title: Count Distinct Occurrences As A Subsequence
signature: function count_distinct_occurrences_as_a_subsequence(s, t)
test_code: using Printf
  
  # SOLUTION CODE
  # ============================================
  PLACEHOLDER_CODE_BODY
  
  # TESTING CODE 
  # ============================================
  function validate_solution(actual, expected)
      return actual == expected
  end
  
  function driver(s, t, expected)
      try
  
          if validate_solution(PLACEHOLDER_FN_NAME(s, t), expected)
              return "PASSED";
          end
          return "FAILED";
      catch exception_obj
          return string(Base.typename(typeof(exception_obj)).wrapper);
      end
  end
  
  function main()
      
      result = driver("banana", "ban", 3);
      @printf("TEST-0...%s\n",result);
      
      result = driver("49597223", "42", 2);
      @printf("TEST-1...%s\n",result);
      
      result = driver("1000010000011", "010", 20);
      @printf("TEST-2...%s\n",result);
      
      result = driver("BTlzufK", "EpsVuzP lf", 0);
      @printf("TEST-3...%s\n",result);
      
      result = driver("3474007", "370", 2);
      @printf("TEST-4...%s\n",result);
      
      result = driver("0010", "00000", 0);
      @printf("TEST-5...%s\n",result);
      
      result = driver("dKHhoTD", "doT", 1);
      @printf("TEST-6...%s\n",result);
      
      result = driver("9123259723", "123", 4);
      @printf("TEST-7...%s\n",result);
      
      result = driver("11001000111110", "0", 6);
      @printf("TEST-8...%s\n",result);
      
      result = driver("iY WJlIZ", "iI", 1);
      @printf("TEST-9...%s\n",result);
      
  end
  
  main()
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 ];
  }