Suppose we have a binary string s we can split s into 3 non-empty strings s1, s2, s3 such that (s1 concatenate s2 concatenate s3 = s). We have to find the number of ways s can be split such that the number of characters '1' is the same in s1, s2, and s3. The answer may be very large so return answer mod 10^9+7.
So, if the input is like s = "11101011", then the output will be 2 because we can split them like "11 | 1010 | 11" and "11 | 101 | 011".
To solve this, we will follow these steps:
- count := count number of 1s in s
- m := 10^9 + 7
- ans := an array of size 2 and fill with 0
- if count mod 3 is not same as 0, then
- return 0
- otherwise when count is same as 0, then
- return (nCr of where n is size of s -1 and r is 2) mod m
- left := 0
- right := size of s - 1
- cum_s := 0, cum_e := 0
- while cum_s <= quotient of count/3 or cum_e <= quotient of count/3, do
- if s[left] is same as "1", then
- cum_s := cum_s + 1
- if s[right] is same as "1", then
- cum_e := cum_e + 1
- if cum_s is same as quotient of count/3, then
- ans[0] := ans[0] + 1
- if cum_e is same as quotient of count/3, then
- ans[1] := ans[1] + 1
- left := left + 1
- right := right - 1
- if s[left] is same as "1", then
- return (ans[0]*ans[1]) mod m
Let us see the following implementation to get better understanding:
Example
def solve(s):
count = s.count("1")
m = 10**9 + 7
ans = [0, 0]
if count % 3 != 0:
return 0
elif count == 0:
return comb(len(s)-1,2) % m
left = 0
right = len(s)-1
cum_s = 0
cum_e = 0
while(cum_s <= count//3 or cum_e <= count//3):
if s[left] == "1":
cum_s+=1
if s[right] == "1":
cum_e+=1
if cum_s == count//3:
ans[0]+=1
if cum_e == count//3:
ans[1]+=1
left += 1
right -= 1
return (ans[0]*ans[1]) % m
s = "11101011"
print(solve(s))Input
"11101011"
Output
2