title: Ways Transforming One String Removing 0 Characters
signature: ways_transforming_one_string_removing_0_characters <- function(a, b) {
test_code: 
  
  
  # SOLUTION CODE
  # ============================================
  PLACEHOLDER_CODE_BODY
  
  # TESTING CODE 
  # ============================================
  validate_solution <- function(actual, expected) {
      toString(actual)==toString(expected)
  }
  
  driver <- function(a, b, expected){
      if (validate_solution(PLACEHOLDER_FN_NAME(a, b), expected)){
          "PASSED"
      } else{
          "FAILED"
      }
  }
  
  main <- function() {
      cat(sprintf("TEST-0...%s\n",driver(
          "abcccdf", "abccdf", 
          3L
      )))
  
      cat(sprintf("TEST-1...%s\n",driver(
          "aabba", "ab", 
          4L
      )))
  
      cat(sprintf("TEST-2...%s\n",driver(
          "aabsdfljk", "aa", 
          1L
      )))
  
      cat(sprintf("TEST-3...%s\n",driver(
          "IONiqV", "XKbBiGZ", 
          0L
      )))
  
      cat(sprintf("TEST-4...%s\n",driver(
          "9667771256770", "50915176", 
          0L
      )))
  
      cat(sprintf("TEST-5...%s\n",driver(
          "10001011", "01", 
          11L
      )))
  
      cat(sprintf("TEST-6...%s\n",driver(
          "fczbDtMDT", "FbX", 
          0L
      )))
  
      cat(sprintf("TEST-7...%s\n",driver(
          "298746088", "29888", 
          1L
      )))
  
      cat(sprintf("TEST-8...%s\n",driver(
          "01100011000", "0", 
          7L
      )))
  
      cat(sprintf("TEST-9...%s\n",driver(
          "Qk", "", 
          1L
      )))
  
  }
  
  main()
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 ];
  }