recursion.md

Recursion

Regular Expression Matching

Given an input string s and a pattern p, implement regular expression matching with support for . and * where:

• . matches any single character.
• * matches zero or more of the preceding element.

The matching should cover the entire input string¹.

#include <string>
#include <vector>

using namespace std;

bool isMatch(string s, string p) {
  int m = s.length(), n = p.length();
  vector<vector<bool>> is_match(2, vector<bool>(n + 1, false));
  is_match[0][0] = true;
  for (int j = 1; j < n; j++)
      if (p[j] == '*')
        is_match[0][j + 1] = is_match[0][j - 1];
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
      if (p[j] == s[i] || p[j] == '.') is_match[1][j + 1] = is_match[0][j];
      else if (p[j] == '*' && j)
        is_match[1][j + 1] = is_match[1][j - 1]
          || (p[j - 1] == s[i] || p[j - 1] == '.')
          && (is_match[0][j + 1] || is_match[1][j] || is_match[1][j + 1]);
      else if (p[j] != '*') is_match[1][j + 1] = false;
    }
    is_match[0] = is_match[1];
  }
  return is_match[0][n];
}

Scramble String

Distinct Subsequences

Given two strings s and t, return the number of distinct subsequences of s which equals t. The answer fits on a 32-bit signed integer.

#include <string>
#include <vector>

using namespace std;

int numDistinct(string s, string t) {
  int m = s.size(), n = t.size();
  if (n > m) return 0;
  vector<unsigned long long> num_distinct(n + 1);
  num_distinct[0] = 1;
  for (int i = 0; i < m; i++)
    for (int j = n; j > 0; j--)
      if (s[i] == t[j - 1])
        num_distinct[j] += num_distinct[j - 1];
  return num_distinct[n];
}

Palindrome Partitioning

Given a string s, partition s such that every substring² of the partition is a palindrome³. Return the minimum cuts needed for a palindrome partitioning of s.

#include <algorithm>
#include <string>
#include <vector>

using namespace std;

int minCut(string s) {
  int n = s.size();
  vector<int> cut(n + 1);
  for (int i = 0; i <= n; i++) cut[i] = i - 1;
  for (int i = 0; i < n; i++)
    for (int k = 0; k < 2; k++) {
      int j = k;
      while (j < min(i + k + 1, n - i) && s[i - j + k] == s[i + j])
        cut[i + j] = min(cut[i - j + k] + 1, cut[i + ++j]);
    }
  return cut[n];
}

Dungeon Game

The demons had captured the princess and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of m * n rooms laid out in a 2-D grid. Our valiant knight was initially positioned in the top-left room and must fight his way through dungeon to rescue the princess. The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately. Some of the rooms are guarded by demons⁴, so the knight loses health upon entering these rooms; other rooms are either empty⁵ or contain magic orbs that increase the knight's health⁶. To reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step. Return the knight's minimum initial health so that he can rescue the princess. Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned.

#include <algorithm>
#include <climits>
#include <vector>

using namespace std;

int calculateMinimumHP(vector<vector<int>>& dungeon) {
  int m = dungeon.size(), n = dungeon[0].size();
  vector<vector<int>> health(2, vector<int>(n + 1, INT_MAX));
  health[0][--n] = 1;
  for (int i = --m; i >= 0; i--) {
    swap(health[0], health[1]);
    for (int j = n; j >= 0; j--)
      health[0][j] = max(min(health[1][j], health[0][j + 1]) - dungeon[i][j],
        1);
  }
  return health[0][0];
}

Best Time to Buy and Sell Stock

You are given an array prices where prices[i] is the price of a given stock on the i^th day. Find the maximum profit you can achieve. You may not engage in multiple transactions simultaneously⁷.

You may complete at most two transactions.

#include <algorithm>
#include <climits>

using namespace std;

int maxProfit(vector<int>& prices) {
  int state[4] = {INT_MAX, 0, INT_MAX, 0};
  for (int price : prices) {
    state[0] = min(state[0], price);
    state[1] = max(state[1], price - state[0]);
    state[2] = min(state[2], price - state[1]);
    state[3] = max(state[3], price - state[2]);
  }
  return state[3];
}

You may complete at most k transactions: i.e. you may buy at most k times and sell at most k times.

import itertools

def maxProfit(k, prices):
    n = len(prices)
    if k < n >> 1:
        dp = [[0] * n for _ in range(2)]
        for j in range(1, k + 1):
            dp[1][0] = 0
            profit = dp[0][0] - prices[0]
            for i in range(1, n):
                dp[1][i] = max(dp[1][i - 1], profit + prices[i])
                profit = max(profit, dp[0][i] - prices[i])
            dp[0] = dp[1][:]
        return dp[0][-1]
    else: return sum(max(b - a, 0) for a, b in itertools.pairwise(prices))

Expression Add Operators

Given a string num that contains only digits and an integer target, return all possibilities to insert the binary operators +, -, and/or * between the digits of num so that the resultant expression evaluates to the target value. Note that operands in the returned expressions should not contain leading zeros.

import collections

operator = '+-*'
State = collections.namedtuple('State',
  ['i', 'expression', 'eval', 'multiplier'])

def addOperators(num, target):
    n = len(num)
    x = int(num[0])
    stack = [State(i=1, expression=num[:1], eval=x, multiplier=x)]
    if num[0] != '0':
        for i in range(2, n + 1):
            expression = num[:i]
            x = int(expression)
            stack.append(State(i, expression, eval=x, multiplier=x))
    possibilities = []
    while stack:
        u = stack.pop()
        for i in range(u.i, n):
            if i > u.i and num[u.i] == '0': break
            i += 1
            u_string = num[u.i:i]
            u_int = int(u_string)
            for operator, x, multiplier in zip(operator,
                [u_int, -u_int, u.multiplier * (u_int - 1)],
                [1, -1, u.multiplier]):
                stack.append(State(i,
                    expression=f'{u.expression}{operator}{u_string}',
                    eval = u.eval + x,
                    multiplier = multiplier * u_int))
        if (u.i, u.eval) == (n, target): possibilities.append(u.expression)
    return possibilities

Zuma Game

Decode Ways

A message containing letters from A-Z can be encoded into numbers using the following mapping:

'A' -> "1"
'B' -> "2"
...
'Z' -> "26"

To decode an encoded message, all the digits must be grouped then mapped back into letters using the reverse of the mapping above⁸.

For example, 11106 can be mapped into:

• AAJF with the grouping (1 1 10 6)
• KJF with the grouping (11 10 6)

The grouping (1 11 06) is invalid because 06 cannot be mapped into F since 6 is different from 06.

In addition to the mapping above, an encoded message may contain the * character, which can represent any digit from 1 to 9⁹. For example, the encoded message 1* may represent any of the encoded messages 11, 12, 13, 14, 15, 16, 17, 18, or 19. Decoding 1* is equivalent to decoding any of the encoded messages it can represent. Given a string s consisting of digits and * characters, return the number of ways to decode it. Since the answer may be very large, return it modulo 10^9 + 7.

import numpy as np

mod = 10**9 + 7

def numDecodings(s):
    ways = np.array([1, 0, 0])
    for x in s:
        if x.isdigit(): y = [[x > '0', 1, x <= '6'], [x == '1', x == '2']]
        else: y = [[9, 9, 6], [1, 1]]
        ways = np.array([1,
          ways[0],
          ways[0]]) * ([sum(ways * y[0]) % mod] + y[1])
    return int(ways[0])

#include <cctype>
#include <numeric>
#include <string>
#include <vector>

using namespace std;

const int mod = pow(10, 9) + 7;

int numDecodings(string s) {
  vector<long> ways = {1, 0, 0};
  for (char x : s) {
    vector<vector<int>> y = isdigit(x) 
      ? vector<vector<int>>{{x > '0', 1, x <= '6'}, {x == '1', x == '2'}}
      : vector<vector<int>>{{9, 9, 6}, {1, 1}};
    ways = {(inner_product(ways.begin(), ways.end(), begin(y[0]), 0L)) % mod,
      (1L * ways[0] * y[1][0]) % mod,
      (1L * ways[0] * y[1][1]) % mod};
  }
  return ways[0];
}

Cherry Pickup

You are given an n * n grid representing a field of cherries, each cell is one of three possible integers.

• 0 means the cell is empty, so you can pass through,
• 1 means the cell contains a cherry that you can pick up and pass through, or
• -1 means the cell contains a thorn that blocks your way.

Return the maximum number of cherries you can collect by following the rules below:

• Starting at the position (0, 0) and reaching (n - 1, n - 1) by moving right or down through valid path cells¹⁰.
• After reaching (n - 1, n - 1), returning to (0, 0) by moving left or up through valid path cells.
• When passing through a path cell containing a cherry, you pick it up, and the cell becomes an empty cell 0.
• If there is no valid path between (0, 0) and (n - 1, n - 1), then no cherries can be collected.

import itertools

def cherryPickup(grid):
    n = len(grid)
    cherries = [[-1] * n for _ in range(n)]
    cherries[0][0] = grid[0][0]
    for k, i1, i2 in itertools.product(range(1, 2 * n - 1),
        reversed(range(n)),
        reversed(range(n))):
        j1, j2 = k - i1, k - i2
        if not 0 <= min(j1,
            j2) <= max(j1,
            j2) < n or -1 in {grid[i1][j1],
            grid[i2][j2]}:
            cherries[i1][i2] = -1
            continue
        if i1: cherries[i1][i2] = max(cherries[i1][i2], cherries[i1 - 1][i2])
        if i2: cherries[i1][i2] = max(cherries[i1][i2], cherries[i1][i2 - 1])
        if i1 and i2:
            cherries[i1][i2] = max(cherries[i1][i2], cherries[i1 - 1][i2 - 1])
        if cherries[i1][i2] >= 0:
            cherries[i1][i2] += grid[i1][j1]
            if i1 != i2: cherries[i1][i2] += grid[i2][j2]
    return max(cherries[-1][-1], 0)

Footnotes

1. not partial ↩

2. A substring is a contiguous non-empty sequence of characters within a string. ↩

3. A palindrome is a string that reads the same forward and backward. ↩

4. represented by negative integers ↩

5. represented as 0 ↩

6. represented by positive integers ↩

7. you must sell the stock before you buy again ↩

8. there may be multiple ways ↩

9. 0 is excluded ↩

10. cells with value 0 or 1 ↩