力扣376周赛

力扣第376场周赛

找出缺失和重复的数字

map模拟

class Solution {
public:
    vector<int> findMissingAndRepeatedValues(vector<vector<int>>& grid) {
        int n = grid.size() , m = grid[0].size();
        map<int,int>mi;
        for(int i = 0 ; i < n ; i ++){
            for(int j = 0 ; j < m ; j ++){
                mi[grid[i][j]] ++;
            }
        }
        vector<int> ans;
        for(int i = 1 ; i <= n * n ; i ++){
            if(mi[i] > 1)ans.push_back(i);
        }
        for(int i = 1 ; i <= n * n ; i ++){
            if(mi[i] == 0)ans.push_back(i);
        }
        return ans;
    }
};

划分数组并满足最大差限制

排序后直接模拟

class Solution {
public:
    vector<vector<int>> divideArray(vector<int>& nums, int k) {
        vector<vector<int> >ans;
        vector<vector<int> >res;
        int n = nums.size();
        ranges::sort(nums);
        bool f = true;
        if(n % 3 != 0)f = false;
        for(int i = 0 ; i + 2 < n ; i += 3){
            int a = nums[i] , c = nums[i + 2];
            if(c - a > k)f = false;
            ans.push_back({a , nums[i + 1] , c});
        }
        if(f == false)return res;
        return ans;
    }
};

使数组成为等数数组的最小代价

找到中位数再二分最近的回文

//代码来自TsReaper
bool inited = false;
vector<int> good;

void init() {
    if (inited) return;
    inited = true;

    for (int i = 1; i < 10; i++) good.push_back(i);

    // 首先枚举回文数一半的长度 len，以及这一半数值的上限 p
    for (int p = 10, len = 1; p <= 1e4; p *= 10, len++) {
        // 枚举回文数的一半具体是什么数
        for (int i = 1; i < p; i++) if (i % 10 != 0) {
            // 把每个数位拆开来
            vector<int> vec;
            for (int x = i, j = len; j > 0; x /= 10, j--) vec.push_back(x % 10);
            
            // 回文数长度是偶数的情况
            int v = 0;
            for (int j = 0; j < len; j++) v = v * 10 + vec[j];
            for (int j = len - 1; j >= 0; j--) v = v * 10 + vec[j];
            good.push_back(v);

            // 回文数长度是奇数的情况，需要枚举中间那一位是什么数
            for (int k = 0; k < 10; k++) {
                v = 0;
                for (int j = 0; j < len; j++) v = v * 10 + vec[j];
                v = v * 10 + k;
                for (int j = len - 1; j >= 0; j--) v = v * 10 + vec[j];
                good.push_back(v);
            }
        }
    }

    sort(good.begin(), good.end());
}

class Solution {
public:
    long long minimumCost(vector<int>& nums) {
        init();

        int n = nums.size();
        sort(nums.begin(), nums.end());

        // 计算所有数都变成 x 的代价
        auto get = [&](int x){
            long long res = 0;
            for(int i = 0 ; i < n ; i ++){
                res += abs(nums[i] - x);
            }
            return res;
        };
        long long ans = 1e18;
        int idx = lower_bound(good.begin(), good.end(), nums[n / 2]) - good.begin();
        ans = min(ans, get(good[idx]));
        if (idx - 1 >= 0) ans = min(ans, get(good[idx - 1]));
        return ans;
    }
};

执行操作使频率分数最大

滑动窗口统计化成中位数的最小步骤

//代码来自TsReaper
class Solution {
public:
    int maxFrequencyScore(vector<int>& nums, long long k) {
        int n = nums.size();
        nums.push_back(0);
        sort(nums.begin(), nums.end());

        // 预处理前缀和
        long long f[n + 1];
        f[0] = 0;
        for (int i = 1; i <= n; i++) f[i] = f[i - 1] + nums[i];

        // 求把 nums[L] 到 nums[R] 全部变成中位数的最小代价
        auto gao = [&](int L, int R) {
            // 找到中位数的下标
            int mid = (L + R) / 2;
            long long l = 1LL * nums[mid] * (mid - L + 1) - (f[mid] - f[L - 1]);
            long long r =  ((f[R] - f[mid]) - 1LL * nums[mid] * (R - mid));
            return l + r;
        };
        
        int ans = 0;
        // 滑动窗口
        //i右端点j左端点
        for (int i = 1, j = 1; i <= n; i++) {
            while (j <= i && gao(j, i) > k) j++;
            ans = max(ans, i - j + 1);
        }
        return ans;
    }
};

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