第 372 场周赛(位运算技巧，跳表 + 二分，线段树）

class Solution:
    def findMinimumOperations(self, s1: str, s2: str, s3: str) -> int:
        cnt = 0
        for a, b, c in zip(s1, s2, s3):
            if not a == b == c:
                break
            cnt += 1
        if cnt == 0:
            return -1
        mi = min(len(s1), len(s2), len(s3))
        return len(s1) + len(s2) + len(s3) - 3 * cnt

class Solution:
    def minimumSteps(self, s: str) -> int:
        ans = cnt1 = 0

        for c in s:
            if c == '0':
                ans += cnt1
            else:
                cnt1 += 1

        return ans

class Solution:
    def maximumXorProduct(self, a: int, b: int, n: int) -> int:
        if a < b:
            a, b = b, a

        mask = (1 << n) - 1 # n个1

        ax = a & ~mask # 无法通过mask修改的部分
        bx = b & ~mask
        # 把低位抠出来
        a &= mask
        b &= mask

        left = a ^ b # 一位为1,一位为0的位
        one = mask ^ left # 全为0的位,全部变为1
        ax |= one
        bx |= one

        if left > 0 and ax == bx:
            high_bit = 1 << (left.bit_length() - 1)
            ax |= high_bit
            left ^= high_bit
        
        bx |= left

        MOD = 10**9 + 7

        return ax * bx % MOD

跳表＋二分

class Solution {
public:
    const static int N = 50005, M = 17;
    int f[N][M];
    
    void init(const vector<int> &a, int n) {
        for(int j = 0; j < M; j ++ ) {
            for(int i = 1; i <= n - (1 << j) + 1; i ++ ) {
                if(!j) f[i][j] = a[i - 1];
                else {
                    f[i][j] = max(f[i][j - 1], f[i + (1 << j - 1)][j - 1]);
                }
            }
        }
    }
    
    int query(int l, int r) {
        int k = __lg(r - l + 1);
        return max(f[l][k], f[r - (1 << k) + 1][k]);
    }
    
    vector<int> leftmostBuildingQueries(vector<int>& heights, vector<vector<int>>& queries) {
        int n = heights.size(), m = queries.size();
        init(heights, n);
        vector<int> res(m);
        for(int i = 0; i < m; i ++ ) {
            int l = queries[i][0] + 1, r = queries[i][1] + 1;
            if(l > r) swap(l, r);
            if(l == r) res[i] = l - 1;
            else if(heights[l - 1] < heights[r - 1]) res[i] = r - 1;
            else {
                int L = r, R = n + 2;
                while(L + 1 < R) {
                    int m = (L + R) >> 1;
                    if(query(r + 1, m) > heights[l - 1]) R = m;
                    else L = m;
                }
                if(L == n + 1) res[i] = -1;
                else res[i] = L;
            }
        }
        
        return res;
    }
};

线段树做法

class Solution {
public:

    vector<int> tr;

    void build(int u, int l, int r, vector<int> &heights) {
        if(l == r) {
            tr[u] = heights[l - 1];
            return;
        }
        int m = l + r >> 1;
        build(u << 1, l, m, heights);
        build(u << 1 | 1, m + 1, r, heights);
        tr[u] = max(tr[u << 1], tr[u << 1 | 1]);
    }

    int query(int u, int l, int r, int L, int v) {
        if(v >= tr[u]) return 0;
        if(l == r) return l;
        int mid = l + r >> 1;
        if(L <= mid) {
            int pos = query(u << 1, l, mid, L, v);
            if(pos > 0) return pos; 
        }

        return query(u << 1 | 1, mid + 1, r, L, v);
    }

    vector<int> leftmostBuildingQueries(vector<int>& heights, vector<vector<int>>& queries) {
        int n = heights.size();
        tr.resize(n * 4);
        build(1, 1, n, heights);

        vector<int> ans;

        for(auto &q: queries) {
            int i = q[0], j = q[1];
            if(i > j) swap(i, j);

            if(i == j || heights[i] < heights[j]) {
                ans.push_back(j);
            } else {
                int pos = query(1, 1, n, j + 1, heights[i]);
                ans.push_back(pos - 1);
            }
        }

        return ans;
    }
};