title: Count Distinct Occurrences As A Subsequence
signature: countDistinctOccurrencesAsASubsequence :: String -> String -> Integer
  countDistinctOccurrencesAsASubsequence s t = 
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 s t expected = do
      if (validateSolution (PLACEHOLDER_FN_NAME s t) expected)
          then "PASSED"
      else "FAILED"
  
  main = do
      printf "TEST-0...%s\n" (driver ("banana") ("ban") (3) )
  
      printf "TEST-1...%s\n" (driver ("49597223") ("42") (2) )
  
      printf "TEST-2...%s\n" (driver ("1000010000011") ("010") (20) )
  
      printf "TEST-3...%s\n" (driver ("BTlzufK") ("EpsVuzP lf") (0) )
  
      printf "TEST-4...%s\n" (driver ("3474007") ("370") (2) )
  
      printf "TEST-5...%s\n" (driver ("0010") ("00000") (0) )
  
      printf "TEST-6...%s\n" (driver ("dKHhoTD") ("doT") (1) )
  
      printf "TEST-7...%s\n" (driver ("9123259723") ("123") (4) )
  
      printf "TEST-8...%s\n" (driver ("11001000111110") ("0") (6) )
  
      printf "TEST-9...%s\n" (driver ("iY WJlIZ") ("iI") (1) )
  
source_py: def count_distinct_occurrences_as_a_subsequence(S, T):
      m = len(T)
      n = len(S)
      if (m > n):
          return 0
      mat = [[0 for _ in range((n + 1))] for __ in range((m + 1))]
      for i in range(1, (m + 1)):
          mat[i][0] = 0
      for j in range((n + 1)):
          mat[0][j] = 1
      for i in range(1, (m + 1)):
          for j in range(1, (n + 1)):
              if (T[(i - 1)] != S[(j - 1)]):
                  mat[i][j] = mat[i][(j - 1)]
              else:
                  mat[i][j] = (mat[i][(j - 1)] + mat[(i - 1)][(j - 1)])
      return mat[m][n]
source_cpp: int count_distinct_occurrences_as_a_subsequence(string S, string T) {
    int m = T . length ( ), n = S . length ( );
    if ( m > n ) return 0;
    int mat [ m + 1 ] [ n + 1 ];
    for ( int i = 1;
    i <= m;
    i ++ ) mat [ i ] [ 0 ] = 0;
    for ( int j = 0;
    j <= n;
    j ++ ) mat [ 0 ] [ j ] = 1;
    for ( int i = 1;
    i <= m;
    i ++ ) {
      for ( int j = 1;
      j <= n;
      j ++ ) {
        if ( T [ i - 1 ] != S [ j - 1 ] ) mat [ i ] [ j ] = mat [ i ] [ j - 1 ];
        else mat [ i ] [ j ] = mat [ i ] [ j - 1 ] + mat [ i - 1 ] [ j - 1 ];
      }
    }
    return mat [ m ] [ n ];
  }