Today I have done Jump Game V, Find Longest Awesome Substring, Max Sum of Rectangle No Larger Than K and leetcode's January LeetCoding Challenge with cpp.

dp is so hard................................

but ICPC guy won't give up on it.

Jump Game V

Description

Given an array of integers arr and an integer d. In one step you can jump from index i to index:

• i + x where: i + x < arr.length and 0 < x <= d.
• i - x where: i - x >= 0 and 0 < x <= d.

In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)).

You can choose any index of the array and start jumping. Return the maximum number of indices you can visit.

Notice that you can not jump outside of the array at any time.

Example 1:

Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown.
Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9.
Similarly You cannot jump from index 3 to index 2 or index 1.

Example 2:

Input: arr = [3,3,3,3,3], d = 3
Output: 1
Explanation: You can start at any index. You always cannot jump to any index.

Example 3:

Input: arr = [7,6,5,4,3,2,1], d = 1
Output: 7
Explanation: Start at index 0. You can visit all the indicies. 

Example 4:

Input: arr = [7,1,7,1,7,1], d = 2
Output: 2

Example 5:

Input: arr = [66], d = 1
Output: 1

Constraints:

• 1 <= arr.length <= 1000
• 1 <= arr[i] <= 10^5
• 1 <= d <= arr.length

Solution

DAG then get max depth

class Solution {
public:
    int maxJumps(vector<int>& arr, int d) {
        int len = arr.size();
        vector<int> degree(len), jump(len);
        vector<vector<int>> edge(len);
        for(int i = 0; i < len; ++i) {
            for(int j = 1; j <= d && (i-j) >= 0 && arr[i-j] < arr[i]; ++j) {
                edge[i].push_back(i-j);
                degree[i-j] += 1;
            }
        }
        for(int i = 0; i < len; ++i) {
            for(int j = 1; j <= d && (i+j) < len && arr[i+j] < arr[i]; ++j) {
                edge[i].push_back(i+j);
                degree[i+j] += 1;
            }
        }
        queue<int> q;
        for(int i = 0; i < len; ++i) {
            if(degree[i] == 0) q.push(i);
        }
        while(!q.empty()) {
            int current = q.front();
            q.pop();
            for(auto i : edge[current]) {
                degree[i] -= 1;
                jump[i] = max(jump[i], jump[current] + 1);
                if(degree[i] == 0) q.push(i);
            }
        }

        return *max_element(jump.begin(), jump.end())+1;
    }
};

Find Longest Awesome Substring

Description

Given a string s. An awesome substring is a non-empty substring of s such that we can make any number of swaps in order to make it palindrome.

Return the length of the maximum length awesome substring of s.

Example 1:

Input: s = "3242415"
Output: 5
Explanation: "24241" is the longest awesome substring, we can form the palindrome "24142" with some swaps.

Example 2:

Input: s = "12345678"
Output: 1

Example 3:

Input: s = "213123"
Output: 6
Explanation: "213123" is the longest awesome substring, we can form the palindrome "231132" with some swaps.

Example 4:

Input: s = "00"
Output: 2

Constraints:

• 1 <= s.length <= 10^5
• s consists only of digits.

Solution

digits dp

class Solution {
public:
    int longestAwesome(string s) {
        vector<int> pos(1024, 100100);
        int len = s.length();
        int current = 0;
        int answer = 1;
        pos[0] = -1;
        for(int i = 0; i < len; ++i) {
            current ^= (1 << (s[i]-'0'));
            pos[current] = min(pos[current], i);
            answer = max(answer, i-pos[current]);
            for(int j = 0; j < 10; ++j) {
                answer = max(answer, i-pos[current^(1<<j)]);
            }
            // cout << i << ' ' << answer << endl;
        }
        return answer;
    }
};

Max Sum of Rectangle No Larger Than K

Description

Given a non-empty 2D matrix matrix and an integer k, find the max sum of a rectangle in the matrix such that its sum is no larger than k.

Example:

Input: matrix = [[1,0,1],[0,-2,3]], k = 2
Output: 2 
Explanation: Because the sum of rectangle [[0, 1], [-2, 3]] is 2,
             and 2 is the max number no larger than k (k = 2).

Note:

1. The rectangle inside the matrix must have an area > 0.
2. What if the number of rows is much larger than the number of columns?

Solution

prefix sum with binary search

class Solution {
public:
    int maxSumSubmatrix(vector<vector<int>>& matrix, int kk) {
        int rows = matrix.size();
        int cols = matrix.front().size();
        vector<vector<int>> prefix(rows+1, vector<int>(cols+1));

        for(int i = 0; i < rows; ++i) {
            for(int j = 0; j < cols; ++j) {
                prefix[i+1][j+1] = matrix[i][j] + prefix[i+1][j] + prefix[i][j+1] - prefix[i][j];
            }
        }

        auto sumRange = [&](int left, int right, int up, int down) {
            return prefix[down][right] - prefix[up][right] - prefix[down][left] + prefix[up][left];
        };

        int answer = INT_MIN;
        for(int i = 0; i < cols; ++i) {
            for(int j = i+1; j <= cols; ++j) {
                vector<int> arr = {0};
                for(int k = 1; k <= rows; ++k) {
                    int s = sumRange(i, j, 0, k);
                    auto it = lower_bound(arr.begin(), arr.end(), s-kk);
                    if(it != arr.end()) {
                        answer = max(answer, s-*it);
                    }
                    arr.insert(lower_bound(arr.begin(), arr.end(), s), s);
                }
            }
        }
        return answer;
    }
};

January LeetCoding Challenge 28

Description

Smallest String With A Given Numeric Value

The numeric value of a lowercase character is defined as its position (1-indexed) in the alphabet, so the numeric value of a is 1, the numeric value of b is 2, the numeric value of c is 3, and so on.

The numeric value of a string consisting of lowercase characters is defined as the sum of its characters' numeric values. For example, the numeric value of the string "abe" is equal to 1 + 2 + 5 = 8.

You are given two integers n and k. Return the lexicographically smallest string with length equal to n and numeric value equal to k.

Note that a string x is lexicographically smaller than string y if x comes before y in dictionary order, that is, either x is a prefix of y, or if i is the first position such that x[i] != y[i], then x[i] comes before y[i] in alphabetic order.

Example 1:

Input: n = 3, k = 27
Output: "aay"
Explanation: The numeric value of the string is 1 + 1 + 25 = 27, and it is the smallest string with such a value and length equal to 3.

Example 2:

Input: n = 5, k = 73
Output: "aaszz"

Constraints:

• $1 \le n \le 10^5$
• n <= k <= 26 * n

Solution

class Solution {
public:
    string getSmallestString(int n, int k) {
        int zs = max(ceil((k - n) / 25.0) - 1, 0.0);
        k -= zs * 26;
        n -= zs;
        return string(n-1, 'a') + string(1, 'a'+ k - n) + string(zs, 'z');
    }
};