ShaysXIA/CodeInsightBench huggingface dataset:
Problem: 2050D - Digital string maximization
Language: python

Solution 1 (runtime: 1171 ms):

class LinkedList:
    def __init__(self, l):
        self.list = l
        self.len = len(l)
        self.next = [i+1 for i in range(self.len)]
        self.next[-1] = None
        self.prev = [i-1 for i in range(self.len)]
        self.prev[0] = None
        self.head = 0
        self.current = self.head
    
    def to_head(self):
        self.current = self.head
        
    def pop(self, i=None):
        if i == None:
            self.current = self.head
        else:
            self.current = i
        next_ = self.next[self.current]
        prev_ = self.prev[self.current]
        if self.len == 1:
            self.head=None  
        elif self.current == self.head:
            self.head = self.next[self.head]
            self.prev[self.head] = None
        elif next_ == None:
            self.next[prev_] = None 
        else:
            self.next[prev_] = next_
            self.prev[next_] = prev_
        self.len -= 1
        
    def to_next(self):
        self.current = self.next[self.current]
        
    def elem(self):
        return self.list[self.current]
    
    def print_(self):
        i = self.head
        data = []
        data_n = []
        while i != None:
            data.append(self.list[i])
            data_n.append(self.next[i])
            i = self.next[i]
        print(data, data_n)
    
        


t = int(input())
for k in range(t):
    s = LinkedList(list(map(int,list(str(input())))))
    res = ''
    n = s.len
    count_masked = 0
    i = 0
    #s.print_()
    while s.len > 0:
        s.to_head()
        cost=1
        max_v=s.elem()
        max_j=-1
        s.to_next()
        while 9 - (s.list[s.head] + cost) > 0 and s.current != None:
            if s.list[s.current] - cost > max_v:
                max_v = s.list[s.current] - cost
                max_j = s.current
            cost+=1
            s.to_next()
        if max_j != -1:
            res+=str(max_v)
            s.pop(max_j)
        else:
            res+=str(s.list[s.head])
            s.pop()
        #s.print_()
    print(res)

Solution 2 (runtime: 234 ms):
n = int(input())

for x in range(n):

s= list(map(int, input()))

g = len(s)

for i in range(1,g):

for j in range(0, s[i]):

if(i-j-1 < 0):

break

if(s[i-j]-1 > s[i-j-1]):

temp = s[i-j-1]

s[i-j-1] = s[i-j]-1

s[i-j] = temp

else:

break

result = ''.join(map(str,s))

print(result)
Problem: 2050D - Digital string maximization
Language: python

Solution 1 (runtime: 1171 ms):

class LinkedList:
    def __init__(self, l):
        self.list = l
        self.len = len(l)
        self.next = [i+1 for i in range(self.len)]
        self.next[-1] = None
        self.prev = [i-1 for i in range(self.len)]
        self.prev[0] = None
        self.head = 0
        self.current = self.head
    
    def to_head(self):
        self.current = self.head
        
    def pop(self, i=None):
        if i == None:
            self.current = self.head
        else:
            self.current = i
        next_ = self.next[self.current]
        prev_ = self.prev[self.current]
        if self.len == 1:
            self.head=None  
        elif self.current == self.head:
            self.head = self.next[self.head]
            self.prev[self.head] = None
        elif next_ == None:
            self.next[prev_] = None 
        else:
            self.next[prev_] = next_
            self.prev[next_] = prev_
        self.len -= 1
        
    def to_next(self):
        self.current = self.next[self.current]
        
    def elem(self):
        return self.list[self.current]
    
    def print_(self):
        i = self.head
        data = []
        data_n = []
        while i != None:
            data.append(self.list[i])
            data_n.append(self.next[i])
            i = self.next[i]
        print(data, data_n)
    
        


t = int(input())
for k in range(t):
    s = LinkedList(list(map(int,list(str(input())))))
    res = ''
    n = s.len
    count_masked = 0
    i = 0
    #s.print_()
    while s.len > 0:
        s.to_head()
        cost=1
        max_v=s.elem()
        max_j=-1
        s.to_next()
        while 9 - (s.list[s.head] + cost) > 0 and s.current != None:
            if s.list[s.current] - cost > max_v:
                max_v = s.list[s.current] - cost
                max_j = s.current
            cost+=1
            s.to_next()
        if max_j != -1:
            res+=str(max_v)
            s.pop(max_j)
        else:
            res+=str(s.list[s.head])
            s.pop()
        #s.print_()
    print(res)

Solution 2 (runtime: 156 ms):
for _ in range(int(input())):

a = input()

s = [int(digit) for digit in a]

n = len(s)

for i in range(n):

best_value = s[i]

best_idx = -1

for j in range(i + 1, min(i + 9, n)):

score = s[j]-(j - i)

if score>best_value:

best_value=score

best_idx=j



if best_idx != -1:

# print(best_value,s[best_idx],i,best_idx)

s[i], s[best_idx] = best_value, s[i]

for j in range(best_idx, i+1, -1):

s[j], s[j - 1] = s[j - 1], s[j]

# print(s)



print("".join(map(str, s)))
Problem: 2050D - Digital string maximization
Language: python

Solution 1 (runtime: 1171 ms):

class LinkedList:
    def __init__(self, l):
        self.list = l
        self.len = len(l)
        self.next = [i+1 for i in range(self.len)]
        self.next[-1] = None
        self.prev = [i-1 for i in range(self.len)]
        self.prev[0] = None
        self.head = 0
        self.current = self.head
    
    def to_head(self):
        self.current = self.head
        
    def pop(self, i=None):
        if i == None:
            self.current = self.head
        else:
            self.current = i
        next_ = self.next[self.current]
        prev_ = self.prev[self.current]
        if self.len == 1:
            self.head=None  
        elif self.current == self.head:
            self.head = self.next[self.head]
            self.prev[self.head] = None
        elif next_ == None:
            self.next[prev_] = None 
        else:
            self.next[prev_] = next_
            self.prev[next_] = prev_
        self.len -= 1
        
    def to_next(self):
        self.current = self.next[self.current]
        
    def elem(self):
        return self.list[self.current]
    
    def print_(self):
        i = self.head
        data = []
        data_n = []
        while i != None:
            data.append(self.list[i])
            data_n.append(self.next[i])
            i = self.next[i]
        print(data, data_n)
    
        


t = int(input())
for k in range(t):
    s = LinkedList(list(map(int,list(str(input())))))
    res = ''
    n = s.len
    count_masked = 0
    i = 0
    #s.print_()
    while s.len > 0:
        s.to_head()
        cost=1
        max_v=s.elem()
        max_j=-1
        s.to_next()
        while 9 - (s.list[s.head] + cost) > 0 and s.current != None:
            if s.list[s.current] - cost > max_v:
                max_v = s.list[s.current] - cost
                max_j = s.current
            cost+=1
            s.to_next()
        if max_j != -1:
            res+=str(max_v)
            s.pop(max_j)
        else:
            res+=str(s.list[s.head])
            s.pop()
        #s.print_()
    print(res)

Solution 2 (runtime: 171 ms):
mod = 1e9 + 7

for _ in range(int(input())):

s = list(map(int,list(input())))

for i in range(len(s)-1):

mxi = -1

mx = 0

for j in range(1,10):

if i + j >= len(s):

break

if int(s[i+j]) - j - int(s[i]) > mx:

mx = int(s[i+j]) - j - int(s[i])

mxi = j

if mxi == -1:

continue



for jj in range(i+mxi,i,-1):

s[jj],s[jj-1] = s[jj-1],s[jj] - 1



print(''.join(list(map(str,s))))
Problem: 2050D - Digital string maximization
Language: python

Solution 1 (runtime: 1171 ms):

class LinkedList:
    def __init__(self, l):
        self.list = l
        self.len = len(l)
        self.next = [i+1 for i in range(self.len)]
        self.next[-1] = None
        self.prev = [i-1 for i in range(self.len)]
        self.prev[0] = None
        self.head = 0
        self.current = self.head
    
    def to_head(self):
        self.current = self.head
        
    def pop(self, i=None):
        if i == None:
            self.current = self.head
        else:
            self.current = i
        next_ = self.next[self.current]
        prev_ = self.prev[self.current]
        if self.len == 1:
            self.head=None  
        elif self.current == self.head:
            self.head = self.next[self.head]
            self.prev[self.head] = None
        elif next_ == None:
            self.next[prev_] = None 
        else:
            self.next[prev_] = next_
            self.prev[next_] = prev_
        self.len -= 1
        
    def to_next(self):
        self.current = self.next[self.current]
        
    def elem(self):
        return self.list[self.current]
    
    def print_(self):
        i = self.head
        data = []
        data_n = []
        while i != None:
            data.append(self.list[i])
            data_n.append(self.next[i])
            i = self.next[i]
        print(data, data_n)
    
        


t = int(input())
for k in range(t):
    s = LinkedList(list(map(int,list(str(input())))))
    res = ''
    n = s.len
    count_masked = 0
    i = 0
    #s.print_()
    while s.len > 0:
        s.to_head()
        cost=1
        max_v=s.elem()
        max_j=-1
        s.to_next()
        while 9 - (s.list[s.head] + cost) > 0 and s.current != None:
            if s.list[s.current] - cost > max_v:
                max_v = s.list[s.current] - cost
                max_j = s.current
            cost+=1
            s.to_next()
        if max_j != -1:
            res+=str(max_v)
            s.pop(max_j)
        else:
            res+=str(s.list[s.head])
            s.pop()
        #s.print_()
    print(res)

Solution 2 (runtime: 218 ms):
t =  int(input())

def solve():

s = [int(i) for i in list(input())]

n = len(s)

for i in range(1,n,1):

for j in range(max(i-10, 0), i):

if s[i] - (i-j) > s[j]:

tmp = s[j:i]

s[j] = s[i] - (i-j)

s[j+1:i+1] = tmp

break

print("".join([str(i) for i in s]))

return

for i in range(t):

solve()
Problem: 2050D - Digital string maximization
Language: python

Solution 1 (runtime: 1171 ms):

class LinkedList:
    def __init__(self, l):
        self.list = l
        self.len = len(l)
        self.next = [i+1 for i in range(self.len)]
        self.next[-1] = None
        self.prev = [i-1 for i in range(self.len)]
        self.prev[0] = None
        self.head = 0
        self.current = self.head
    
    def to_head(self):
        self.current = self.head
        
    def pop(self, i=None):
        if i == None:
            self.current = self.head
        else:
            self.current = i
        next_ = self.next[self.current]
        prev_ = self.prev[self.current]
        if self.len == 1:
            self.head=None  
        elif self.current == self.head:
            self.head = self.next[self.head]
            self.prev[self.head] = None
        elif next_ == None:
            self.next[prev_] = None 
        else:
            self.next[prev_] = next_
            self.prev[next_] = prev_
        self.len -= 1
        
    def to_next(self):
        self.current = self.next[self.current]
        
    def elem(self):
        return self.list[self.current]
    
    def print_(self):
        i = self.head
        data = []
        data_n = []
        while i != None:
            data.append(self.list[i])
            data_n.append(self.next[i])
            i = self.next[i]
        print(data, data_n)
    
        


t = int(input())
for k in range(t):
    s = LinkedList(list(map(int,list(str(input())))))
    res = ''
    n = s.len
    count_masked = 0
    i = 0
    #s.print_()
    while s.len > 0:
        s.to_head()
        cost=1
        max_v=s.elem()
        max_j=-1
        s.to_next()
        while 9 - (s.list[s.head] + cost) > 0 and s.current != None:
            if s.list[s.current] - cost > max_v:
                max_v = s.list[s.current] - cost
                max_j = s.current
            cost+=1
            s.to_next()
        if max_j != -1:
            res+=str(max_v)
            s.pop(max_j)
        else:
            res+=str(s.list[s.head])
            s.pop()
        #s.print_()
    print(res)

Solution 2 (runtime: 906 ms):
def solve():

s = list(input())

n = len(s)

while True:

changed = False

for i in range(1, n):

if s[i] != '0' and int(s[i]) - 1 > int(s[i-1]):

num = int(s[i])

s[i] = str(int(s[i-1]))

s[i-1] = str(num - 1)

changed = True

if not changed:

break

print("".join(s))



t = int(input())

for _ in range(t):

solve()
Problem: 2050D - Digital string maximization
Language: python

Solution 1 (runtime: 1171 ms):

class LinkedList:
    def __init__(self, l):
        self.list = l
        self.len = len(l)
        self.next = [i+1 for i in range(self.len)]
        self.next[-1] = None
        self.prev = [i-1 for i in range(self.len)]
        self.prev[0] = None
        self.head = 0
        self.current = self.head
    
    def to_head(self):
        self.current = self.head
        
    def pop(self, i=None):
        if i == None:
            self.current = self.head
        else:
            self.current = i
        next_ = self.next[self.current]
        prev_ = self.prev[self.current]
        if self.len == 1:
            self.head=None  
        elif self.current == self.head:
            self.head = self.next[self.head]
            self.prev[self.head] = None
        elif next_ == None:
            self.next[prev_] = None 
        else:
            self.next[prev_] = next_
            self.prev[next_] = prev_
        self.len -= 1
        
    def to_next(self):
        self.current = self.next[self.current]
        
    def elem(self):
        return self.list[self.current]
    
    def print_(self):
        i = self.head
        data = []
        data_n = []
        while i != None:
            data.append(self.list[i])
            data_n.append(self.next[i])
            i = self.next[i]
        print(data, data_n)
    
        


t = int(input())
for k in range(t):
    s = LinkedList(list(map(int,list(str(input())))))
    res = ''
    n = s.len
    count_masked = 0
    i = 0
    #s.print_()
    while s.len > 0:
        s.to_head()
        cost=1
        max_v=s.elem()
        max_j=-1
        s.to_next()
        while 9 - (s.list[s.head] + cost) > 0 and s.current != None:
            if s.list[s.current] - cost > max_v:
                max_v = s.list[s.current] - cost
                max_j = s.current
            cost+=1
            s.to_next()
        if max_j != -1:
            res+=str(max_v)
            s.pop(max_j)
        else:
            res+=str(s.list[s.head])
            s.pop()
        #s.print_()
    print(res)

Solution 2 (runtime: 265 ms):
for _ in range(int(input())):

s = list(map(int,input()))

for i in range(9):

for j in range(len(s)-1):

if s[j] < s[j+1] - 1:

s[j],s[j+1] = s[j+1]-1,s[j]

print(*s,sep= '')
Problem: 2050D - Digital string maximization
Language: python

Solution 1 (runtime: 718 ms):
t=int(input())
for _ in range(t):
    s=input()
    l=list(map(int,s))
    for i in range(1,len(s)):
        while i>0 and l[i]>l[i-1]+1:
            l[i],l[i-1]=l[i-1],l[i]-1
            i-=1
    print(''.join(map(str,l)))


Solution 2 (runtime: 1202 ms):
t = int(input())



for _ in range(t):

    word = input()

    s = [int(digit) for digit in word]

    l = len(s)



    for i in range(l):

        findMax= s[i]

        maxPos = i

        for j in range(10):

            if s[i+j]>findMax+((j+i)-maxPos):

                findMax = s[i+j]

                maxPos = i+j

            

            if l-(i+j)-1==0:

                break



        for j in range(maxPos-i):

            temp = s[maxPos-1]

            s[maxPos-1] = s[maxPos]-1

            s[maxPos] = temp

            maxPos-=1



    print("".join([str(digit) for digit in s]))
Problem: 2050D - Digital string maximization
Language: python

Solution 1 (runtime: 718 ms):
t=int(input())
for _ in range(t):
    s=input()
    l=list(map(int,s))
    for i in range(1,len(s)):
        while i>0 and l[i]>l[i-1]+1:
            l[i],l[i-1]=l[i-1],l[i]-1
            i-=1
    print(''.join(map(str,l)))


Solution 2 (runtime: 203 ms):
t= int(input())
for k in range(t):    
    a=list(input())
    b= len(a)
    for i in range(b):
        t= min(i+9,b)
        m= int(a[i])
        n=i
        for j in range(i,t):
            if (int(a[j])-j+i)> m:
                m= int(a[j])-j+i
                n= j
        while n>i:
            a[n],a[n-1]= a[n-1],str(int(a[n])-1)
            n-=1
    print("".join(a))
Problem: 2050D - Digital string maximization
Language: python

Solution 1 (runtime: 718 ms):
t=int(input())
for _ in range(t):
    s=input()
    l=list(map(int,s))
    for i in range(1,len(s)):
        while i>0 and l[i]>l[i-1]+1:
            l[i],l[i-1]=l[i-1],l[i]-1
            i-=1
    print(''.join(map(str,l)))


Solution 2 (runtime: 702 ms):
n = int(input())

res = []

for _ in range(n):

    s = input()

    temp = []

    for ch in s:

        temp.append(int(ch))

    for i in range(1, len(s)):

        j = i - 1

        while j >= 0:

            r = temp[j+1]

            l = temp[j]

            if r > l + 1:

                temp[j] = r - 1

                temp[j+1] = l

            else:

                break

            j -= 1

    r = ''

    for n in temp:

        r += str(n)

    res.append(r)



for val in res:

    print(val)
Problem: 2050D - Digital string maximization
Language: python

Solution 1 (runtime: 718 ms):
t=int(input())
for _ in range(t):
    s=input()
    l=list(map(int,s))
    for i in range(1,len(s)):
        while i>0 and l[i]>l[i-1]+1:
            l[i],l[i-1]=l[i-1],l[i]-1
            i-=1
    print(''.join(map(str,l)))


Solution 2 (runtime: 218 ms):
for _ in range(int(input())):

    s=list(input())

    #print(s)

    for i in range(len(s)):

        val=str(s[i])

        index=i



        for j in range(i+1,min(i+10,len(s))):

            if (int(s[j])-(j-i))>int(val):

                val=str(int(s[j])-(j-i))

                index=j 

        

        if index!=i:

            prev=s[i]

            for j in range(i+1,index+1):

                x=s[j]

                s[j]=prev

                prev=x

                

            s[i]=val

    print("".join(s))
Problem: 2050D - Digital string maximization
Language: python

Solution 1 (runtime: 718 ms):
t=int(input())
for _ in range(t):
    s=input()
    l=list(map(int,s))
    for i in range(1,len(s)):
        while i>0 and l[i]>l[i-1]+1:
            l[i],l[i-1]=l[i-1],l[i]-1
            i-=1
    print(''.join(map(str,l)))


Solution 2 (runtime: 312 ms):
import math
from decimal import Decimal, getcontext
import itertools
import queue
from collections import Counter, deque
import bisect
import sys


def list_read(t: type = int):
    return list(map(t, input().split()))


def params_read(t: type = int):
    return map(t, input().split())


def pprint(A):
    if len(A) > 0:
        if type(A[0]) != list:
            A = [A]
    for i in range(len(A)):
        print(*A[i])


def sieve(N):
    primes = [i for i in range(N + 1)]
    primes[1] = 0
    i = 2
    while i <= N:
        if primes[i] != 0:
            j = i + i
            while j <= N:
                primes[j] = 0
                j = j + i
        i += 1
    primes = [i for i in primes if i != 0]

    return primes


def sum_of_the_digits(a):
    return sum(map(int, str(a)))


def binpow(a, n, mod):
    res = 1
    while n != 0:
        if n & 1:
            res = res * a % mod
        a = a * a % mod
        n >>= 1
    return res


def get_next_symbol(a, p=26):
    ABC = list(".abcdefghijklmnopqrstuvwxyz{")
    a = ord(a)
    a += 1
    a = chr(a)
    p = ABC[p + 1]
    if a == p:
        a = "a"
    return a


def Cnk(n, k, MOD):
    if k > n:
        return 0
    if k == 0 or k == n:
        return 1
    k = min(k, n - k)
    c = 1
    for i in range(k):
        c = c * (n - i) // (i + 1)
        c %= MOD
    return c


def alg_sum(a, b, n):
    return int((a + b) / 2 * n)  # int(Decimal(a + b)/2 * n)


def get_factors(n):
    factors = []

    i = 1
    while i * i < n:
        if n % i == 0:
            factors.append(i)
            factors.append(n // i)
        i += 1

    if i * i == n:
        factors.append(i)

    return factors


def factorize(n):
    primes = []
    while n % 2 == 0:
        primes.append(2)
        n //= 2

    i = 3
    while i * i <= n:
        while n % i == 0:
            primes.append(i)
            n //= i
        i += 2

    if n > 1:
        primes.append(n)

    return primes


def fact_for_binom(n, MOD):
    # (f[N] * invf[K] * invf[N - K]) % MOD

    f = [1]
    invf = [1]

    for i in range(1, n):
        f.append(f[-1] * i % MOD)
        f[-1] %= MOD

        invf.append(binpow(f[-1], MOD - 2, MOD))
        invf[-1] %= MOD
    return f, invf


# n = int(input())
# MOD = 1000000007
def solve():
    A = list(input())

    for i in range(len(A)):
        curr_d = int(A[i])
        pos = i
        for j in range(i+1, min(len(A), i + 10)):
            new_d = int(A[j]) - j + i
            if new_d > curr_d:
                curr_d = new_d
                pos = j
        A[pos] = str(curr_d)
        for j in range(pos-1, i-1, -1):
            A[j], A[j+1] = A[j+1], A[j]
    print("".join(A))

def main():
    t = 1
    t = int(input())
    for i in range(t):
        solve()


main()
Problem: 2050D - Digital string maximization
Language: python

Solution 1 (runtime: 718 ms):
t=int(input())
for _ in range(t):
    s=input()
    l=list(map(int,s))
    for i in range(1,len(s)):
        while i>0 and l[i]>l[i-1]+1:
            l[i],l[i-1]=l[i-1],l[i]-1
            i-=1
    print(''.join(map(str,l)))


Solution 2 (runtime: 624 ms):
def solve():

    s = list(input().strip())  # Convert string to list for mutability

    n = len(s)



    for i in range(1, n):

        j, curr = i - 1, i

        while j >= 0 and s[curr] > '0' and s[curr] > chr(ord(s[j]) + 1):

            s[curr] = chr(ord(s[curr]) - 1)  # Decrease digit

            s[curr], s[j] = s[j], s[curr]  # Swap

            curr -= 1

            j = curr - 1



    print("".join(s))



# Handling multiple test cases

t = int(input())

for _ in range(t):

    solve()
Problem: 2050D - Digital string maximization
Language: python

Solution 1 (runtime: 718 ms):
t=int(input())
for _ in range(t):
    s=input()
    l=list(map(int,s))
    for i in range(1,len(s)):
        while i>0 and l[i]>l[i-1]+1:
            l[i],l[i-1]=l[i-1],l[i]-1
            i-=1
    print(''.join(map(str,l)))


Solution 2 (runtime: 234 ms):
n = int(input())

for x in range(n):

s= list(map(int, input()))

g = len(s)

for i in range(1,g):

for j in range(0, s[i]):

if(i-j-1 < 0):

break

if(s[i-j]-1 > s[i-j-1]):

temp = s[i-j-1]

s[i-j-1] = s[i-j]-1

s[i-j] = temp

else:

break

result = ''.join(map(str,s))

print(result)
Problem: 2050D - Digital string maximization
Language: python

Solution 1 (runtime: 718 ms):
t=int(input())
for _ in range(t):
    s=input()
    l=list(map(int,s))
    for i in range(1,len(s)):
        while i>0 and l[i]>l[i-1]+1:
            l[i],l[i-1]=l[i-1],l[i]-1
            i-=1
    print(''.join(map(str,l)))


Solution 2 (runtime: 156 ms):
for _ in range(int(input())):

a = input()

s = [int(digit) for digit in a]

n = len(s)

for i in range(n):

best_value = s[i]

best_idx = -1

for j in range(i + 1, min(i + 9, n)):

score = s[j]-(j - i)

if score>best_value:

best_value=score

best_idx=j



if best_idx != -1:

# print(best_value,s[best_idx],i,best_idx)

s[i], s[best_idx] = best_value, s[i]

for j in range(best_idx, i+1, -1):

s[j], s[j - 1] = s[j - 1], s[j]

# print(s)



print("".join(map(str, s)))