動機

不同的DP不同的難易度

Problem

We can scramble a string s to get a string t using the following algorithm:

  1. If the length of the string is 1, stop.
  2. If the length of the string is > 1, do the following:
    • Split the string into two non-empty substrings at a random index, i.e., if the string is s, divide it to x and y where s = x + y.
    • Randomly decide to swap the two substrings or to keep them in the same order. i.e., after this step, s may become s = x + y or s = y + x.
    • Apply step 1 recursively on each of the two substrings x and y.

Given two strings s1 and s2 of the same length, return true if s2 is a scrambled string of s1, otherwise, return false.

 

Example 1:

Input: s1 = great, s2 = rgeatOutput: trueExplanation: One possible scenario applied on s1 is:great --> gr/eat // divide at random index.gr/eat --> gr/eat // random decision is not to swap the two substrings and keep them in order.gr/eat --> g/r / e/at // apply the same algorithm recursively on both substrings. divide at ranom index each of them.g/r / e/at --> r/g / e/at // random decision was to swap the first substring and to keep the second substring in the same order.r/g / e/at --> r/g / e/ a/t // again apply the algorithm recursively, divide at to a/t.r/g / e/ a/t --> r/g / e/ a/t // random decision is to keep both substrings in the same order.The algorithm stops now and the result string is rgeat which is s2.As there is one possible scenario that led s1 to be scrambled to s2, we return true.

Example 2:

Input: s1 = abcde, s2 = caebdOutput: false

Example 3:

Input: s1 = a, s2 = aOutput: true

 

Constraints:

  • s1.length == s2.length
  • 1 <= s1.length <= 30
  • s1 and s2 consist of lower-case English letters.

Ver1:WA

def allOcuurLenFromEnd(s,c,i,j):
    #print('find',c,i,j)
    ans = []
    while i < j:
        i = s.find(c,i,j+1)
        if i != -1:
            ans.append(j-i+1)
            i += 1
        else:
            break
    return ans
dp = {}
def f(s1,s2,i,j,l):
    #print(i,j,l)
    if (i,j,l) not in dp:
        if l is 0:
            dp[(i,j,l)] = True
        elif l is 1:
            dp[(i,j,l)] = s1[i] == s2[j]
        else:
            if s1[i] == s2[j]:
                dp[(i,j,l)] = f(s1,s2,i-1,j-1,l-1)
            else:
                ii = allOcuurLenFromEnd(s2,s1[i],j+1-l,j)
                #print(ii)
                for x in ii:
                    ret = f(s1,s2,i-x,j-x,l-x) and f(s1,s2,i-1,j,x-1)
                    #print("ret",ret,(i-x,j-x,l-x),(i-1,j,x-1))
                    if ret:
                        dp[(i,j,l)] = ret
                        return ret
                dp[(i,j,l)] = False
    #print((i,j,l),dp[(i,j,l)])
    return dp[(i,j,l)]
        
class Solution:
    def isScramble(self, s1: str, s2: str) -> bool:
        dp.clear()
        return f(s1,s2,len(s1)-1,len(s2)-1,len(s1))

Ver2: WA

def allOcuurLenFromEnd(s,c,i,j):
    #print('find',c,i,j)
    ans = []
    while i < j:
        i = s.find(c,i,j+1)
        if i != -1:
            ans.append(j-i+1)
            i += 1
        else:
            break
    return ans
def getll(s1,s2,i,j,limit):
    ll = 0
    while 0 <= i and 0 <= j and s1[i] == s2[j] and limit > 0:
        ll += 1
        i -= 1
        j -= 1
        limit -= 1
    return ll
dp = {}
def f(s1,s2,i,j,l):
    print(i,j,l)
    if (i,j,l) not in dp:
        if l is 0:
            dp[(i,j,l)] = True
        elif l is 1:
            dp[(i,j,l)] = s1[i] == s2[j]
        else:
            if s1[i] == s2[j]:
                dp[(i,j,l)] = f(s1,s2,i-1,j-1,l-1)
            else:
                s2_start_index = j+1-l
                s2_end_index = j
                lens = allOcuurLenFromEnd(s2,s1[i],s2_start_index,s2_end_index)
                print("lens",lens)
                for x in lens:
                    occur_in_s2 = j-x+1
                    pos_to_find_ll_in_s2_from = occur_in_s2-1
                    pos_to_find_ll_in_s1_from = i-1
                    ll = getll(s1,s2,pos_to_find_ll_in_s1_from,pos_to_find_ll_in_s2_from,l-1)
                    lhs_len = l-x-ll
                    print("ll",ll)
                    lhs_s1_end = i-x-ll
                    lhs_s2_end = j-x-ll
                    rhs_len = x - 1 
                    rhs_s1_end = i-1-ll
                    rhs_s2_end = j
                    ret = f(s1,s2,lhs_s1_end,lhs_s2_end,lhs_len) and f(s1,s2,rhs_s1_end,rhs_s2_end,rhs_len)
                    print("ret",ret,(i-x,j-x,l-x),(i-1,j,x-1))
                    if ret:
                        dp[(i,j,l)] = ret
                        return ret
                dp[(i,j,l)] = False
    print((i,j,l),dp[(i,j,l)])
    return dp[(i,j,l)]
        
class Solution:
    def isScramble(self, s1: str, s2: str) -> bool:
        dp.clear()
        return f(s1,s2,len(s1)-1,len(s2)-1,len(s1))

Ver2: AC

case 有點多

  1. 交叉
  2. 直的
  3. 前或後有一樣的字
class Solution:
    def isScramble(self, s1: str, s2: str) -> bool:
        dp = {}
        if len(s1) != len(s2):
            return False
        #s1, s2 = removePrefix_postfix(s1,s2)
        #if not s1 and not s2:
        #    return True
        for i in range(0,len(s1)):
            for j in range(0,len(s1)):
                dp[(i,j,1)] = s1[i] == s2[j]
        for l in range(2,len(s1)+1):
            for i in range(0,len(s1)):
                for j in range(0,len(s1)):
                    if i-l < 0 or j-l < 0:
                        dp[(i,j,l)] = False
        for l in range(2,len(s1)+1):
            for x in range(1,l):
                for i in range(1,len(s1)):
                    if i-x >= 0:
                        for j in range(1,len(s1)):
                            if j-(l-x) >= 0:
                                if (i,j,l) not in dp:
                                    dp[(i,j,l)] = False
                                dp[(i,j,l)] = dp[(i,j,l)] or (i-l+1,j-l+1,1) in dp and dp[(i-l+1,j-l+1,1)] and dp[(i,j,l-1)]
                                dp[(i,j,l)] = dp[(i,j,l)] or (dp[(i,j-(l-x),x)] and dp[(i-x,j,l-x)]) or (dp[(i,j,1)] and dp[(i-1,j-1,l-1)]) or (i-(l-x) >= 0 and j-(l-x) >= 0 and dp[(i,j,(l-x))] and dp[(i-(l-x),j-(l-x),x)])
                                #print((i,j,l),dp[(i,j,l)],(i,j-(l-x),x), dp[(i,j-(l-x),x)] ,(i-x,j,l-x), dp[(i-x,j,l-x)])
        return dp[(len(s1)-1,len(s2)-1,len(s1))]

Sol

照題目走

def cache_computing(f,is_computing=False):
    mem = {}
    @functools.wraps(f)
    def wrapper(*args,**kwds):
        if args not in mem:
            mem[args] = is_computing
            mem[args] = f(*args,**kwds)
        return mem[args]
    return wrapper
class Solution:
    @cache_computing
    def isScramble(self, s1: str, s2: str) -> bool:
        #print(s1,s2)
        if not s1 and not s2:
            return True
        elif s1[0] == s2[0]:
            return self.isScramble(s1[1:],s2[1:])
        elif len(s1) == 1:
            return False
        else:
            return any([(self.isScramble(s1[k:],s2[k:]) and self.isScramble(s1[:k],s2[:k])) or (self.isScramble(s1[k:],s2[:len(s2)-k]) and self.isScramble(s1[:k],s2[len(s2)-k:])) for k in range(1,len(s1))])