Math
String
Description
Given a binary string s
, you can split s
into 3 non-empty strings s1
, s2
, and s3
where s1 + s2 + s3 = s
.
Return the number of ways s
can be split such that the number of ones is the same in s1
, s2
, and s3
. Since the answer may be too large, return it modulo 109 + 7
.
Example 1:
Input: s = "10101"
Output: 4
Explanation: There are four ways to split s in 3 parts where each part contain the same number of letters '1'.
"1|010|1"
"1|01|01"
"10|10|1"
"10|1|01"
Example 2:
Input: s = "1001"
Output: 0
Example 3:
Input: s = "0000"
Output: 3
Explanation: There are three ways to split s in 3 parts.
"0|0|00"
"0|00|0"
"00|0|0"
Constraints:
3 <= s.length <= 105
s[i]
is either '0'
or '1'
.
Solutions
Solution 1
Python3 Java C++ Go
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19 class Solution :
def numWays ( self , s : str ) -> int :
def find ( x ):
t = 0
for i , c in enumerate ( s ):
t += int ( c == '1' )
if t == x :
return i
cnt , m = divmod ( sum ( c == '1' for c in s ), 3 )
if m :
return 0
n = len ( s )
mod = 10 ** 9 + 7
if cnt == 0 :
return (( n - 1 ) * ( n - 2 ) // 2 ) % mod
i1 , i2 = find ( cnt ), find ( cnt + 1 )
j1 , j2 = find ( cnt * 2 ), find ( cnt * 2 + 1 )
return ( i2 - i1 ) * ( j2 - j1 ) % ( 10 ** 9 + 7 )
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36 class Solution {
private String s ;
public int numWays ( String s ) {
this . s = s ;
int cnt = 0 ;
int n = s . length ();
for ( int i = 0 ; i < n ; ++ i ) {
if ( s . charAt ( i ) == '1' ) {
++ cnt ;
}
}
int m = cnt % 3 ;
if ( m != 0 ) {
return 0 ;
}
final int mod = ( int ) 1e9 + 7 ;
if ( cnt == 0 ) {
return ( int ) ((( n - 1L ) * ( n - 2 ) / 2 ) % mod );
}
cnt /= 3 ;
long i1 = find ( cnt ), i2 = find ( cnt + 1 );
long j1 = find ( cnt * 2 ), j2 = find ( cnt * 2 + 1 );
return ( int ) (( i2 - i1 ) * ( j2 - j1 ) % mod );
}
private int find ( int x ) {
int t = 0 ;
for ( int i = 0 ;; ++ i ) {
t += s . charAt ( i ) == '1' ? 1 : 0 ;
if ( t == x ) {
return i ;
}
}
}
}
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31 class Solution {
public :
int numWays ( string s ) {
int cnt = 0 ;
for ( char & c : s ) {
cnt += c == '1' ;
}
int m = cnt % 3 ;
if ( m ) {
return 0 ;
}
const int mod = 1e9 + 7 ;
int n = s . size ();
if ( cnt == 0 ) {
return ( n - 1L L ) * ( n - 2 ) / 2 % mod ;
}
cnt /= 3 ;
auto find = [ & ]( int x ) {
int t = 0 ;
for ( int i = 0 ;; ++ i ) {
t += s [ i ] == '1' ;
if ( t == x ) {
return i ;
}
}
};
int i1 = find ( cnt ), i2 = find ( cnt + 1 );
int j1 = find ( cnt * 2 ), j2 = find ( cnt * 2 + 1 );
return ( 1L L * ( i2 - i1 ) * ( j2 - j1 )) % mod ;
}
};
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32 func numWays ( s string ) int {
cnt := 0
for _ , c := range s {
if c == '1' {
cnt ++
}
}
m := cnt % 3
if m != 0 {
return 0
}
const mod = 1e9 + 7
n := len ( s )
if cnt == 0 {
return ( n - 1 ) * ( n - 2 ) / 2 % mod
}
cnt /= 3
find := func ( x int ) int {
t := 0
for i := 0 ; ; i ++ {
if s [ i ] == '1' {
t ++
if t == x {
return i
}
}
}
}
i1 , i2 := find ( cnt ), find ( cnt + 1 )
j1 , j2 := find ( cnt * 2 ), find ( cnt * 2 + 1 )
return ( i2 - i1 ) * ( j2 - j1 ) % mod
}
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