLimits) {
            res += v;
            if (res >= (ans + 1) * (ans + 2) / 2) {
                ans++;
            }
        }
        return ans;
    }
};