title: Ways Transforming One String Removing 0 Characters
signature: waysTransformingOneStringRemoving0Characters :: String -> String -> Integer
  waysTransformingOneStringRemoving0Characters a b = 
test_code: import Text.Printf
  import Control.Exception (catch, SomeException)
  import Data.Typeable
  import qualified Data.List as List
  import qualified Data.Map as Map
  import qualified Data.Set as Set
  
  -- SOLUTION CODE
  -- ============================================
  
  PLACEHOLDER_CODE_BODY
  
  -- TESTING CODE 
  -- ============================================
  validateSolution :: Integer -> Integer -> Bool
  validateSolution actual expected = do
      actual == expected
  
  
  driver :: String -> String -> Integer -> String
  driver a b expected = do
      if (validateSolution (PLACEHOLDER_FN_NAME a b) expected)
          then "PASSED"
      else "FAILED"
  
  main = do
      printf "TEST-0...%s\n" (driver ("abcccdf") ("abccdf") (3) )
  
      printf "TEST-1...%s\n" (driver ("aabba") ("ab") (4) )
  
      printf "TEST-2...%s\n" (driver ("aabsdfljk") ("aa") (1) )
  
      printf "TEST-3...%s\n" (driver ("IONiqV") ("XKbBiGZ") (0) )
  
      printf "TEST-4...%s\n" (driver ("9667771256770") ("50915176") (0) )
  
      printf "TEST-5...%s\n" (driver ("10001011") ("01") (11) )
  
      printf "TEST-6...%s\n" (driver ("fczbDtMDT") ("FbX") (0) )
  
      printf "TEST-7...%s\n" (driver ("298746088") ("29888") (1) )
  
      printf "TEST-8...%s\n" (driver ("01100011000") ("0") (7) )
  
      printf "TEST-9...%s\n" (driver ("Qk") ("") (1) )
  
source_py: def ways_transforming_one_string_removing_0_characters(a, b):
      n = len(a)
      m = len(b)
      if (m == 0):
          return 1
      dp = [([0] * (n + 1)) for _ in range((m + 1))]
      for i in range(m):
          for j in range(i, n):
              if (i == 0):
                  if (j == 0):
                      if (a[j] == b[i]):
                          dp[i][j] = 1
                      else:
                          dp[i][j] = 0
                  elif (a[j] == b[i]):
                      dp[i][j] = (dp[i][(j - 1)] + 1)
                  else:
                      dp[i][j] = dp[i][(j - 1)]
              elif (a[j] == b[i]):
                  dp[i][j] = (dp[i][(j - 1)] + dp[(i - 1)][(j - 1)])
              else:
                  dp[i][j] = dp[i][(j - 1)]
      return dp[(m - 1)][(n - 1)]
source_cpp: int ways_transforming_one_string_removing_0_characters(string a, string b) {
    int n = a . size ( ), m = b . size ( );
    if ( m == 0 ) return 1;
    int dp [ m + 1 ] [ n + 1 ];
    memset ( dp, 0, sizeof ( dp ) );
    for ( int i = 0;
    i < m;
    i ++ ) {
      for ( int j = i;
      j < n;
      j ++ ) {
        if ( i == 0 ) {
          if ( j == 0 ) dp [ i ] [ j ] = ( a [ j ] == b [ i ] ) ? 1 : 0;
          else if ( a [ j ] == b [ i ] ) dp [ i ] [ j ] = dp [ i ] [ j - 1 ] + 1;
          else dp [ i ] [ j ] = dp [ i ] [ j - 1 ];
        }
        else {
          if ( a [ j ] == b [ i ] ) dp [ i ] [ j ] = dp [ i ] [ j - 1 ] + dp [ i - 1 ] [ j - 1 ];
          else dp [ i ] [ j ] = dp [ i ] [ j - 1 ];
        }
      }
    }
    return dp [ m - 1 ] [ n - 1 ];
  }