Program to find number of ways to split a string in Python

PythonServer Side ProgrammingProgramming

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
  • 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
raja
Published on 04-Oct-2021 08:02:11
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