message stringlengths 2 23.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 97 109k | cluster float64 0 0 | __index_level_0__ int64 194 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a positive even number N.
Find the number of strings s of length N consisting of `A`, `B`, and `C` that satisfy the following condition:
* s can be converted to the empty string by repeating the following operation:
* Choose two consecutive characters in s and erase them. However, choosing `AB` or `BA` is not allowed.
For example, `ABBC` satisfies the condition for N=4, because we can convert it as follows: `ABBC` → (erase `BB`) → `AC` → (erase `AC`) → `(empty)`.
The answer can be enormous, so compute the count modulo 998244353.
Constraints
* 2 \leq N \leq 10^7
* N is an even number.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of strings that satisfy the conditions, modulo 998244353.
Examples
Input
2
Output
7
Input
10
Output
50007
Input
1000000
Output
210055358
Submitted Solution:
```
n=int(input());l=[0,1];a=0;b=c=1;p=998244353
for i in range(2,n):l+=[l[p%i]*(p-p//i)%p]
for i in range(n,n//2,-1):a+=b*c;b+=b%p;c=c*i*l[n+1-i]%p
print((pow(3,n,p)-2*a)%p)
``` | instruction | 0 | 89,907 | 0 | 179,814 |
Yes | output | 1 | 89,907 | 0 | 179,815 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a positive even number N.
Find the number of strings s of length N consisting of `A`, `B`, and `C` that satisfy the following condition:
* s can be converted to the empty string by repeating the following operation:
* Choose two consecutive characters in s and erase them. However, choosing `AB` or `BA` is not allowed.
For example, `ABBC` satisfies the condition for N=4, because we can convert it as follows: `ABBC` → (erase `BB`) → `AC` → (erase `AC`) → `(empty)`.
The answer can be enormous, so compute the count modulo 998244353.
Constraints
* 2 \leq N \leq 10^7
* N is an even number.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of strings that satisfy the conditions, modulo 998244353.
Examples
Input
2
Output
7
Input
10
Output
50007
Input
1000000
Output
210055358
Submitted Solution:
```
"""
Writer: SPD_9X2
https://atcoder.jp/contests/agc040/tasks/agc040_c
奇数文字目と偶数文字目がセットで消える
奇数文字目のABを反転すると、AB or BA or AC or BC or CC で消えていいことになる
すなわち、異なる文字の切れ目では必ず消せる
異なる切れ目は必ず存在するので、Cを適当にABに割り振った時に全て消せるかどうかが問題になる
A,Bのみの時に全て消せる条件は 両者の数が等しい事
Cを適切に割り振った時に両者の数を等しくできる必要十分条件は max(a,b) <= N//2
あとはこれを数えればよい→どうやって？
全ての並び方は 3**N ここから補集合を引く？
Aが半数を超える(k個)の時のおき方は、 NCk * (2**(N-k))で求まる
Bの時も同様に。
"""
def modfac(n, MOD):
f = 1
factorials = [1]
for m in range(1, n + 1):
f *= m
f %= MOD
factorials.append(f)
inv = pow(f, MOD - 2, MOD)
invs = [1] * (n + 1)
invs[n] = inv
for m in range(n, 1, -1):
inv *= m
inv %= MOD
invs[m - 1] = inv
return factorials, invs
def modnCr(n,r,mod,fac,inv):
return fac[n] * inv[n-r] * inv[r] % mod
N = int(input())
mod = 998244353
fac,inv = modfac(N+10,mod)
ans = pow(3,N,mod)
tpow = [1]
for i in range(N//2+10):
tpow.append(tpow[-1]*2%mod)
for k in range(N//2+1,N+1):
now = 2 * modnCr(N,k,mod,fac,inv) * tpow[N-k]
ans -= now
ans %= mod
print (ans)
``` | instruction | 0 | 89,908 | 0 | 179,816 |
Yes | output | 1 | 89,908 | 0 | 179,817 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a positive even number N.
Find the number of strings s of length N consisting of `A`, `B`, and `C` that satisfy the following condition:
* s can be converted to the empty string by repeating the following operation:
* Choose two consecutive characters in s and erase them. However, choosing `AB` or `BA` is not allowed.
For example, `ABBC` satisfies the condition for N=4, because we can convert it as follows: `ABBC` → (erase `BB`) → `AC` → (erase `AC`) → `(empty)`.
The answer can be enormous, so compute the count modulo 998244353.
Constraints
* 2 \leq N \leq 10^7
* N is an even number.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of strings that satisfy the conditions, modulo 998244353.
Examples
Input
2
Output
7
Input
10
Output
50007
Input
1000000
Output
210055358
Submitted Solution:
```
import numpy as np
def prepare(n, MOD):
f = 1
for m in range(1, n + 1):
f = f * m % MOD
fn = f
inv = pow(f, MOD - 2, MOD)
invs = [1] * (n + 1)
invs[n] = inv
for m in range(n, 1, -1):
inv = invs[m - 1] = inv * m % MOD
invs = np.array(invs, dtype=np.int64)
return fn, invs
n = int(input())
MOD = 998244353
fn, invs = prepare(n, MOD)
n2 = n // 2
# mul = 1
# muls = [0] * n2
# for i in range(n2):
# mul = muls[i] = mul * 2 % MOD
# mul = 1
# muls = np.zeros(n2, dtype=np.int64)
# for i in range(n2):
# mul = muls[i] = mul * 2 % MOD
muls = np.zeros(1, dtype=np.int64)
muls[0] = 2
while muls.size < n2:
muls = np.append(muls, muls * muls[-1] % MOD)
muls = muls[:n2]
impossibles = invs[:n2] * invs[n:n2:-1] % MOD
impossibles = impossibles * muls % MOD
impossible = sum(list(impossibles)) % MOD * fn
print((pow(3, n, MOD) - impossible) % MOD)
``` | instruction | 0 | 89,909 | 0 | 179,818 |
No | output | 1 | 89,909 | 0 | 179,819 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a positive even number N.
Find the number of strings s of length N consisting of `A`, `B`, and `C` that satisfy the following condition:
* s can be converted to the empty string by repeating the following operation:
* Choose two consecutive characters in s and erase them. However, choosing `AB` or `BA` is not allowed.
For example, `ABBC` satisfies the condition for N=4, because we can convert it as follows: `ABBC` → (erase `BB`) → `AC` → (erase `AC`) → `(empty)`.
The answer can be enormous, so compute the count modulo 998244353.
Constraints
* 2 \leq N \leq 10^7
* N is an even number.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of strings that satisfy the conditions, modulo 998244353.
Examples
Input
2
Output
7
Input
10
Output
50007
Input
1000000
Output
210055358
Submitted Solution:
```
N = int(input())
MOD = 998244353
ans = 0
d = 1
b = 1
for i in range((N//2)):
ans += d * b
b *= 2
b %= MOD
ans %= MOD
d = ((N-i)*d)%MOD
d = (d*pow(i+1,MOD-2,MOD))%MOD
print((pow(3,N,MOD)-ans*2)%MOD)
``` | instruction | 0 | 89,910 | 0 | 179,820 |
No | output | 1 | 89,910 | 0 | 179,821 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a positive even number N.
Find the number of strings s of length N consisting of `A`, `B`, and `C` that satisfy the following condition:
* s can be converted to the empty string by repeating the following operation:
* Choose two consecutive characters in s and erase them. However, choosing `AB` or `BA` is not allowed.
For example, `ABBC` satisfies the condition for N=4, because we can convert it as follows: `ABBC` → (erase `BB`) → `AC` → (erase `AC`) → `(empty)`.
The answer can be enormous, so compute the count modulo 998244353.
Constraints
* 2 \leq N \leq 10^7
* N is an even number.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of strings that satisfy the conditions, modulo 998244353.
Examples
Input
2
Output
7
Input
10
Output
50007
Input
1000000
Output
210055358
Submitted Solution:
```
#!/usr/bin/env python3
import sys
sys.setrecursionlimit(10**8)
input = sys.stdin.readline
MOD = 998244353
MAX_N = 10**7
fac = [1] + [0] * MAX_N
fac_inv = [1] + [0] * MAX_N
mod_pow2_n = [1] + [0] * MAX_N
f = 1
finv = 1
p2 = 1
for i in range(1, MAX_N+1):
# fac[i] = fac[i-1] * i % MOD
f *= i
f %= MOD
fac[i] = f
# Fermat's little theorem says
# a**(p-1) mod p == 1
# then, a * a**(p-2) mod p == 1
# it means a**(p-2) is inverse element
# fac_inv[i] = fac_inv[i-1] * pow(i, MOD-2, MOD) % MOD
finv *= pow(i, MOD-2, MOD)
finv %= MOD
fac_inv[i] = finv
p2 *= 2
p2 %= MOD
mod_pow2_n[i] = p2
def mod_nCr(n, r):
if n < r or n < 0 or r < 0:
return 0
tmp = fac_inv[n-r] * fac_inv[r] % MOD
return tmp * fac[n] % MOD
def single_mod_nCr(n, r):
if n < r or n < 0 or r < 0:
return 0
if r > n - r:
r = n - r
ret = 1
for i in range(r):
ret *= n - i
ret *= pow(i+1, MOD-2, MOD)
ret %= MOD
return ret
n = int(input())
ans = 0
for i in range(n//2+1, n+1):
ans += mod_nCr(n, i) * mod_pow2_n[n-i]
ans %= MOD
print((pow(3, n, MOD) - ans * 2 + MOD)%MOD)
``` | instruction | 0 | 89,911 | 0 | 179,822 |
No | output | 1 | 89,911 | 0 | 179,823 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a positive even number N.
Find the number of strings s of length N consisting of `A`, `B`, and `C` that satisfy the following condition:
* s can be converted to the empty string by repeating the following operation:
* Choose two consecutive characters in s and erase them. However, choosing `AB` or `BA` is not allowed.
For example, `ABBC` satisfies the condition for N=4, because we can convert it as follows: `ABBC` → (erase `BB`) → `AC` → (erase `AC`) → `(empty)`.
The answer can be enormous, so compute the count modulo 998244353.
Constraints
* 2 \leq N \leq 10^7
* N is an even number.
Input
Input is given from Standard Input in the following format:
N
Output
Print the number of strings that satisfy the conditions, modulo 998244353.
Examples
Input
2
Output
7
Input
10
Output
50007
Input
1000000
Output
210055358
Submitted Solution:
```
N=int(input())
mod=998244353
FACT=[1]
for i in range(1,N+1):
FACT.append(FACT[-1]*i%mod)
FACT_INV=[pow(FACT[-1],mod-2,mod)]
for i in range(N,0,-1):
FACT_INV.append(FACT_INV[-1]*i%mod)
FACT_INV.reverse()
def Combi(a,b):
if 0<=b<=a:
return FACT[a]*FACT_INV[b]*FACT_INV[a-b]%mod
SC=0
for i in range(N//2+1,N+1):
SC+=Combi(N,i)*pow(2,N-i,mod)
print((pow(3,N,mod)-SC*2)%mod)
``` | instruction | 0 | 89,912 | 0 | 179,824 |
No | output | 1 | 89,912 | 0 | 179,825 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Shubham has a binary string s. A binary string is a string containing only characters "0" and "1".
He can perform the following operation on the string any amount of times:
* Select an index of the string, and flip the character at that index. This means, if the character was "0", it becomes "1", and vice versa.
A string is called good if it does not contain "010" or "101" as a subsequence — for instance, "1001" contains "101" as a subsequence, hence it is not a good string, while "1000" doesn't contain neither "010" nor "101" as subsequences, so it is a good string.
What is the minimum number of operations he will have to perform, so that the string becomes good? It can be shown that with these operations we can make any string good.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Input
The first line of the input contains a single integer t (1≤ t ≤ 100) — the number of test cases.
Each of the next t lines contains a binary string s (1 ≤ |s| ≤ 1000).
Output
For every string, output the minimum number of operations required to make it good.
Example
Input
7
001
100
101
010
0
1
001100
Output
0
0
1
1
0
0
2
Note
In test cases 1, 2, 5, 6 no operations are required since they are already good strings.
For the 3rd test case: "001" can be achieved by flipping the first character — and is one of the possible ways to get a good string.
For the 4th test case: "000" can be achieved by flipping the second character — and is one of the possible ways to get a good string.
For the 7th test case: "000000" can be achieved by flipping the third and fourth characters — and is one of the possible ways to get a good string. | instruction | 0 | 90,306 | 0 | 180,612 |
Tags: implementation, strings
Correct Solution:
```
import sys,math
for _ in range(int(input())):
s=list(map(int,list(input())))
cost=0
cost1=0
mi=10**9
n=len(s)
for i in range(n):
cost=0
cost1=0
for j in range(n):
if j<=i:
if (s[j]==0):
cost+=1
else:
cost1+=1
else:
if (s[j]==1):
cost+=1
else:
cost1+=1
#print("cost = ",cost," cost1 = ",cost1)
mi=min(mi,min(cost,cost1))
print(mi)
``` | output | 1 | 90,306 | 0 | 180,613 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Shubham has a binary string s. A binary string is a string containing only characters "0" and "1".
He can perform the following operation on the string any amount of times:
* Select an index of the string, and flip the character at that index. This means, if the character was "0", it becomes "1", and vice versa.
A string is called good if it does not contain "010" or "101" as a subsequence — for instance, "1001" contains "101" as a subsequence, hence it is not a good string, while "1000" doesn't contain neither "010" nor "101" as subsequences, so it is a good string.
What is the minimum number of operations he will have to perform, so that the string becomes good? It can be shown that with these operations we can make any string good.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Input
The first line of the input contains a single integer t (1≤ t ≤ 100) — the number of test cases.
Each of the next t lines contains a binary string s (1 ≤ |s| ≤ 1000).
Output
For every string, output the minimum number of operations required to make it good.
Example
Input
7
001
100
101
010
0
1
001100
Output
0
0
1
1
0
0
2
Note
In test cases 1, 2, 5, 6 no operations are required since they are already good strings.
For the 3rd test case: "001" can be achieved by flipping the first character — and is one of the possible ways to get a good string.
For the 4th test case: "000" can be achieved by flipping the second character — and is one of the possible ways to get a good string.
For the 7th test case: "000000" can be achieved by flipping the third and fourth characters — and is one of the possible ways to get a good string. | instruction | 0 | 90,307 | 0 | 180,614 |
Tags: implementation, strings
Correct Solution:
```
n=int(input())
for i in range(n):
j=input()
if list(j)==sorted(j) or list(j)==sorted(j)[::-1]:
print(0)
else:
x, y=j.count("1"), j.count("0")
a, b, c, d=0, y, 0, x
new=[y, x]
for k in j:
if k=="1":
a+=1
d-=1
else:
b-=1
c+=1
new.extend([a+b, c+d])
print(min(new))
``` | output | 1 | 90,307 | 0 | 180,615 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Shubham has a binary string s. A binary string is a string containing only characters "0" and "1".
He can perform the following operation on the string any amount of times:
* Select an index of the string, and flip the character at that index. This means, if the character was "0", it becomes "1", and vice versa.
A string is called good if it does not contain "010" or "101" as a subsequence — for instance, "1001" contains "101" as a subsequence, hence it is not a good string, while "1000" doesn't contain neither "010" nor "101" as subsequences, so it is a good string.
What is the minimum number of operations he will have to perform, so that the string becomes good? It can be shown that with these operations we can make any string good.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Input
The first line of the input contains a single integer t (1≤ t ≤ 100) — the number of test cases.
Each of the next t lines contains a binary string s (1 ≤ |s| ≤ 1000).
Output
For every string, output the minimum number of operations required to make it good.
Example
Input
7
001
100
101
010
0
1
001100
Output
0
0
1
1
0
0
2
Note
In test cases 1, 2, 5, 6 no operations are required since they are already good strings.
For the 3rd test case: "001" can be achieved by flipping the first character — and is one of the possible ways to get a good string.
For the 4th test case: "000" can be achieved by flipping the second character — and is one of the possible ways to get a good string.
For the 7th test case: "000000" can be achieved by flipping the third and fourth characters — and is one of the possible ways to get a good string. | instruction | 0 | 90,308 | 0 | 180,616 |
Tags: implementation, strings
Correct Solution:
```
## necessary imports
import sys
input = sys.stdin.readline
from math import ceil, floor;
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp
## gcd function
def gcd(a,b):
if a == 0:
return b
return gcd(b%a, a)
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if(k > n - k):
k = n - k
res = 1
for i in range(k):
res = res * (n - i)
res = res / (i + 1)
return int(res)
## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0):
hi = len(a)
while lo < hi:
mid = (lo+hi)//2
if a[mid] < x:
lo = mid+1
else:
hi = mid
return lo
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b
# find function with path compression included (recursive)
# def find(x, link):
# if link[x] == x:
# return x
# link[x] = find(link[x], link);
# return link[x];
# find function with path compression (ITERATIVE)
def find(x, link):
p = x;
while( p != link[p]):
p = link[p];
while( x != p):
nex = link[x];
link[x] = p;
x = nex;
return p;
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve()
def factoriazation(x):
ret = {};
while x != 1:
ret[spf[x]] = ret.get(spf[x], 0) + 1;
x = x//spf[x]
return ret
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
## taking integer array input
def int_array():
return list(map(int, input().strip().split()))
## taking string array input
def str_array():
return input().strip().split();
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
for _ in range(int(input())):
s = input().strip(); n = len(s);
zero = []; one = [];
z = o = 0;
for i in s:
if i == '1':
o += 1;
else:
z += 1;
zero.append(z); one.append(o);
ans = min(o, z);
for i in range(n):
a = one[i] + (z - zero[i]);
b = zero[i] + (o - one[i]);
ans = min(ans, a, b);
print(ans);
``` | output | 1 | 90,308 | 0 | 180,617 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Shubham has a binary string s. A binary string is a string containing only characters "0" and "1".
He can perform the following operation on the string any amount of times:
* Select an index of the string, and flip the character at that index. This means, if the character was "0", it becomes "1", and vice versa.
A string is called good if it does not contain "010" or "101" as a subsequence — for instance, "1001" contains "101" as a subsequence, hence it is not a good string, while "1000" doesn't contain neither "010" nor "101" as subsequences, so it is a good string.
What is the minimum number of operations he will have to perform, so that the string becomes good? It can be shown that with these operations we can make any string good.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Input
The first line of the input contains a single integer t (1≤ t ≤ 100) — the number of test cases.
Each of the next t lines contains a binary string s (1 ≤ |s| ≤ 1000).
Output
For every string, output the minimum number of operations required to make it good.
Example
Input
7
001
100
101
010
0
1
001100
Output
0
0
1
1
0
0
2
Note
In test cases 1, 2, 5, 6 no operations are required since they are already good strings.
For the 3rd test case: "001" can be achieved by flipping the first character — and is one of the possible ways to get a good string.
For the 4th test case: "000" can be achieved by flipping the second character — and is one of the possible ways to get a good string.
For the 7th test case: "000000" can be achieved by flipping the third and fourth characters — and is one of the possible ways to get a good string. | instruction | 0 | 90,309 | 0 | 180,618 |
Tags: implementation, strings
Correct Solution:
```
for _ in range(int(input())):
s = input()
ch0 = 0
ch1 = 0
ch = [[] for _ in range(1001)]
for i in range(len(s)):
ch[i] = [ch0, ch1]
if int(s[i]) == 0:
ch0 += 1
else:
ch1 += 1
ch[len(s)] = [ch0, ch1]
minn = []
for j in range(len(s) + 1):
now = ch[j]
c0 = now[0]
c1 = now[1]
minn.append(ch1 - c1 + c0)
minn.append(ch0 - c0 + c1)
print(min(minn))
``` | output | 1 | 90,309 | 0 | 180,619 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Shubham has a binary string s. A binary string is a string containing only characters "0" and "1".
He can perform the following operation on the string any amount of times:
* Select an index of the string, and flip the character at that index. This means, if the character was "0", it becomes "1", and vice versa.
A string is called good if it does not contain "010" or "101" as a subsequence — for instance, "1001" contains "101" as a subsequence, hence it is not a good string, while "1000" doesn't contain neither "010" nor "101" as subsequences, so it is a good string.
What is the minimum number of operations he will have to perform, so that the string becomes good? It can be shown that with these operations we can make any string good.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Input
The first line of the input contains a single integer t (1≤ t ≤ 100) — the number of test cases.
Each of the next t lines contains a binary string s (1 ≤ |s| ≤ 1000).
Output
For every string, output the minimum number of operations required to make it good.
Example
Input
7
001
100
101
010
0
1
001100
Output
0
0
1
1
0
0
2
Note
In test cases 1, 2, 5, 6 no operations are required since they are already good strings.
For the 3rd test case: "001" can be achieved by flipping the first character — and is one of the possible ways to get a good string.
For the 4th test case: "000" can be achieved by flipping the second character — and is one of the possible ways to get a good string.
For the 7th test case: "000000" can be achieved by flipping the third and fourth characters — and is one of the possible ways to get a good string. | instruction | 0 | 90,310 | 0 | 180,620 |
Tags: implementation, strings
Correct Solution:
```
class Solution:
def __init__(self):
self.s = input()
def solve_and_print(self):
right_1 = self.s.count('1')
left_0, left_1, right_0 = 0, 0, len(self.s)-right_1
min_op = len(self.s)
for c in self.s:
min_op = min(min_op, left_0+right_1, left_1+right_0)
if c == '1':
left_1 += 1
right_1 -= 1
else:
left_0 += 1
right_0 -= 1
print(min_op)
if __name__ == "__main__":
for _ in range(int(input())):
Solution().solve_and_print()
``` | output | 1 | 90,310 | 0 | 180,621 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Shubham has a binary string s. A binary string is a string containing only characters "0" and "1".
He can perform the following operation on the string any amount of times:
* Select an index of the string, and flip the character at that index. This means, if the character was "0", it becomes "1", and vice versa.
A string is called good if it does not contain "010" or "101" as a subsequence — for instance, "1001" contains "101" as a subsequence, hence it is not a good string, while "1000" doesn't contain neither "010" nor "101" as subsequences, so it is a good string.
What is the minimum number of operations he will have to perform, so that the string becomes good? It can be shown that with these operations we can make any string good.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Input
The first line of the input contains a single integer t (1≤ t ≤ 100) — the number of test cases.
Each of the next t lines contains a binary string s (1 ≤ |s| ≤ 1000).
Output
For every string, output the minimum number of operations required to make it good.
Example
Input
7
001
100
101
010
0
1
001100
Output
0
0
1
1
0
0
2
Note
In test cases 1, 2, 5, 6 no operations are required since they are already good strings.
For the 3rd test case: "001" can be achieved by flipping the first character — and is one of the possible ways to get a good string.
For the 4th test case: "000" can be achieved by flipping the second character — and is one of the possible ways to get a good string.
For the 7th test case: "000000" can be achieved by flipping the third and fourth characters — and is one of the possible ways to get a good string. | instruction | 0 | 90,311 | 0 | 180,622 |
Tags: implementation, strings
Correct Solution:
```
for _ in range(int(input())):
s=input()
sum=0
for i in s:
if i=='1':
sum+=1
if sum==0 or sum==len(s):
print(0)
else:
ans_now=100000000000000
l=0
r=sum
for i in range(len(s)):
if s[i]=='1':
l+=1
r-=1
ans_now=min(i+1-l+r,ans_now)
ans_now=min(l+len(s)-i-1-r,ans_now)
print(ans_now)
``` | output | 1 | 90,311 | 0 | 180,623 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Shubham has a binary string s. A binary string is a string containing only characters "0" and "1".
He can perform the following operation on the string any amount of times:
* Select an index of the string, and flip the character at that index. This means, if the character was "0", it becomes "1", and vice versa.
A string is called good if it does not contain "010" or "101" as a subsequence — for instance, "1001" contains "101" as a subsequence, hence it is not a good string, while "1000" doesn't contain neither "010" nor "101" as subsequences, so it is a good string.
What is the minimum number of operations he will have to perform, so that the string becomes good? It can be shown that with these operations we can make any string good.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Input
The first line of the input contains a single integer t (1≤ t ≤ 100) — the number of test cases.
Each of the next t lines contains a binary string s (1 ≤ |s| ≤ 1000).
Output
For every string, output the minimum number of operations required to make it good.
Example
Input
7
001
100
101
010
0
1
001100
Output
0
0
1
1
0
0
2
Note
In test cases 1, 2, 5, 6 no operations are required since they are already good strings.
For the 3rd test case: "001" can be achieved by flipping the first character — and is one of the possible ways to get a good string.
For the 4th test case: "000" can be achieved by flipping the second character — and is one of the possible ways to get a good string.
For the 7th test case: "000000" can be achieved by flipping the third and fourth characters — and is one of the possible ways to get a good string. | instruction | 0 | 90,312 | 0 | 180,624 |
Tags: implementation, strings
Correct Solution:
```
for _ in range(int(input())):
s = input()
ans = 10000000
total = 0
for i in s:
if i=='1':
total+=1
l = len(s)
ans = min(ans, total, l-total)
preZeroes = 0
for i in range(l):
if s[i]=='0':
preZeroes += 1
preOnes = i+1-preZeroes
postOnes = total - preOnes
postZeroes = l-i-1-postOnes
# print(postOnes,postZeroes,preZeroes,preOnes)
flip = min(preOnes+postZeroes, postOnes+preZeroes)
ans = min(ans, flip)
print(ans)
``` | output | 1 | 90,312 | 0 | 180,625 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Shubham has a binary string s. A binary string is a string containing only characters "0" and "1".
He can perform the following operation on the string any amount of times:
* Select an index of the string, and flip the character at that index. This means, if the character was "0", it becomes "1", and vice versa.
A string is called good if it does not contain "010" or "101" as a subsequence — for instance, "1001" contains "101" as a subsequence, hence it is not a good string, while "1000" doesn't contain neither "010" nor "101" as subsequences, so it is a good string.
What is the minimum number of operations he will have to perform, so that the string becomes good? It can be shown that with these operations we can make any string good.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Input
The first line of the input contains a single integer t (1≤ t ≤ 100) — the number of test cases.
Each of the next t lines contains a binary string s (1 ≤ |s| ≤ 1000).
Output
For every string, output the minimum number of operations required to make it good.
Example
Input
7
001
100
101
010
0
1
001100
Output
0
0
1
1
0
0
2
Note
In test cases 1, 2, 5, 6 no operations are required since they are already good strings.
For the 3rd test case: "001" can be achieved by flipping the first character — and is one of the possible ways to get a good string.
For the 4th test case: "000" can be achieved by flipping the second character — and is one of the possible ways to get a good string.
For the 7th test case: "000000" can be achieved by flipping the third and fourth characters — and is one of the possible ways to get a good string. | instruction | 0 | 90,313 | 0 | 180,626 |
Tags: implementation, strings
Correct Solution:
```
T=int(input())
for i in range(T):
s=input()
l=[]
ans=99999999999
n=len(s)
for i in range(n):
c=s[i]
ls=s[:i]
rs=s[i+1:]
lsz=ls.count("0")
lso=ls.count("1")
rsz=rs.count("0")
rso=rs.count("1")
ans=min(ans,lsz+rso,lso+rsz)
print(ans)
``` | output | 1 | 90,313 | 0 | 180,627 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Shubham has a binary string s. A binary string is a string containing only characters "0" and "1".
He can perform the following operation on the string any amount of times:
* Select an index of the string, and flip the character at that index. This means, if the character was "0", it becomes "1", and vice versa.
A string is called good if it does not contain "010" or "101" as a subsequence — for instance, "1001" contains "101" as a subsequence, hence it is not a good string, while "1000" doesn't contain neither "010" nor "101" as subsequences, so it is a good string.
What is the minimum number of operations he will have to perform, so that the string becomes good? It can be shown that with these operations we can make any string good.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Input
The first line of the input contains a single integer t (1≤ t ≤ 100) — the number of test cases.
Each of the next t lines contains a binary string s (1 ≤ |s| ≤ 1000).
Output
For every string, output the minimum number of operations required to make it good.
Example
Input
7
001
100
101
010
0
1
001100
Output
0
0
1
1
0
0
2
Note
In test cases 1, 2, 5, 6 no operations are required since they are already good strings.
For the 3rd test case: "001" can be achieved by flipping the first character — and is one of the possible ways to get a good string.
For the 4th test case: "000" can be achieved by flipping the second character — and is one of the possible ways to get a good string.
For the 7th test case: "000000" can be achieved by flipping the third and fourth characters — and is one of the possible ways to get a good string.
Submitted Solution:
```
import sys, math
#from bisect import bisect_left as bl, bisect_right as br, insort
#from heapq import heapify, heappush, heappop
from collections import defaultdict as dd, deque, Counter
#from itertools import permutations,combinations
#from decimal import Decimal
def data(): return sys.stdin.readline().strip()
def mdata(): return list(map(int, data().split()))
def outl(var) : sys.stdout.write(' '.join(map(str, var))+'\n')
def out(var) : sys.stdout.write(str(var)+'\n')
sys.setrecursionlimit(100000)
INF = float('inf')
mod = int(1e9)+7
def main():
for t in range(int(data())):
s=data()
n=len(s)
dp1=[0]*n
dp2=[0]*n
for i in range(n):
if s[i]=='1':
dp1[i]=dp1[i-1]+1
dp2[i]=dp2[i-1]
else:
dp2[i] = dp2[i - 1] + 1
dp1[i] = dp1[i - 1]
m=dp1[-1]
for i in range(n):
m=min(m,dp1[i]+dp2[-1]-dp2[i],dp2[i]+dp1[-1]-dp1[i])
out(m)
if __name__ == '__main__':
main()
``` | instruction | 0 | 90,314 | 0 | 180,628 |
Yes | output | 1 | 90,314 | 0 | 180,629 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Shubham has a binary string s. A binary string is a string containing only characters "0" and "1".
He can perform the following operation on the string any amount of times:
* Select an index of the string, and flip the character at that index. This means, if the character was "0", it becomes "1", and vice versa.
A string is called good if it does not contain "010" or "101" as a subsequence — for instance, "1001" contains "101" as a subsequence, hence it is not a good string, while "1000" doesn't contain neither "010" nor "101" as subsequences, so it is a good string.
What is the minimum number of operations he will have to perform, so that the string becomes good? It can be shown that with these operations we can make any string good.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Input
The first line of the input contains a single integer t (1≤ t ≤ 100) — the number of test cases.
Each of the next t lines contains a binary string s (1 ≤ |s| ≤ 1000).
Output
For every string, output the minimum number of operations required to make it good.
Example
Input
7
001
100
101
010
0
1
001100
Output
0
0
1
1
0
0
2
Note
In test cases 1, 2, 5, 6 no operations are required since they are already good strings.
For the 3rd test case: "001" can be achieved by flipping the first character — and is one of the possible ways to get a good string.
For the 4th test case: "000" can be achieved by flipping the second character — and is one of the possible ways to get a good string.
For the 7th test case: "000000" can be achieved by flipping the third and fourth characters — and is one of the possible ways to get a good string.
Submitted Solution:
```
#qn given an arr of size n for every i position find a min j such that j>i and arr[j]>arr[i]
#it prints the position of the nearest .
import sys
input = sys.stdin.readline
#qn given an arr of size n for every i position find a min j such that j>i and arr[j]>arr[i]
#it prints the position of the nearest .
# import sys
import heapq
import copy
import math
import decimal
# from decimal import *
#heapq.heapify(li)
#
#heapq.heappush(li,4)
#
#heapq.heappop(li)
#
# & Bitwise AND Operator 10 & 7 = 2
# | Bitwise OR Operator 10 | 7 = 15
# ^ Bitwise XOR Operator 10 ^ 7 = 13
# << Bitwise Left Shift operator 10<<2 = 40
# >> Bitwise Right Shift Operator
# '''############ ---- Input Functions ---- #######Start#####'''
class Node:
def _init_(self,val):
self.data = val
self.left = None
self.right = None
##to initialize wire : object_name= node(val)##to create a new node
## can also be used to create linked list
class fen_tree:
"""Implementation of a Binary Indexed Tree (Fennwick Tree)"""
#def __init__(self, list):
# """Initialize BIT with list in O(n*log(n))"""
# self.array = [0] * (len(list) + 1)
# for idx, val in enumerate(list):
# self.update(idx, val)
def __init__(self, list):
""""Initialize BIT with list in O(n)"""
self.array = [0] + list
for idx in range(1, len(self.array)):
idx2 = idx + (idx & -idx)
if idx2 < len(self.array):
self.array[idx2] += self.array[idx]
def prefix_query(self, idx):
"""Computes prefix sum of up to including the idx-th element"""
# idx += 1
result = 0
while idx:
result += self.array[idx]
idx -= idx & -idx
return result
def prints(self):
print(self.array)
return
# for i in self.array:
# print(i,end = " ")
# return
def range_query(self, from_idx, to_idx):
"""Computes the range sum between two indices (both inclusive)"""
return self.prefix_query(to_idx) - self.prefix_query(from_idx - 1)
def update(self, idx, add):
"""Add a value to the idx-th element"""
# idx += 1
while idx < len(self.array):
self.array[idx] += add
idx += idx & -idx
def pre_sum(arr):
#"""returns the prefix sum inclusive ie ith position in ans represent sum from 0 to ith position"""
p = [0]
for i in arr:
p.append(p[-1] + i)
p.pop(0)
return p
def pre_back(arr):
#"""returns the prefix sum inclusive ie ith position in ans represent sum from 0 to ith position"""
p = [0]
for i in arr:
p.append(p[-1] + i)
p.pop(0)
return p
def bin_search(arr,l,r,val):#strickly greater
if arr[r] <= val:
return r+1
if r-l < 2:
if arr[l]>val:
return l
else:
return r
mid = int((l+r)/2)
if arr[mid] <= val:
return bin_search(arr,mid,r,val)
else:
return bin_search(arr,l,mid,val)
def search_leftmost(arr,val):
def helper(arr,l,r,val):
# print(arr)
print(l,r)
if arr[l] == val:
return l
if r -l <=1:
if arr[r] == val:
return r
else:
print("not found")
return
mid = int((r+l)/2)
if arr[mid] >= val:
return helper(arr,l,mid,val)
else:
return helper(arr,mid,r,val)
return helper(arr,0,len(arr)-1,val)
def search_rightmost(arr,val):
def helper(arr,l,r,val):
# print(arr)
print(l,r)
if arr[r] == val:
return r
if r -l <=1:
if arr[l] == val:
return r
else:
print("not found")
return
mid = int((r+l)/2)
if arr[mid] > val:
return helper(arr,l,mid,val)
else:
return helper(arr,mid,r,val)
return helper(arr,0,len(arr)-1,val)
def preorder_postorder(root,paths,parent,left,level):
# if len(paths[node]) ==1 and node !=1 :
# return [parent,left,level]
l = 0
queue = [root]
while(queue!=[]):
node = queue[-1]
# print(queue,node)
flag = 0
for i in paths[node]:
if i!=parent[node] and level[i]== sys.maxsize:
parent[i] = node
level[i] = level[node]+1
flag = 1
queue.append(i)
if flag == 0:
left[parent[node]] +=(1+left[node])
queue.pop()
# [parent,left,level] = helper(i,paths,parent,left,level)
# l = left[i] + l +1
# left[node] = l
return [parent,left,level]
def nCr(n, r, p):
# initialize numerator
# and denominator
if r == 0:
return 1
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,p - 2, p)) % p
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
############ ---- Input Functions ---- #######End
# #####
def pr_list(a):
print(*a, sep=" ")
def main():
tests = inp()
# tests = 1
mod = 1000000007
limit = 10**18
for test in range(tests):
s = insr()
r = [0]
l= [0]
n = len(s)
for i in s:
if i == "0":
l.append(l[-1])
else:
l.append(l[-1]+1)
l.pop()
for i in range(n-1,-1,-1):
if s[i] == "0":
r.append(r[-1])
else:
r.append(r[-1]+1)
r.pop()
r.reverse()
# print(l,r)
ans = min(l[-1]+int(s[-1]), n - (l[-1]+int(s[-1])))
for i in range(0,n):
if i>0 or i < n-1:
l0 = l[i]
r1 = n-i-1 - r[i]
l1 = i-l[i]
r0 = r[i]
ans = min (ans,l0+r1,l1+r0)
elif i == 0:
ans = min (ans,r[0],n-r[0]-1)
else:
ans = min (ans,l[0],n-l[0]-1)
print(ans)
if __name__== "__main__":
main()
``` | instruction | 0 | 90,315 | 0 | 180,630 |
Yes | output | 1 | 90,315 | 0 | 180,631 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Shubham has a binary string s. A binary string is a string containing only characters "0" and "1".
He can perform the following operation on the string any amount of times:
* Select an index of the string, and flip the character at that index. This means, if the character was "0", it becomes "1", and vice versa.
A string is called good if it does not contain "010" or "101" as a subsequence — for instance, "1001" contains "101" as a subsequence, hence it is not a good string, while "1000" doesn't contain neither "010" nor "101" as subsequences, so it is a good string.
What is the minimum number of operations he will have to perform, so that the string becomes good? It can be shown that with these operations we can make any string good.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Input
The first line of the input contains a single integer t (1≤ t ≤ 100) — the number of test cases.
Each of the next t lines contains a binary string s (1 ≤ |s| ≤ 1000).
Output
For every string, output the minimum number of operations required to make it good.
Example
Input
7
001
100
101
010
0
1
001100
Output
0
0
1
1
0
0
2
Note
In test cases 1, 2, 5, 6 no operations are required since they are already good strings.
For the 3rd test case: "001" can be achieved by flipping the first character — and is one of the possible ways to get a good string.
For the 4th test case: "000" can be achieved by flipping the second character — and is one of the possible ways to get a good string.
For the 7th test case: "000000" can be achieved by flipping the third and fourth characters — and is one of the possible ways to get a good string.
Submitted Solution:
```
a = int(input())
for x in range(a):
c = input()
m = []
for y in range(0,len(c)):
f = 0
f += c[:y].count("1")
f += c[y:].count("0")
j = 0
j += c[:y].count("0")
j += c[y:].count("1")
m.append(f)
m.append(j)
print(min(m))
``` | instruction | 0 | 90,316 | 0 | 180,632 |
Yes | output | 1 | 90,316 | 0 | 180,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Shubham has a binary string s. A binary string is a string containing only characters "0" and "1".
He can perform the following operation on the string any amount of times:
* Select an index of the string, and flip the character at that index. This means, if the character was "0", it becomes "1", and vice versa.
A string is called good if it does not contain "010" or "101" as a subsequence — for instance, "1001" contains "101" as a subsequence, hence it is not a good string, while "1000" doesn't contain neither "010" nor "101" as subsequences, so it is a good string.
What is the minimum number of operations he will have to perform, so that the string becomes good? It can be shown that with these operations we can make any string good.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Input
The first line of the input contains a single integer t (1≤ t ≤ 100) — the number of test cases.
Each of the next t lines contains a binary string s (1 ≤ |s| ≤ 1000).
Output
For every string, output the minimum number of operations required to make it good.
Example
Input
7
001
100
101
010
0
1
001100
Output
0
0
1
1
0
0
2
Note
In test cases 1, 2, 5, 6 no operations are required since they are already good strings.
For the 3rd test case: "001" can be achieved by flipping the first character — and is one of the possible ways to get a good string.
For the 4th test case: "000" can be achieved by flipping the second character — and is one of the possible ways to get a good string.
For the 7th test case: "000000" can be achieved by flipping the third and fourth characters — and is one of the possible ways to get a good string.
Submitted Solution:
```
for _ in range(int(input())):
s = input()
n = len(s)
p = [0]
for c in s:
p.append((int(c) + p[-1]))
best = n
for i in range(1, n+1):
prez = p[i]
preo = i - prez
suffz = p[n] - p[i]
suffo = (n - i) - suffz
#print('i:', i, 'prez', prez, 'preo', preo, 'suffz', suffz, 'suffo', suffo)
#print('curr', i, prez + suffo, preo + suffz, prez + suffz, preo + suffo)
best = min(best, prez + suffo, preo + suffz, prez + suffz, preo + suffo)
print(best)
``` | instruction | 0 | 90,317 | 0 | 180,634 |
Yes | output | 1 | 90,317 | 0 | 180,635 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Shubham has a binary string s. A binary string is a string containing only characters "0" and "1".
He can perform the following operation on the string any amount of times:
* Select an index of the string, and flip the character at that index. This means, if the character was "0", it becomes "1", and vice versa.
A string is called good if it does not contain "010" or "101" as a subsequence — for instance, "1001" contains "101" as a subsequence, hence it is not a good string, while "1000" doesn't contain neither "010" nor "101" as subsequences, so it is a good string.
What is the minimum number of operations he will have to perform, so that the string becomes good? It can be shown that with these operations we can make any string good.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Input
The first line of the input contains a single integer t (1≤ t ≤ 100) — the number of test cases.
Each of the next t lines contains a binary string s (1 ≤ |s| ≤ 1000).
Output
For every string, output the minimum number of operations required to make it good.
Example
Input
7
001
100
101
010
0
1
001100
Output
0
0
1
1
0
0
2
Note
In test cases 1, 2, 5, 6 no operations are required since they are already good strings.
For the 3rd test case: "001" can be achieved by flipping the first character — and is one of the possible ways to get a good string.
For the 4th test case: "000" can be achieved by flipping the second character — and is one of the possible ways to get a good string.
For the 7th test case: "000000" can be achieved by flipping the third and fourth characters — and is one of the possible ways to get a good string.
Submitted Solution:
```
n=int(input())
l=[]
for i in range(n):
z=0
o=0
s=input()
for j in range(len(s)):
if s[j]=='0':
z+=1
else:
o+=1
if s[0]=='1' or s[-1]=='1':
o-=1
if o>z:
mini=z
else:
mini=o
l.append(mini)
for i in l:
print(i)
``` | instruction | 0 | 90,318 | 0 | 180,636 |
No | output | 1 | 90,318 | 0 | 180,637 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Shubham has a binary string s. A binary string is a string containing only characters "0" and "1".
He can perform the following operation on the string any amount of times:
* Select an index of the string, and flip the character at that index. This means, if the character was "0", it becomes "1", and vice versa.
A string is called good if it does not contain "010" or "101" as a subsequence — for instance, "1001" contains "101" as a subsequence, hence it is not a good string, while "1000" doesn't contain neither "010" nor "101" as subsequences, so it is a good string.
What is the minimum number of operations he will have to perform, so that the string becomes good? It can be shown that with these operations we can make any string good.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Input
The first line of the input contains a single integer t (1≤ t ≤ 100) — the number of test cases.
Each of the next t lines contains a binary string s (1 ≤ |s| ≤ 1000).
Output
For every string, output the minimum number of operations required to make it good.
Example
Input
7
001
100
101
010
0
1
001100
Output
0
0
1
1
0
0
2
Note
In test cases 1, 2, 5, 6 no operations are required since they are already good strings.
For the 3rd test case: "001" can be achieved by flipping the first character — and is one of the possible ways to get a good string.
For the 4th test case: "000" can be achieved by flipping the second character — and is one of the possible ways to get a good string.
For the 7th test case: "000000" can be achieved by flipping the third and fourth characters — and is one of the possible ways to get a good string.
Submitted Solution:
```
t=int(input())
for _ in range(t):
s=input()
if(len(s)<3):
print(0)
else:
t3=0
if(s[0]=='1'):
i=1
q=0
t2=1
while(i<len(s)):
if(s[i]=='0'):
q=1
else:
if(q==1):
t3+=t2
q=0
t2=1
else:
t2+=1
i+=1
else:
i=1
q=0
q1=0
t2=0
while(i<len(s)):
if(s[i]=='1'):
if(q1==1):
t3-=t2
break
else:
t2+=1
q=1
else:
if(q==1):
q1=1
t3+=t2
t2=0
i+=1
print(t3)
``` | instruction | 0 | 90,319 | 0 | 180,638 |
No | output | 1 | 90,319 | 0 | 180,639 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Shubham has a binary string s. A binary string is a string containing only characters "0" and "1".
He can perform the following operation on the string any amount of times:
* Select an index of the string, and flip the character at that index. This means, if the character was "0", it becomes "1", and vice versa.
A string is called good if it does not contain "010" or "101" as a subsequence — for instance, "1001" contains "101" as a subsequence, hence it is not a good string, while "1000" doesn't contain neither "010" nor "101" as subsequences, so it is a good string.
What is the minimum number of operations he will have to perform, so that the string becomes good? It can be shown that with these operations we can make any string good.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Input
The first line of the input contains a single integer t (1≤ t ≤ 100) — the number of test cases.
Each of the next t lines contains a binary string s (1 ≤ |s| ≤ 1000).
Output
For every string, output the minimum number of operations required to make it good.
Example
Input
7
001
100
101
010
0
1
001100
Output
0
0
1
1
0
0
2
Note
In test cases 1, 2, 5, 6 no operations are required since they are already good strings.
For the 3rd test case: "001" can be achieved by flipping the first character — and is one of the possible ways to get a good string.
For the 4th test case: "000" can be achieved by flipping the second character — and is one of the possible ways to get a good string.
For the 7th test case: "000000" can be achieved by flipping the third and fourth characters — and is one of the possible ways to get a good string.
Submitted Solution:
```
t=int(input())
for i in range(0,t):
s=input()
if len(s)<=2:
print(0)
else:
n=len(s)
p=s[0]
c1=0
c2=0
for j in range(0,len(s)):
if s[j]=='0':
c1+=1
else:
c2+=1
ans=min(c1,c2)
c=0
f=0
for j in range(1,n):
if s[j]!=p and f==0:
f=1
p=s[j]
elif s[j]!=p:
c+=1
ans=min(ans,c)
p=s[len(s)-1]
c=0
f=0
for j in range(n-1,-1,-1):
if s[j]!=p and f==0:
f=1
p=s[j]
elif s[j]!=p:
c+=1
ans=min(ans,c)
print(ans)
``` | instruction | 0 | 90,320 | 0 | 180,640 |
No | output | 1 | 90,320 | 0 | 180,641 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Shubham has a binary string s. A binary string is a string containing only characters "0" and "1".
He can perform the following operation on the string any amount of times:
* Select an index of the string, and flip the character at that index. This means, if the character was "0", it becomes "1", and vice versa.
A string is called good if it does not contain "010" or "101" as a subsequence — for instance, "1001" contains "101" as a subsequence, hence it is not a good string, while "1000" doesn't contain neither "010" nor "101" as subsequences, so it is a good string.
What is the minimum number of operations he will have to perform, so that the string becomes good? It can be shown that with these operations we can make any string good.
A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters.
Input
The first line of the input contains a single integer t (1≤ t ≤ 100) — the number of test cases.
Each of the next t lines contains a binary string s (1 ≤ |s| ≤ 1000).
Output
For every string, output the minimum number of operations required to make it good.
Example
Input
7
001
100
101
010
0
1
001100
Output
0
0
1
1
0
0
2
Note
In test cases 1, 2, 5, 6 no operations are required since they are already good strings.
For the 3rd test case: "001" can be achieved by flipping the first character — and is one of the possible ways to get a good string.
For the 4th test case: "000" can be achieved by flipping the second character — and is one of the possible ways to get a good string.
For the 7th test case: "000000" can be achieved by flipping the third and fourth characters — and is one of the possible ways to get a good string.
Submitted Solution:
```
"""
~~ Author : Bhaskar
~~ Dated : 01~06~2020
"""
import sys
from bisect import *
from math import floor,sqrt,ceil,factorial as F,gcd,pi
from itertools import chain,combinations,permutations,accumulate
from collections import Counter,defaultdict,OrderedDict,deque
from array import array
INT_MAX = sys.maxsize
INT_MIN = -(sys.maxsize)-1
mod = 1000000007
ch = "abcdefghijklmnopqrstuvwxyz"
lcm = lambda a,b : (a*b)//gcd(a,b)
setbit = lambda x : bin(x)[2:].count("1")
def solve():
T = int(sys.stdin.readline())
for _ in range(T):
s = str(sys.stdin.readline()).replace('\n',"")
one = s.count("1")
zero = len(s)-one
if one == 0 or zero == 0:
print(0)
else:
if s[0] == s[-1] == "0":
print(s[1:-1].count("1"))
elif s[0] == s[-1] == "1":
print(s[1:-1].count("0"))
else:
if s[0] == "0" and s[-1] == "1":
grab = s[:-1]
print(min(grab.count("1"),grab.count("0")))
if s[0] == "1" and s[-1] == "0":
grab = s[1:]
print(min(grab.count("1"),grab.count("0")))
# print(one,zero)
if __name__ == "__main__":
try:
sys.stdin = open("input.txt","r")
except :
pass
solve()
``` | instruction | 0 | 90,321 | 0 | 180,642 |
No | output | 1 | 90,321 | 0 | 180,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s, consisting only of characters '0' or '1'. Let |s| be the length of s.
You are asked to choose some integer k (k > 0) and find a sequence a of length k such that:
* 1 ≤ a_1 < a_2 < ... < a_k ≤ |s|;
* a_{i-1} + 1 < a_i for all i from 2 to k.
The characters at positions a_1, a_2, ..., a_k are removed, the remaining characters are concatenated without changing the order. So, in other words, the positions in the sequence a should not be adjacent.
Let the resulting string be s'. s' is called sorted if for all i from 2 to |s'| s'_{i-1} ≤ s'_i.
Does there exist such a sequence a that the resulting string s' is sorted?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains a string s (2 ≤ |s| ≤ 100). Each character is either '0' or '1'.
Output
For each testcase print "YES" if there exists a sequence a such that removing the characters at positions a_1, a_2, ..., a_k and concatenating the parts without changing the order produces a sorted string.
Otherwise, print "NO".
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
10101011011
0000
11111
110
1100
Output
YES
YES
YES
YES
NO
Note
In the first testcase you can choose a sequence a=[1,3,6,9]. Removing the underlined letters from "10101011011" will produce a string "0011111", which is sorted.
In the second and the third testcases the sequences are already sorted.
In the fourth testcase you can choose a sequence a=[3]. s'= "11", which is sorted.
In the fifth testcase there is no way to choose a sequence a such that s' is sorted. | instruction | 0 | 90,364 | 0 | 180,728 |
Tags: brute force, dp, greedy, implementation
Correct Solution:
```
t=int(input())
for _ in range(t):
s=input()
n=len(s)
a,b=-1,-1
if s.count('1')==n or s.count('0')==n:
print("YES")
else:
a=s.find("11")
b=s.rfind("00")
if a==-1 or b==-1:
print("YES")
else:
if a<b:
print("NO")
else:
print("YES")
``` | output | 1 | 90,364 | 0 | 180,729 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s, consisting only of characters '0' or '1'. Let |s| be the length of s.
You are asked to choose some integer k (k > 0) and find a sequence a of length k such that:
* 1 ≤ a_1 < a_2 < ... < a_k ≤ |s|;
* a_{i-1} + 1 < a_i for all i from 2 to k.
The characters at positions a_1, a_2, ..., a_k are removed, the remaining characters are concatenated without changing the order. So, in other words, the positions in the sequence a should not be adjacent.
Let the resulting string be s'. s' is called sorted if for all i from 2 to |s'| s'_{i-1} ≤ s'_i.
Does there exist such a sequence a that the resulting string s' is sorted?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains a string s (2 ≤ |s| ≤ 100). Each character is either '0' or '1'.
Output
For each testcase print "YES" if there exists a sequence a such that removing the characters at positions a_1, a_2, ..., a_k and concatenating the parts without changing the order produces a sorted string.
Otherwise, print "NO".
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
10101011011
0000
11111
110
1100
Output
YES
YES
YES
YES
NO
Note
In the first testcase you can choose a sequence a=[1,3,6,9]. Removing the underlined letters from "10101011011" will produce a string "0011111", which is sorted.
In the second and the third testcases the sequences are already sorted.
In the fourth testcase you can choose a sequence a=[3]. s'= "11", which is sorted.
In the fifth testcase there is no way to choose a sequence a such that s' is sorted. | instruction | 0 | 90,365 | 0 | 180,730 |
Tags: brute force, dp, greedy, implementation
Correct Solution:
```
tc = int(input())
for t in range(tc):
seq = input()
rmost00 = seq.rfind("00")
lmost11 = seq.find("11")
if rmost00 >= 0 and lmost11 >= 0 and rmost00 > lmost11:
print("NO")
else:
print("YES")
# prev = seq[0]
# for i in range(1, len(seq)):
# item = seq[i]
# if item <
``` | output | 1 | 90,365 | 0 | 180,731 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s, consisting only of characters '0' or '1'. Let |s| be the length of s.
You are asked to choose some integer k (k > 0) and find a sequence a of length k such that:
* 1 ≤ a_1 < a_2 < ... < a_k ≤ |s|;
* a_{i-1} + 1 < a_i for all i from 2 to k.
The characters at positions a_1, a_2, ..., a_k are removed, the remaining characters are concatenated without changing the order. So, in other words, the positions in the sequence a should not be adjacent.
Let the resulting string be s'. s' is called sorted if for all i from 2 to |s'| s'_{i-1} ≤ s'_i.
Does there exist such a sequence a that the resulting string s' is sorted?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains a string s (2 ≤ |s| ≤ 100). Each character is either '0' or '1'.
Output
For each testcase print "YES" if there exists a sequence a such that removing the characters at positions a_1, a_2, ..., a_k and concatenating the parts without changing the order produces a sorted string.
Otherwise, print "NO".
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
10101011011
0000
11111
110
1100
Output
YES
YES
YES
YES
NO
Note
In the first testcase you can choose a sequence a=[1,3,6,9]. Removing the underlined letters from "10101011011" will produce a string "0011111", which is sorted.
In the second and the third testcases the sequences are already sorted.
In the fourth testcase you can choose a sequence a=[3]. s'= "11", which is sorted.
In the fifth testcase there is no way to choose a sequence a such that s' is sorted. | instruction | 0 | 90,366 | 0 | 180,732 |
Tags: brute force, dp, greedy, implementation
Correct Solution:
```
#!/usr/bin/env python
from __future__ import division, print_function
import os
import sys
sys.setrecursionlimit(1000000)
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
def main():
pass
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
'''
b = list(map(int, input().split()))
arr.sort(key=lambda x :x[0]-x[1])
'''
import math
import sys
import bisect
from collections import defaultdict
for t in range(int(input())):
# n, k1, k2 = map(int, input().split())
s = input()
flag = True
idx = -1
idx0 = -1
for i in range(len(s)-1):
if s[i] == '1' and s[i+1] == '1' and flag:
idx = i
flag = False
if s[i] == '0' and s[i+1] == '0':
idx0 = i
if idx == -1 or idx0 == -1:
print("YES")
continue
if idx < idx0:
print("NO")
else:
print("YES")
if __name__ == "__main__":
main()
``` | output | 1 | 90,366 | 0 | 180,733 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s, consisting only of characters '0' or '1'. Let |s| be the length of s.
You are asked to choose some integer k (k > 0) and find a sequence a of length k such that:
* 1 ≤ a_1 < a_2 < ... < a_k ≤ |s|;
* a_{i-1} + 1 < a_i for all i from 2 to k.
The characters at positions a_1, a_2, ..., a_k are removed, the remaining characters are concatenated without changing the order. So, in other words, the positions in the sequence a should not be adjacent.
Let the resulting string be s'. s' is called sorted if for all i from 2 to |s'| s'_{i-1} ≤ s'_i.
Does there exist such a sequence a that the resulting string s' is sorted?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains a string s (2 ≤ |s| ≤ 100). Each character is either '0' or '1'.
Output
For each testcase print "YES" if there exists a sequence a such that removing the characters at positions a_1, a_2, ..., a_k and concatenating the parts without changing the order produces a sorted string.
Otherwise, print "NO".
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
10101011011
0000
11111
110
1100
Output
YES
YES
YES
YES
NO
Note
In the first testcase you can choose a sequence a=[1,3,6,9]. Removing the underlined letters from "10101011011" will produce a string "0011111", which is sorted.
In the second and the third testcases the sequences are already sorted.
In the fourth testcase you can choose a sequence a=[3]. s'= "11", which is sorted.
In the fifth testcase there is no way to choose a sequence a such that s' is sorted. | instruction | 0 | 90,367 | 0 | 180,734 |
Tags: brute force, dp, greedy, implementation
Correct Solution:
```
t = int(input())
results = []
def find_dob(s):
return [(i, s[i]) for i in range(len(s)-1) if not(s[i]^s[i+1])]
for i in range(t):
s = input()
s = [int(c) for c in s]
dobs = find_dob(s)
dobs_z = [a for (a, b) in dobs if b==0]
dobs_o = [a for (a, b) in dobs if b==1]
if (len(dobs_z)==0) or len(dobs_o)==0:
results.append('yes')
else:
if dobs_z[-1]<dobs_o[0]:
results.append('yes')
else:
results.append('no')
for i in range(t):
print(results.pop(0))
``` | output | 1 | 90,367 | 0 | 180,735 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s, consisting only of characters '0' or '1'. Let |s| be the length of s.
You are asked to choose some integer k (k > 0) and find a sequence a of length k such that:
* 1 ≤ a_1 < a_2 < ... < a_k ≤ |s|;
* a_{i-1} + 1 < a_i for all i from 2 to k.
The characters at positions a_1, a_2, ..., a_k are removed, the remaining characters are concatenated without changing the order. So, in other words, the positions in the sequence a should not be adjacent.
Let the resulting string be s'. s' is called sorted if for all i from 2 to |s'| s'_{i-1} ≤ s'_i.
Does there exist such a sequence a that the resulting string s' is sorted?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains a string s (2 ≤ |s| ≤ 100). Each character is either '0' or '1'.
Output
For each testcase print "YES" if there exists a sequence a such that removing the characters at positions a_1, a_2, ..., a_k and concatenating the parts without changing the order produces a sorted string.
Otherwise, print "NO".
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
10101011011
0000
11111
110
1100
Output
YES
YES
YES
YES
NO
Note
In the first testcase you can choose a sequence a=[1,3,6,9]. Removing the underlined letters from "10101011011" will produce a string "0011111", which is sorted.
In the second and the third testcases the sequences are already sorted.
In the fourth testcase you can choose a sequence a=[3]. s'= "11", which is sorted.
In the fifth testcase there is no way to choose a sequence a such that s' is sorted. | instruction | 0 | 90,368 | 0 | 180,736 |
Tags: brute force, dp, greedy, implementation
Correct Solution:
```
for _ in range(int(input())):
s=input()
n=len(s)
f=True
o,z = s.count('1'),s.count('0')
if o==n or z==n:
print('YES')
continue
for i in range(n-1):
if s[i]=='1' and s[i+1]=='1':
for i in range(i+2,n-1):
if s[i]=='0' and s[i+1]=='0':
f=False
if f:
print('YES')
else:
print('NO')
``` | output | 1 | 90,368 | 0 | 180,737 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s, consisting only of characters '0' or '1'. Let |s| be the length of s.
You are asked to choose some integer k (k > 0) and find a sequence a of length k such that:
* 1 ≤ a_1 < a_2 < ... < a_k ≤ |s|;
* a_{i-1} + 1 < a_i for all i from 2 to k.
The characters at positions a_1, a_2, ..., a_k are removed, the remaining characters are concatenated without changing the order. So, in other words, the positions in the sequence a should not be adjacent.
Let the resulting string be s'. s' is called sorted if for all i from 2 to |s'| s'_{i-1} ≤ s'_i.
Does there exist such a sequence a that the resulting string s' is sorted?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains a string s (2 ≤ |s| ≤ 100). Each character is either '0' or '1'.
Output
For each testcase print "YES" if there exists a sequence a such that removing the characters at positions a_1, a_2, ..., a_k and concatenating the parts without changing the order produces a sorted string.
Otherwise, print "NO".
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
10101011011
0000
11111
110
1100
Output
YES
YES
YES
YES
NO
Note
In the first testcase you can choose a sequence a=[1,3,6,9]. Removing the underlined letters from "10101011011" will produce a string "0011111", which is sorted.
In the second and the third testcases the sequences are already sorted.
In the fourth testcase you can choose a sequence a=[3]. s'= "11", which is sorted.
In the fifth testcase there is no way to choose a sequence a such that s' is sorted. | instruction | 0 | 90,369 | 0 | 180,738 |
Tags: brute force, dp, greedy, implementation
Correct Solution:
```
import sys
import math
import bisect
import functools
from functools import lru_cache
from sys import stdin, stdout
from math import gcd, floor, sqrt, log, ceil
from heapq import heappush, heappop, heapify
from collections import defaultdict as dd
from collections import Counter as cc
from bisect import bisect_left as bl
from bisect import bisect_right as br
def lcm(a, b):
return abs(a*b) // math.gcd(a, b)
'''
testcase = sys.argv[1]
f = open(testcase, "r")
input = f.readline
'''
sys.setrecursionlimit(100000000)
intinp = lambda: int(input().strip())
stripinp = lambda: input().strip()
fltarr = lambda: list(map(float,input().strip().split()))
intarr = lambda: list(map(int,input().strip().split()))
ceildiv = lambda x,d: x//d if(x%d==0) else x//d+1
MOD=1_000_000_007
@lru_cache(None)
def help(need, left):
if left == "":
#print(sorted(curr), curr, s)
return True
if len(left) > 1:
if left[0] == "0" and need and left[1] == "1":
return help(need, left[2:])
elif left[1] == "0" and need and left[0] == "1":
return help(need, left[1:])
elif need and left[1] == "0" and left[0] == "0":
return False
else:
return help(need or int(left[1]), left[2:]) or help(need or int(left[0]), left[1:])
else:
if left[0] == "0" and need:
return help(need, left[1:])
else:
return help(need or int(left[0]), left[1:]) or help(need, left[1:])
num_cases = intinp()
#num_cases = 1
for _ in range(num_cases):
#n, k = intarr()
#n = intinp()
s = stripinp()
#arr = intarr()
if help(0, s):
print("YES")
else:
print("NO")
``` | output | 1 | 90,369 | 0 | 180,739 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s, consisting only of characters '0' or '1'. Let |s| be the length of s.
You are asked to choose some integer k (k > 0) and find a sequence a of length k such that:
* 1 ≤ a_1 < a_2 < ... < a_k ≤ |s|;
* a_{i-1} + 1 < a_i for all i from 2 to k.
The characters at positions a_1, a_2, ..., a_k are removed, the remaining characters are concatenated without changing the order. So, in other words, the positions in the sequence a should not be adjacent.
Let the resulting string be s'. s' is called sorted if for all i from 2 to |s'| s'_{i-1} ≤ s'_i.
Does there exist such a sequence a that the resulting string s' is sorted?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains a string s (2 ≤ |s| ≤ 100). Each character is either '0' or '1'.
Output
For each testcase print "YES" if there exists a sequence a such that removing the characters at positions a_1, a_2, ..., a_k and concatenating the parts without changing the order produces a sorted string.
Otherwise, print "NO".
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
10101011011
0000
11111
110
1100
Output
YES
YES
YES
YES
NO
Note
In the first testcase you can choose a sequence a=[1,3,6,9]. Removing the underlined letters from "10101011011" will produce a string "0011111", which is sorted.
In the second and the third testcases the sequences are already sorted.
In the fourth testcase you can choose a sequence a=[3]. s'= "11", which is sorted.
In the fifth testcase there is no way to choose a sequence a such that s' is sorted. | instruction | 0 | 90,370 | 0 | 180,740 |
Tags: brute force, dp, greedy, implementation
Correct Solution:
```
import sys
def solve(s):
n = len(s)
ans = False
for i in range(n):
local_ans = True
removal = []
for j in range(i):
if s[j] == "1":
removal.append(j)
for j in range(i, n):
if s[j] == "0":
removal.append(j)
for i in range(1, len(removal)):
if removal[i-1] + 1 >= removal[i]:
local_ans = False
ans |= local_ans
return 'YES' if ans else 'NO'
if 1 == 2:
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
else:
input = sys.stdin.readline
T = int(input())
for _ in range(T):
s = input()
print(solve(s))
``` | output | 1 | 90,370 | 0 | 180,741 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s, consisting only of characters '0' or '1'. Let |s| be the length of s.
You are asked to choose some integer k (k > 0) and find a sequence a of length k such that:
* 1 ≤ a_1 < a_2 < ... < a_k ≤ |s|;
* a_{i-1} + 1 < a_i for all i from 2 to k.
The characters at positions a_1, a_2, ..., a_k are removed, the remaining characters are concatenated without changing the order. So, in other words, the positions in the sequence a should not be adjacent.
Let the resulting string be s'. s' is called sorted if for all i from 2 to |s'| s'_{i-1} ≤ s'_i.
Does there exist such a sequence a that the resulting string s' is sorted?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains a string s (2 ≤ |s| ≤ 100). Each character is either '0' or '1'.
Output
For each testcase print "YES" if there exists a sequence a such that removing the characters at positions a_1, a_2, ..., a_k and concatenating the parts without changing the order produces a sorted string.
Otherwise, print "NO".
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
10101011011
0000
11111
110
1100
Output
YES
YES
YES
YES
NO
Note
In the first testcase you can choose a sequence a=[1,3,6,9]. Removing the underlined letters from "10101011011" will produce a string "0011111", which is sorted.
In the second and the third testcases the sequences are already sorted.
In the fourth testcase you can choose a sequence a=[3]. s'= "11", which is sorted.
In the fifth testcase there is no way to choose a sequence a such that s' is sorted. | instruction | 0 | 90,371 | 0 | 180,742 |
Tags: brute force, dp, greedy, implementation
Correct Solution:
```
import sys
input = sys.stdin.readline
for _ in range(int(input())):
s = input()
try:
i = s.index('11') + 2
if '00' in s[i:]:
print("NO")
else:
print("YES")
except:
print("YES")
``` | output | 1 | 90,371 | 0 | 180,743 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s, consisting only of characters '0' or '1'. Let |s| be the length of s.
You are asked to choose some integer k (k > 0) and find a sequence a of length k such that:
* 1 ≤ a_1 < a_2 < ... < a_k ≤ |s|;
* a_{i-1} + 1 < a_i for all i from 2 to k.
The characters at positions a_1, a_2, ..., a_k are removed, the remaining characters are concatenated without changing the order. So, in other words, the positions in the sequence a should not be adjacent.
Let the resulting string be s'. s' is called sorted if for all i from 2 to |s'| s'_{i-1} ≤ s'_i.
Does there exist such a sequence a that the resulting string s' is sorted?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains a string s (2 ≤ |s| ≤ 100). Each character is either '0' or '1'.
Output
For each testcase print "YES" if there exists a sequence a such that removing the characters at positions a_1, a_2, ..., a_k and concatenating the parts without changing the order produces a sorted string.
Otherwise, print "NO".
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
10101011011
0000
11111
110
1100
Output
YES
YES
YES
YES
NO
Note
In the first testcase you can choose a sequence a=[1,3,6,9]. Removing the underlined letters from "10101011011" will produce a string "0011111", which is sorted.
In the second and the third testcases the sequences are already sorted.
In the fourth testcase you can choose a sequence a=[3]. s'= "11", which is sorted.
In the fifth testcase there is no way to choose a sequence a such that s' is sorted.
Submitted Solution:
```
t= int(input())
for p in range(t):
n= input()
if n.find("11") == -1 or n.find("00")== -1:
print("YES")
else:
if n.find("00", n.find("11")+ 1) == -1:
print("YES")
else:
print("NO")
``` | instruction | 0 | 90,372 | 0 | 180,744 |
Yes | output | 1 | 90,372 | 0 | 180,745 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s, consisting only of characters '0' or '1'. Let |s| be the length of s.
You are asked to choose some integer k (k > 0) and find a sequence a of length k such that:
* 1 ≤ a_1 < a_2 < ... < a_k ≤ |s|;
* a_{i-1} + 1 < a_i for all i from 2 to k.
The characters at positions a_1, a_2, ..., a_k are removed, the remaining characters are concatenated without changing the order. So, in other words, the positions in the sequence a should not be adjacent.
Let the resulting string be s'. s' is called sorted if for all i from 2 to |s'| s'_{i-1} ≤ s'_i.
Does there exist such a sequence a that the resulting string s' is sorted?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains a string s (2 ≤ |s| ≤ 100). Each character is either '0' or '1'.
Output
For each testcase print "YES" if there exists a sequence a such that removing the characters at positions a_1, a_2, ..., a_k and concatenating the parts without changing the order produces a sorted string.
Otherwise, print "NO".
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
10101011011
0000
11111
110
1100
Output
YES
YES
YES
YES
NO
Note
In the first testcase you can choose a sequence a=[1,3,6,9]. Removing the underlined letters from "10101011011" will produce a string "0011111", which is sorted.
In the second and the third testcases the sequences are already sorted.
In the fourth testcase you can choose a sequence a=[3]. s'= "11", which is sorted.
In the fifth testcase there is no way to choose a sequence a such that s' is sorted.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
import math
BUFSIZE = 8192
from collections import Counter
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now---------------------------------------------------
# t=1
t=(int)(input())
for _ in range(t):
# n=(int)(input())
# l=list(map(int,input().split()))
# a,b=map(int,input().split())
s=input()
# p=-1
# if '1' not in s:
# p=-1
# else:
# p=s.index('1')
# if p==len(s)-1 or p==-1:
# print("YES")
# else:
# f=True
# for i in range(p+1,len(s)):
# if s[i]==s[i-1]=='0':
# f=False
# break
# if f:
# print("YES")
# else:
# print("NO")
f=0
for i in range (len(s)):
r=1
for j in range (1,i):
if s[j]==s[j-1]=='1':
r=0
break
for j in range (i+2,len(s)):
if s[j]==s[j-1]=='0':
r=0
break
if r==1:
f=1
break
if f:
print("YES")
else:
print("NO")
``` | instruction | 0 | 90,373 | 0 | 180,746 |
Yes | output | 1 | 90,373 | 0 | 180,747 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s, consisting only of characters '0' or '1'. Let |s| be the length of s.
You are asked to choose some integer k (k > 0) and find a sequence a of length k such that:
* 1 ≤ a_1 < a_2 < ... < a_k ≤ |s|;
* a_{i-1} + 1 < a_i for all i from 2 to k.
The characters at positions a_1, a_2, ..., a_k are removed, the remaining characters are concatenated without changing the order. So, in other words, the positions in the sequence a should not be adjacent.
Let the resulting string be s'. s' is called sorted if for all i from 2 to |s'| s'_{i-1} ≤ s'_i.
Does there exist such a sequence a that the resulting string s' is sorted?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains a string s (2 ≤ |s| ≤ 100). Each character is either '0' or '1'.
Output
For each testcase print "YES" if there exists a sequence a such that removing the characters at positions a_1, a_2, ..., a_k and concatenating the parts without changing the order produces a sorted string.
Otherwise, print "NO".
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
10101011011
0000
11111
110
1100
Output
YES
YES
YES
YES
NO
Note
In the first testcase you can choose a sequence a=[1,3,6,9]. Removing the underlined letters from "10101011011" will produce a string "0011111", which is sorted.
In the second and the third testcases the sequences are already sorted.
In the fourth testcase you can choose a sequence a=[3]. s'= "11", which is sorted.
In the fifth testcase there is no way to choose a sequence a such that s' is sorted.
Submitted Solution:
```
import sys
lines = sys.stdin.readlines()
T = int(lines[0].strip())
for t in range(T):
s = lines[t+1].strip()
L = len(s)
satisfy = True
pairOne = False
ind = -1
pairZero = False
for i in range(1, L):
if s[i] == '1' and s[i-1] == '1':
pairOne = True
ind = i
break
if pairOne:
for i in range(ind+2, L):
if s[i] == '0' and s[i-1] == '0':
pairZero = True
break
if pairZero:
print("NO")
else:
print("YES")
``` | instruction | 0 | 90,374 | 0 | 180,748 |
Yes | output | 1 | 90,374 | 0 | 180,749 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s, consisting only of characters '0' or '1'. Let |s| be the length of s.
You are asked to choose some integer k (k > 0) and find a sequence a of length k such that:
* 1 ≤ a_1 < a_2 < ... < a_k ≤ |s|;
* a_{i-1} + 1 < a_i for all i from 2 to k.
The characters at positions a_1, a_2, ..., a_k are removed, the remaining characters are concatenated without changing the order. So, in other words, the positions in the sequence a should not be adjacent.
Let the resulting string be s'. s' is called sorted if for all i from 2 to |s'| s'_{i-1} ≤ s'_i.
Does there exist such a sequence a that the resulting string s' is sorted?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains a string s (2 ≤ |s| ≤ 100). Each character is either '0' or '1'.
Output
For each testcase print "YES" if there exists a sequence a such that removing the characters at positions a_1, a_2, ..., a_k and concatenating the parts without changing the order produces a sorted string.
Otherwise, print "NO".
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
10101011011
0000
11111
110
1100
Output
YES
YES
YES
YES
NO
Note
In the first testcase you can choose a sequence a=[1,3,6,9]. Removing the underlined letters from "10101011011" will produce a string "0011111", which is sorted.
In the second and the third testcases the sequences are already sorted.
In the fourth testcase you can choose a sequence a=[3]. s'= "11", which is sorted.
In the fifth testcase there is no way to choose a sequence a such that s' is sorted.
Submitted Solution:
```
from collections import deque, defaultdict
from math import sqrt, ceil, factorial, floor, inf, log2, sqrt
import bisect
import copy
from itertools import combinations
import sys
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def input(): return sys.stdin.readline().strip()
for _ in range(int(input())):
##n=int(input())
s=input()
flag=0
for i in range(len(s)):
ind=i
c,d=[],[]
for j in range(ind):
if s[j]=='1':
c.append(j)
for j in range(ind+1,len(s)):
if s[j]=='0':
d.append(j)
m=0
for i in range(1,len(c)):
if c[i]-c[i-1]<2:
m=1
break
for i in range(1,len(d)):
if d[i]-d[i-1]<2:
m=1
break
if m==0:
flag=1
break
if flag==0:
print("NO")
else:
print("YES")
``` | instruction | 0 | 90,375 | 0 | 180,750 |
Yes | output | 1 | 90,375 | 0 | 180,751 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s, consisting only of characters '0' or '1'. Let |s| be the length of s.
You are asked to choose some integer k (k > 0) and find a sequence a of length k such that:
* 1 ≤ a_1 < a_2 < ... < a_k ≤ |s|;
* a_{i-1} + 1 < a_i for all i from 2 to k.
The characters at positions a_1, a_2, ..., a_k are removed, the remaining characters are concatenated without changing the order. So, in other words, the positions in the sequence a should not be adjacent.
Let the resulting string be s'. s' is called sorted if for all i from 2 to |s'| s'_{i-1} ≤ s'_i.
Does there exist such a sequence a that the resulting string s' is sorted?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains a string s (2 ≤ |s| ≤ 100). Each character is either '0' or '1'.
Output
For each testcase print "YES" if there exists a sequence a such that removing the characters at positions a_1, a_2, ..., a_k and concatenating the parts without changing the order produces a sorted string.
Otherwise, print "NO".
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
10101011011
0000
11111
110
1100
Output
YES
YES
YES
YES
NO
Note
In the first testcase you can choose a sequence a=[1,3,6,9]. Removing the underlined letters from "10101011011" will produce a string "0011111", which is sorted.
In the second and the third testcases the sequences are already sorted.
In the fourth testcase you can choose a sequence a=[3]. s'= "11", which is sorted.
In the fifth testcase there is no way to choose a sequence a such that s' is sorted.
Submitted Solution:
```
import sys
from math import gcd,ceil,sqrt
INF = float('inf')
MOD = 998244353
mod = 10**9+7
from collections import Counter,defaultdict as df
from functools import reduce
def counter(l): return dict(Counter(l))
def so(x): return {k: v for k, v in sorted(x.items(), key=lambda item: item[1])}
def rev(l): return list(reversed(l))
def ini(): return int(sys.stdin.readline())
def inp(): return map(int, sys.stdin.readline().strip().split())
def li(): return list(map(int, sys.stdin.readline().strip().split()))
def input(): return sys.stdin.readline().strip()
def inputm(n,m):return [li() for i in range(m)]
for i in range(ini()):
s=input()
n=len(s)
c1=-1
c2=-1
for i in range(n-1):
if s[i]==s[i+1]:
if s[i]=='1':
c1=i
else:
c2=i
if c1==-1 or c2==-1 or c1>c2:
print('YES')
else:
print('NO')
``` | instruction | 0 | 90,376 | 0 | 180,752 |
No | output | 1 | 90,376 | 0 | 180,753 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s, consisting only of characters '0' or '1'. Let |s| be the length of s.
You are asked to choose some integer k (k > 0) and find a sequence a of length k such that:
* 1 ≤ a_1 < a_2 < ... < a_k ≤ |s|;
* a_{i-1} + 1 < a_i for all i from 2 to k.
The characters at positions a_1, a_2, ..., a_k are removed, the remaining characters are concatenated without changing the order. So, in other words, the positions in the sequence a should not be adjacent.
Let the resulting string be s'. s' is called sorted if for all i from 2 to |s'| s'_{i-1} ≤ s'_i.
Does there exist such a sequence a that the resulting string s' is sorted?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains a string s (2 ≤ |s| ≤ 100). Each character is either '0' or '1'.
Output
For each testcase print "YES" if there exists a sequence a such that removing the characters at positions a_1, a_2, ..., a_k and concatenating the parts without changing the order produces a sorted string.
Otherwise, print "NO".
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
10101011011
0000
11111
110
1100
Output
YES
YES
YES
YES
NO
Note
In the first testcase you can choose a sequence a=[1,3,6,9]. Removing the underlined letters from "10101011011" will produce a string "0011111", which is sorted.
In the second and the third testcases the sequences are already sorted.
In the fourth testcase you can choose a sequence a=[3]. s'= "11", which is sorted.
In the fifth testcase there is no way to choose a sequence a such that s' is sorted.
Submitted Solution:
```
import collections
import math
n = int(input())
for i in range(n):
s = input()
ans1 = []
ans2 = []
for i in range(len(s)):
if s[i] == "0":
ans1.append(i)
else:
ans2.append(i)
if len(ans1) <= 1 or ans1 == [i for i in range(len(s))] or len(ans2)<=1 or ans2 == [i for i in range(len(s))]:
print('YES')
else:
flag1 = 1
for i in range(1,len(ans1)):
if ans1[i] > ans1[i-1]+1:
pass
else:
flag1 = 0
flag2 = 1
for i in range(1,len(ans2)):
if ans2[i] > ans2[i-1]+1:
pass
else:
flag2 = 0
if flag1 or flag2:
print("Yes")
else:
print('nO')
``` | instruction | 0 | 90,377 | 0 | 180,754 |
No | output | 1 | 90,377 | 0 | 180,755 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s, consisting only of characters '0' or '1'. Let |s| be the length of s.
You are asked to choose some integer k (k > 0) and find a sequence a of length k such that:
* 1 ≤ a_1 < a_2 < ... < a_k ≤ |s|;
* a_{i-1} + 1 < a_i for all i from 2 to k.
The characters at positions a_1, a_2, ..., a_k are removed, the remaining characters are concatenated without changing the order. So, in other words, the positions in the sequence a should not be adjacent.
Let the resulting string be s'. s' is called sorted if for all i from 2 to |s'| s'_{i-1} ≤ s'_i.
Does there exist such a sequence a that the resulting string s' is sorted?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains a string s (2 ≤ |s| ≤ 100). Each character is either '0' or '1'.
Output
For each testcase print "YES" if there exists a sequence a such that removing the characters at positions a_1, a_2, ..., a_k and concatenating the parts without changing the order produces a sorted string.
Otherwise, print "NO".
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
10101011011
0000
11111
110
1100
Output
YES
YES
YES
YES
NO
Note
In the first testcase you can choose a sequence a=[1,3,6,9]. Removing the underlined letters from "10101011011" will produce a string "0011111", which is sorted.
In the second and the third testcases the sequences are already sorted.
In the fourth testcase you can choose a sequence a=[3]. s'= "11", which is sorted.
In the fifth testcase there is no way to choose a sequence a such that s' is sorted.
Submitted Solution:
```
t=int(input())
for i in range(t):
s=input()
l=[]
flag=0
last=0
for t in s:
l.append(int(t))
lim=len(l)
i=0
while(i<lim):
if(l[i]==1 and flag==0):
if(last+1==i):
flag=1
else:
last=i
if(l[i]==0 and flag==1):
if(last+1==i):
flag=-2
break;
else:
last=i
i+=1
if(flag==-2):
print("No");
else:
print("YES");
l.clear()
flag=0
last=0
``` | instruction | 0 | 90,378 | 0 | 180,756 |
No | output | 1 | 90,378 | 0 | 180,757 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s, consisting only of characters '0' or '1'. Let |s| be the length of s.
You are asked to choose some integer k (k > 0) and find a sequence a of length k such that:
* 1 ≤ a_1 < a_2 < ... < a_k ≤ |s|;
* a_{i-1} + 1 < a_i for all i from 2 to k.
The characters at positions a_1, a_2, ..., a_k are removed, the remaining characters are concatenated without changing the order. So, in other words, the positions in the sequence a should not be adjacent.
Let the resulting string be s'. s' is called sorted if for all i from 2 to |s'| s'_{i-1} ≤ s'_i.
Does there exist such a sequence a that the resulting string s' is sorted?
Input
The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
Then the descriptions of t testcases follow.
The only line of each testcase contains a string s (2 ≤ |s| ≤ 100). Each character is either '0' or '1'.
Output
For each testcase print "YES" if there exists a sequence a such that removing the characters at positions a_1, a_2, ..., a_k and concatenating the parts without changing the order produces a sorted string.
Otherwise, print "NO".
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES are all recognized as positive answer).
Example
Input
5
10101011011
0000
11111
110
1100
Output
YES
YES
YES
YES
NO
Note
In the first testcase you can choose a sequence a=[1,3,6,9]. Removing the underlined letters from "10101011011" will produce a string "0011111", which is sorted.
In the second and the third testcases the sequences are already sorted.
In the fourth testcase you can choose a sequence a=[3]. s'= "11", which is sorted.
In the fifth testcase there is no way to choose a sequence a such that s' is sorted.
Submitted Solution:
```
for i in range(int(input())):
s=input()
k=list(s)
k.sort()
k="".join(k)
if k==s:
print("YES")
else:
s=s.replace("1","",1)
if "100" in s:
print("NO")
else:
print("YES")
``` | instruction | 0 | 90,379 | 0 | 180,758 |
No | output | 1 | 90,379 | 0 | 180,759 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string S of length n with each character being one of the first m lowercase English letters.
Calculate how many different strings T of length n composed from the first m lowercase English letters exist such that the length of LCS (longest common subsequence) between S and T is n - 1.
Recall that LCS of two strings S and T is the longest string C such that C both in S and T as a subsequence.
Input
The first line contains two numbers n and m denoting the length of string S and number of first English lowercase characters forming the character set for strings (1 ≤ n ≤ 100 000, 2 ≤ m ≤ 26).
The second line contains string S.
Output
Print the only line containing the answer.
Examples
Input
3 3
aaa
Output
6
Input
3 3
aab
Output
11
Input
1 2
a
Output
1
Input
10 9
abacadefgh
Output
789
Note
For the first sample, the 6 possible strings T are: aab, aac, aba, aca, baa, caa.
For the second sample, the 11 possible strings T are: aaa, aac, aba, abb, abc, aca, acb, baa, bab, caa, cab.
For the third sample, the only possible string T is b. | instruction | 0 | 90,531 | 0 | 181,062 |
Tags: dp, greedy
Correct Solution:
```
def main():
n, m = map(int, input().split())
s = input()
k = sum(s[i] != s[i - 1] for i in range(1, n)) + 1
x = i = 0
while i < n - 1:
if s[i] != s[i + 1]:
j = i
while i + 2 < n and s[i] == s[i + 2]:
i += 1
j = (i - j) + 2
x += j * (j - 1) // 2
i += 1
ans = k * (m * n - n) - x
print(ans)
main()
``` | output | 1 | 90,531 | 0 | 181,063 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings a and b. You have to remove the minimum possible number of consecutive (standing one after another) characters from string b in such a way that it becomes a subsequence of string a. It can happen that you will not need to remove any characters at all, or maybe you will have to remove all of the characters from b and make it empty.
Subsequence of string s is any such string that can be obtained by erasing zero or more characters (not necessarily consecutive) from string s.
Input
The first line contains string a, and the second line — string b. Both of these strings are nonempty and consist of lowercase letters of English alphabet. The length of each string is no bigger than 105 characters.
Output
On the first line output a subsequence of string a, obtained from b by erasing the minimum number of consecutive characters.
If the answer consists of zero characters, output «-» (a minus sign).
Examples
Input
hi
bob
Output
-
Input
abca
accepted
Output
ac
Input
abacaba
abcdcba
Output
abcba
Note
In the first example strings a and b don't share any symbols, so the longest string that you can get is empty.
In the second example ac is a subsequence of a, and at the same time you can obtain it by erasing consecutive symbols cepted from string b. | instruction | 0 | 90,580 | 0 | 181,160 |
Tags: binary search, hashing, strings, two pointers
Correct Solution:
```
a = list(input())
b = list(input())
s = [-1]
i = 0
j = 0
while i < len(a) and j < len(b):
if a[i] == b[j]:
s.append(i)
j += 1
i += 1
e = [10**6]
i = len(a) - 1
j = len(b) - 1
while i >= 0 and j >= 0:
if a[i] == b[j]:
e.append(i)
j -= 1
i -= 1
ans_len = 0
ans = ['-']
i = len(s) - 1
j = 0
while i >= 0 and j < len(e):
if s[i] < e[j] and i + j - 1< len(b):
new_len = j + i
if new_len > ans_len:
ans_len = new_len
if j > 0:
ans = b[:i] + b[-j:]
else:
ans = b[:i]
j += 1
else:
i -= 1
print(''.join(ans))
``` | output | 1 | 90,580 | 0 | 181,161 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings a and b. You have to remove the minimum possible number of consecutive (standing one after another) characters from string b in such a way that it becomes a subsequence of string a. It can happen that you will not need to remove any characters at all, or maybe you will have to remove all of the characters from b and make it empty.
Subsequence of string s is any such string that can be obtained by erasing zero or more characters (not necessarily consecutive) from string s.
Input
The first line contains string a, and the second line — string b. Both of these strings are nonempty and consist of lowercase letters of English alphabet. The length of each string is no bigger than 105 characters.
Output
On the first line output a subsequence of string a, obtained from b by erasing the minimum number of consecutive characters.
If the answer consists of zero characters, output «-» (a minus sign).
Examples
Input
hi
bob
Output
-
Input
abca
accepted
Output
ac
Input
abacaba
abcdcba
Output
abcba
Note
In the first example strings a and b don't share any symbols, so the longest string that you can get is empty.
In the second example ac is a subsequence of a, and at the same time you can obtain it by erasing consecutive symbols cepted from string b. | instruction | 0 | 90,581 | 0 | 181,162 |
Tags: binary search, hashing, strings, two pointers
Correct Solution:
```
import math
def prefixIds(a, b):
prefSubsId = [math.inf] * len(b)
# print(a)
# print(b)
bId = 0
aId = 0
while aId < len(a):
if bId == len(b):
break
if a[aId] == b[bId]:
prefSubsId[bId] = aId + 1
bId += 1
aId += 1
else:
aId += 1
return prefSubsId
a = input()
b = input()
# print(a)
# print(b)
n = len(b)
prefLens = prefixIds(a, b)
suffLens = prefixIds(a[::-1], b[::-1])[::-1]
# for i in range(n):
# if suffLens[i] != math.inf:
# suffLens[i] = len(a) - suffLens[i]
# print(*prefLens, sep='\t')
# print(*suffLens, sep='\t')
prefLen = 0
suffLen = 0
minCutLen = n
lBorder = -1
rBorder = n
while suffLen < n and suffLens[suffLen] == math.inf:
suffLen += 1
curCutLen = suffLen
# print(curCutLen)
if curCutLen < minCutLen:
minCutLen = curCutLen
rBorder = suffLen
while prefLen < suffLen and prefLens[prefLen] != math.inf:
while suffLen < n and prefLens[prefLen] + suffLens[suffLen] > len(a):
# print(suffLen)
suffLen += 1
# print(prefLen)
# print(suffLen)
curCutLen = suffLen - prefLen - 1
# print(curCutLen)
if curCutLen < minCutLen:
minCutLen = curCutLen
lBorder = prefLen
rBorder = suffLen
prefLen += 1
# print(prefLens[prefLen])
# print(suffLens[suffLen])
# # print()
# print("pref, suff")
# print(prefLen)
# print(suffLen)
# print(minCutLen)
# print(n)
# print(lBorder)
# print(rBorder)
if minCutLen == n:
print('-')
elif minCutLen == 0:
print(b)
else:
print(b[:lBorder + 1] + b[rBorder:])
# print(maxPrefLen)
# print(maxSuffLen)
``` | output | 1 | 90,581 | 0 | 181,163 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings a and b. You have to remove the minimum possible number of consecutive (standing one after another) characters from string b in such a way that it becomes a subsequence of string a. It can happen that you will not need to remove any characters at all, or maybe you will have to remove all of the characters from b and make it empty.
Subsequence of string s is any such string that can be obtained by erasing zero or more characters (not necessarily consecutive) from string s.
Input
The first line contains string a, and the second line — string b. Both of these strings are nonempty and consist of lowercase letters of English alphabet. The length of each string is no bigger than 105 characters.
Output
On the first line output a subsequence of string a, obtained from b by erasing the minimum number of consecutive characters.
If the answer consists of zero characters, output «-» (a minus sign).
Examples
Input
hi
bob
Output
-
Input
abca
accepted
Output
ac
Input
abacaba
abcdcba
Output
abcba
Note
In the first example strings a and b don't share any symbols, so the longest string that you can get is empty.
In the second example ac is a subsequence of a, and at the same time you can obtain it by erasing consecutive symbols cepted from string b. | instruction | 0 | 90,582 | 0 | 181,164 |
Tags: binary search, hashing, strings, two pointers
Correct Solution:
```
from sys import stdin
def main():
t = stdin.readline()
s = stdin.readline()
n = len(s) - 1
m = len(t) - 1
post = [-1] * n
ss = n - 1
st = m - 1
while st >= 0 and ss >= 0:
if t[st] == s[ss]:
post[ss] = st
ss -= 1
st -= 1
pre = [-1] * n
ss = 0
st = 0
while st < m and ss < n:
if t[st] == s[ss]:
pre[ss] = st
ss += 1
st += 1
low = 0
high = n
min_ans = n
start = -1
end = -1
while low < high:
mid = (low + high) >> 1
ok = False
if post[mid] != -1:
if mid < min_ans:
min_ans = mid
start = 0
end = mid - 1
ok = True
for i in range(1, n - mid):
if pre[i - 1] != -1 and post[i + mid] != -1 and post[i + mid] > pre[i - 1]:
if mid < min_ans:
min_ans = mid
start = i
end = i + mid - 1
ok = True
if pre[n - mid - 1] != -1:
if mid < min_ans:
min_ans = mid
start = n - mid
end = n - 1
ok = True
if not ok:
low = mid + 1
else:
high = mid
ans = []
for i in range(n):
if start <= i <= end:
continue
ans.append(s[i])
if min_ans == n:
print('-')
else:
print(''.join(ans))
if __name__ == '__main__':
main()
``` | output | 1 | 90,582 | 0 | 181,165 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings a and b. You have to remove the minimum possible number of consecutive (standing one after another) characters from string b in such a way that it becomes a subsequence of string a. It can happen that you will not need to remove any characters at all, or maybe you will have to remove all of the characters from b and make it empty.
Subsequence of string s is any such string that can be obtained by erasing zero or more characters (not necessarily consecutive) from string s.
Input
The first line contains string a, and the second line — string b. Both of these strings are nonempty and consist of lowercase letters of English alphabet. The length of each string is no bigger than 105 characters.
Output
On the first line output a subsequence of string a, obtained from b by erasing the minimum number of consecutive characters.
If the answer consists of zero characters, output «-» (a minus sign).
Examples
Input
hi
bob
Output
-
Input
abca
accepted
Output
ac
Input
abacaba
abcdcba
Output
abcba
Note
In the first example strings a and b don't share any symbols, so the longest string that you can get is empty.
In the second example ac is a subsequence of a, and at the same time you can obtain it by erasing consecutive symbols cepted from string b. | instruction | 0 | 90,583 | 0 | 181,166 |
Tags: binary search, hashing, strings, two pointers
Correct Solution:
```
a = input()
b = input()
prefix = [-1] * len(b)
postfix = [-1] * len(b)
prefix[0] = a.find(b[0])
postfix[len(b) - 1] = a.rfind(b[len(b) - 1])
for i in range(1, len(b)):
prefix[i] = a.find(b[i], prefix[i - 1] + 1)
if prefix[i] == -1:
break
for i in range(len(b) - 2, -1, -1):
postfix[i] = a.rfind(b[i], 0, postfix[i+1])
if postfix[i] == -1:
break
best_left = -1
best_right = len(b)
left = -1
while left + 1 < len(b) and prefix[left + 1] != -1:
left += 1
if left > -1:
best_left = left
best_right = len(b)
right = len(b)
while right - 1 >= 0 and postfix[right - 1] != -1:
right -= 1
if right < len(b) and right + 1 < best_right - best_left:
best_left = -1
best_right = right
left = 0
right = len(b)
while left < right and postfix[right - 1] != -1 and postfix[right - 1] > prefix[left]:
right -= 1
while prefix[left] != -1 and left < right < len(b):
while right < len(b) and postfix[right] <= prefix[left]:
right += 1
if right >= len(b):
break
if right - left < best_right - best_left:
best_left = left
best_right = right
left += 1
if left == right:
right += 1
res = b[:best_left + 1] + b[best_right:]
if res == "":
print("-")
else:
print(res)
``` | output | 1 | 90,583 | 0 | 181,167 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings a and b. You have to remove the minimum possible number of consecutive (standing one after another) characters from string b in such a way that it becomes a subsequence of string a. It can happen that you will not need to remove any characters at all, or maybe you will have to remove all of the characters from b and make it empty.
Subsequence of string s is any such string that can be obtained by erasing zero or more characters (not necessarily consecutive) from string s.
Input
The first line contains string a, and the second line — string b. Both of these strings are nonempty and consist of lowercase letters of English alphabet. The length of each string is no bigger than 105 characters.
Output
On the first line output a subsequence of string a, obtained from b by erasing the minimum number of consecutive characters.
If the answer consists of zero characters, output «-» (a minus sign).
Examples
Input
hi
bob
Output
-
Input
abca
accepted
Output
ac
Input
abacaba
abcdcba
Output
abcba
Note
In the first example strings a and b don't share any symbols, so the longest string that you can get is empty.
In the second example ac is a subsequence of a, and at the same time you can obtain it by erasing consecutive symbols cepted from string b. | instruction | 0 | 90,584 | 0 | 181,168 |
Tags: binary search, hashing, strings, two pointers
Correct Solution:
```
import sys
s, t = input(), '*'+input()
n, m = len(s), len(t)-1
inf = 10**9
pre, suf = [-1] + [inf]*(m+1), [-1]*(m+1) + [n]
i = 0
for j in range(1, m+1):
while i < n and s[i] != t[j]:
i += 1
if i == n:
break
pre[j] = i
i += 1
i = n-1
for j in range(m, 0, -1):
while 0 <= i and s[i] != t[j]:
i -= 1
if i == -1:
break
suf[j] = i
i -= 1
max_len, best_l, best_r = 0, 0, 0
j = 1
for i in range(m+1):
j = max(j, i+1)
while j <= m and pre[i] >= suf[j]:
j += 1
if pre[i] == inf:
break
if max_len < i + m + 1 - j:
max_len = i + m + 1 - j
best_l, best_r = i, j
pre_s = t[1:best_l+1]
suf_s = t[best_r:]
print(pre_s + suf_s if max_len else '-')
``` | output | 1 | 90,584 | 0 | 181,169 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings a and b. You have to remove the minimum possible number of consecutive (standing one after another) characters from string b in such a way that it becomes a subsequence of string a. It can happen that you will not need to remove any characters at all, or maybe you will have to remove all of the characters from b and make it empty.
Subsequence of string s is any such string that can be obtained by erasing zero or more characters (not necessarily consecutive) from string s.
Input
The first line contains string a, and the second line — string b. Both of these strings are nonempty and consist of lowercase letters of English alphabet. The length of each string is no bigger than 105 characters.
Output
On the first line output a subsequence of string a, obtained from b by erasing the minimum number of consecutive characters.
If the answer consists of zero characters, output «-» (a minus sign).
Examples
Input
hi
bob
Output
-
Input
abca
accepted
Output
ac
Input
abacaba
abcdcba
Output
abcba
Note
In the first example strings a and b don't share any symbols, so the longest string that you can get is empty.
In the second example ac is a subsequence of a, and at the same time you can obtain it by erasing consecutive symbols cepted from string b. | instruction | 0 | 90,585 | 0 | 181,170 |
Tags: binary search, hashing, strings, two pointers
Correct Solution:
```
import sys
input = sys.stdin.readline
def solve():
a = input().strip()
b = input().strip()
n = len(a)
m = len(b)
c = [0]*m
d = [0]*m
p = 0
for i in range(m):
for j in range(p, n):
if a[j] == b[i]:
c[i] = j
p = j + 1
break
else:
for j in range(i,m):
c[j] = n
break
p = n-1
for i in range(m-1,-1,-1):
for j in range(p,-1,-1):
if a[j] == b[i]:
d[i] = j
p = j - 1
break
else:
for j in range(i,-1,-1):
d[j] = -1
break
j = 0
while j < m and d[j] < 0:
j += 1
res = (m-j, 0, j)
for i in range(m):
p = c[i]
if p == n:
break
while j < m and (j <= i or d[j] <= p):
j += 1
res = max(res,(i+1+m-j,i+1,j))
if res[0] == 0:
print('-')
else:
print(b[:res[1]]+b[res[2]:])
solve()
``` | output | 1 | 90,585 | 0 | 181,171 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings a and b. You have to remove the minimum possible number of consecutive (standing one after another) characters from string b in such a way that it becomes a subsequence of string a. It can happen that you will not need to remove any characters at all, or maybe you will have to remove all of the characters from b and make it empty.
Subsequence of string s is any such string that can be obtained by erasing zero or more characters (not necessarily consecutive) from string s.
Input
The first line contains string a, and the second line — string b. Both of these strings are nonempty and consist of lowercase letters of English alphabet. The length of each string is no bigger than 105 characters.
Output
On the first line output a subsequence of string a, obtained from b by erasing the minimum number of consecutive characters.
If the answer consists of zero characters, output «-» (a minus sign).
Examples
Input
hi
bob
Output
-
Input
abca
accepted
Output
ac
Input
abacaba
abcdcba
Output
abcba
Note
In the first example strings a and b don't share any symbols, so the longest string that you can get is empty.
In the second example ac is a subsequence of a, and at the same time you can obtain it by erasing consecutive symbols cepted from string b. | instruction | 0 | 90,586 | 0 | 181,172 |
Tags: binary search, hashing, strings, two pointers
Correct Solution:
```
a, b = str(input()), str(input())
p, s, pp, sp = [0] * len(b), [0] * len(b), 0, len(a) - 1
p[0] = a.find(b[0])
while sp >= 0 and a[sp] != b[-1]:
sp -= 1
s[len(b) - 1] = sp
pp, sp = p[0] + 1, sp - 1
for i in range(1, len(b)):
if p[i - 1] == -1:
p[i] = -1
else:
while pp < len(a) and a[pp] != b[i]:
pp += 1
p[i] = (pp if pp < len(a) else -1)
pp += 1
for i in range(len(b) - 2, -1, -1):
if s[i + 1] == -1:
s[i] = -1
else:
while sp >= 0 and a[sp] != b[i]:
sp -= 1
s[i] = (sp if sp >= 0 else -1)
sp -= 1
anss, ans = 0, (-1, -1, -1)
for i in range(0, len(b)):
if p[i] == -1:
break
while anss < len(b) and (s[anss] <= p[i] or s[anss] == -1) or anss <= i:
anss += 1
ln = (i + 1 if p[i] != -1 else 0) + (len(b) - anss if anss < len(b) and s[anss] != -1 else 0)
if ln > ans[0]:
ans = (ln, i, anss)
only_s = len(b) - 1
while only_s - 1 >= 0 and s[only_s - 1] >= 0:
only_s -= 1
if only_s >= 0 and s[-1] >= 0 and len(b) - only_s > ans[0]:
print(b[only_s:])
elif ans[0] == -1:
print("-")
else:
print(b[:ans[1] + 1] + (b[ans[2]:] if ans[2] < len(b) else ""))
``` | output | 1 | 90,586 | 0 | 181,173 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings a and b. You have to remove the minimum possible number of consecutive (standing one after another) characters from string b in such a way that it becomes a subsequence of string a. It can happen that you will not need to remove any characters at all, or maybe you will have to remove all of the characters from b and make it empty.
Subsequence of string s is any such string that can be obtained by erasing zero or more characters (not necessarily consecutive) from string s.
Input
The first line contains string a, and the second line — string b. Both of these strings are nonempty and consist of lowercase letters of English alphabet. The length of each string is no bigger than 105 characters.
Output
On the first line output a subsequence of string a, obtained from b by erasing the minimum number of consecutive characters.
If the answer consists of zero characters, output «-» (a minus sign).
Examples
Input
hi
bob
Output
-
Input
abca
accepted
Output
ac
Input
abacaba
abcdcba
Output
abcba
Note
In the first example strings a and b don't share any symbols, so the longest string that you can get is empty.
In the second example ac is a subsequence of a, and at the same time you can obtain it by erasing consecutive symbols cepted from string b. | instruction | 0 | 90,587 | 0 | 181,174 |
Tags: binary search, hashing, strings, two pointers
Correct Solution:
```
a=input()
b=input()
pre=[len(a) for i in range(len(b))]
suf=[-1 for i in range(len(b))]
temp=0
whole=0
for i in range(len(a)):
if b[temp]==a[i]:
pre[temp]=i
temp+=1
if temp==len(b):
whole=1
break
temp=len(b)-1
for i in range(len(a)-1,-1,-1):
if b[temp]==a[i]:
suf[temp]=i
temp-=1
if temp==-1:
whole=1
break
ans=[]
index=0
for i in range(len(b)):
temp=pre[i]
index=max(i+1,index)
start=index
for j in range(start,len(b)):
if suf[j]>pre[i]:
index=j
break
else:
index=len(b)
if index!=len(b) and pre[i]!=len(a):
ans+=[len(b)+1+i-index]
elif pre[i]==len(a):
ans+=[0]
elif index==len(b) and pre[i]!=len(a):
ans+=[i+1]
else:
ans+=[0]
MAXsufANS=0
MAXsufindex=len(b)
for i in range(len(b)-1,-1,-1):
if suf[i]!=-1:
MAXsufANS=len(b)-i
MAXsufindex=i
MAX=0
MAXindex=-1
for i in range(len(b)):
if ans[i]>MAX:
MAXindex=i
MAX=ans[i]
usesuf=0
if MAXsufANS>MAX:
usesuf=1
if whole==1:
print(b)
else:
if max(MAX,MAXsufANS)==0:
print('-')
else:
if usesuf==1:
anss=b[MAXsufindex:]
print(anss)
else:
m=MAXindex
L=MAX
anss=b[:m+1]+b[len(b)-(MAX-(m+1)):]
print(anss)
``` | output | 1 | 90,587 | 0 | 181,175 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings a and b. You have to remove the minimum possible number of consecutive (standing one after another) characters from string b in such a way that it becomes a subsequence of string a. It can happen that you will not need to remove any characters at all, or maybe you will have to remove all of the characters from b and make it empty.
Subsequence of string s is any such string that can be obtained by erasing zero or more characters (not necessarily consecutive) from string s.
Input
The first line contains string a, and the second line — string b. Both of these strings are nonempty and consist of lowercase letters of English alphabet. The length of each string is no bigger than 105 characters.
Output
On the first line output a subsequence of string a, obtained from b by erasing the minimum number of consecutive characters.
If the answer consists of zero characters, output «-» (a minus sign).
Examples
Input
hi
bob
Output
-
Input
abca
accepted
Output
ac
Input
abacaba
abcdcba
Output
abcba
Note
In the first example strings a and b don't share any symbols, so the longest string that you can get is empty.
In the second example ac is a subsequence of a, and at the same time you can obtain it by erasing consecutive symbols cepted from string b.
Submitted Solution:
```
#import sys
#sys.stdin = open('in', 'r')
#n = int(input())
#a = [int(x) for x in input().split()]
#n,m = map(int, input().split())
s1 = input()
s2 = input()
l1 = len(s1)
l2 = len(s2)
dl = {}
dr = {}
i1 = 0
i2 = 0
while i1 < l1 and i2 < l2:
while i1 < l1 and s1[i1] != s2[i2]:
i1 += 1
if i1 < l1:
dl[i2] = i1
i2 += 1
i1 += 1
lmax = i2
if lmax == l2:
print(s2)
else:
i1 = l1 - 1
i2 = l2 - 1
while i1 >= 0 and i2 >= 0:
while i1 >= 0 and s1[i1] != s2[i2]:
i1 -= 1
if i1 >= 0:
dr[i2] = i1
i2 -= 1
i1 -= 1
rmax = i2
le = -1
re = -1
if l2 - lmax < rmax + 1:
rcnt = l2 - lmax
ls = 0
rs = lmax
else:
rcnt = rmax + 1
ls = rmax + 1
rs = l2
rr = rmax + 1
for ll in range(lmax):
while rr < l2 and (rr <= ll or dl[ll] >= dr[rr]):
rr += 1
if rr < l2:
dif = rr - ll - 1
if dif < rcnt:
rcnt = dif
ls = 0
rs = ll + 1
le = rr
re = l2
result = s2[ls:rs]
if le != -1:
result += s2[le:re]
print(result if len(result) > 0 else '-')
``` | instruction | 0 | 90,588 | 0 | 181,176 |
Yes | output | 1 | 90,588 | 0 | 181,177 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings a and b. You have to remove the minimum possible number of consecutive (standing one after another) characters from string b in such a way that it becomes a subsequence of string a. It can happen that you will not need to remove any characters at all, or maybe you will have to remove all of the characters from b and make it empty.
Subsequence of string s is any such string that can be obtained by erasing zero or more characters (not necessarily consecutive) from string s.
Input
The first line contains string a, and the second line — string b. Both of these strings are nonempty and consist of lowercase letters of English alphabet. The length of each string is no bigger than 105 characters.
Output
On the first line output a subsequence of string a, obtained from b by erasing the minimum number of consecutive characters.
If the answer consists of zero characters, output «-» (a minus sign).
Examples
Input
hi
bob
Output
-
Input
abca
accepted
Output
ac
Input
abacaba
abcdcba
Output
abcba
Note
In the first example strings a and b don't share any symbols, so the longest string that you can get is empty.
In the second example ac is a subsequence of a, and at the same time you can obtain it by erasing consecutive symbols cepted from string b.
Submitted Solution:
```
from macpath import curdir
import bisect
a=input()
b=input()
la=len(a)
lb=len(b)
forb=[la]*lb
cur=0
for i in range(lb):
while cur<la and a[cur]!=b[i]:
cur+=1
if cur<la:
forb[i]=cur
cur+=1
else:
break
cur=la-1
bacb=[-1]*lb
for i in range(lb-1,-1,-1):
while cur>=0 and a[cur]!=b[i]:
cur-=1
if cur>=0:
bacb[i]=cur
cur-=1
else:
break
res=lb
start=0
end=lb
for i in range(lb):
low=-1
if i>0 :
low=forb[i-1]
if low>=la:
break
j=bisect.bisect_right(bacb,low)
if res>j-i>=0:
res=j-i
start=i
end=j
if res==lb:
print('-')
else:
print(b[:start]+b[end:])
``` | instruction | 0 | 90,589 | 0 | 181,178 |
Yes | output | 1 | 90,589 | 0 | 181,179 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings a and b. You have to remove the minimum possible number of consecutive (standing one after another) characters from string b in such a way that it becomes a subsequence of string a. It can happen that you will not need to remove any characters at all, or maybe you will have to remove all of the characters from b and make it empty.
Subsequence of string s is any such string that can be obtained by erasing zero or more characters (not necessarily consecutive) from string s.
Input
The first line contains string a, and the second line — string b. Both of these strings are nonempty and consist of lowercase letters of English alphabet. The length of each string is no bigger than 105 characters.
Output
On the first line output a subsequence of string a, obtained from b by erasing the minimum number of consecutive characters.
If the answer consists of zero characters, output «-» (a minus sign).
Examples
Input
hi
bob
Output
-
Input
abca
accepted
Output
ac
Input
abacaba
abcdcba
Output
abcba
Note
In the first example strings a and b don't share any symbols, so the longest string that you can get is empty.
In the second example ac is a subsequence of a, and at the same time you can obtain it by erasing consecutive symbols cepted from string b.
Submitted Solution:
```
# cook your dish here
a = list(input())
b = list(input())
m = len(a)
n = len(b)
s = [-1]
i = 0
j = 0
while i<m and j<n:
if(a[i]==b[j]):
s.append(i)
j+=1
i+=1
e = [1000000]
i = m-1
j= n-1
while i>=0 and j>=0:
if(a[i]==b[j]):
e.append(i)
j-=1
i-=1
i = len(s)-1
j = 0
ans = "-"
ctr = 0
while j<len(e) and i>=0:
if(s[i]<e[j]):
if(i+j > ctr and i+j<=len(b)):
ctr = i+j
if(j>0):
ans = b[:i] + b[-j:]
else:
ans = b[:i]
j+=1
else:
i -= 1
print(''.join(ans))
``` | instruction | 0 | 90,590 | 0 | 181,180 |
Yes | output | 1 | 90,590 | 0 | 181,181 |
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