title: Minimum Number Of Bracket Reversals Needed To Make An Expression Balanced
signature: function minimum_number_of_bracket_reversals_needed_to_make_an_expression_balanced(expr)
test_code: using Printf
  
  # SOLUTION CODE
  # ============================================
  PLACEHOLDER_CODE_BODY
  
  # TESTING CODE 
  # ============================================
  function validate_solution(actual, expected)
      return actual == expected
  end
  
  function driver(expr, expected)
      try
  
          if validate_solution(PLACEHOLDER_FN_NAME(expr), expected)
              return "PASSED";
          end
          return "FAILED";
      catch exception_obj
          return string(Base.typename(typeof(exception_obj)).wrapper);
      end
  end
  
  function main()
      
      result = driver("}{", 1);
      @printf("TEST-0...%s\n",result);
      
      result = driver("{{{", -1);
      @printf("TEST-1...%s\n",result);
      
      result = driver("{{{{", 2);
      @printf("TEST-2...%s\n",result);
      
      result = driver("{{{{}}", 3);
      @printf("TEST-3...%s\n",result);
      
      result = driver("}{{}}{{{", 4);
      @printf("TEST-4...%s\n",result);
      
      result = driver("{}", 1);
      @printf("TEST-5...%s\n",result);
      
      result = driver("", 0);
      @printf("TEST-6...%s\n",result);
      
      result = driver("8", -1);
      @printf("TEST-7...%s\n",result);
      
      result = driver("01111000", 4);
      @printf("TEST-8...%s\n",result);
      
      result = driver("XPkERzHcpId", -1);
      @printf("TEST-9...%s\n",result);
      
  end
  
  main()
source_py: def minimum_number_of_bracket_reversals_needed_to_make_an_expression_balanced(expr):
      lenn = len(expr)
      if (lenn % 2):
          return (- 1)
      s = []
      for i in range(lenn):
          if ((expr[i] == '') and len(s)):
              if (s[0] == ''):
                  s.pop(0)
              else:
                  s.insert(0, expr[i])
          else:
              s.insert(0, expr[i])
      red_len = len(s)
      n = 0
      while (len(s) and (s[0] == '')):
          s.pop(0)
          n += 1
      return ((red_len // 2) + (n % 2))
source_cpp: int minimum_number_of_bracket_reversals_needed_to_make_an_expression_balanced(string expr) {
    int len = expr . length ( );
    if ( len % 2 ) return - 1;
    stack < char > s;
    for ( int i = 0;
    i < len;
    i ++ ) {
      if ( expr [ i ] == '}' && ! s . empty ( ) ) {
        if ( s . top ( ) == '{' ) s . pop ( );
        else s . push ( expr [ i ] );
      }
      else s . push ( expr [ i ] );
    }
    int red_len = s . size ( );
    int n = 0;
    while ( ! s . empty ( ) && s . top ( ) == '{' ) {
      s . pop ( );
      n ++;
    }
    return ( red_len / 2 + n % 2 );
  }