Python Code [Intuitive]
\nclass Solution:\n    def solve(self,a,b, A, B):\n            # a is the character with count A (may or maynot be  \'a\')\n\t\t\t# b is the character with coun
```\nclass Solution:\n    def solve(self,a,b, A, B):\n            # a is the character with count A (may or maynot be  \'a\')\n\t\t\t# b is the character with count B (may or maynot be  \'b\')\n\t\t\t# Note : A >= B (check how we\'re calling the function)\n\t\t\tans = []\n            idx = 2\n            numB = B\n            while A and B:\n                if A >= 2*B:\n                    ans.append(a)\n                    A-=1\n                    if A:\n                        ans.append(a)\n                        A-=1\n                    ans.append(b)\n                    B-=1\n                else:\n                    ans.append(a)\n                    A-=1\n                    if A:\n                        ans.append(a)\n                        A-=1\n                    ans.append(b)\n                    B-=1\n                    if B:\n                        ans.append(b)\n                        B-=1\n            while A:\n                ans.append(a)\n                A-=1\n            while B:\n                ans.append(b)\n                B-=1\n            return ans\n    def strWithout3a3b(self, A: int, B: int) -> str:\n                \n        ans = []\n        if A < 3 and B < 3:\n            return \'a\'*A + \'b\'*B\n        if B < A:\n            return \'\'.join(self.solve(\'a\',\'b\', A, B))\n        else:\n            return \'\'.join(self.solve(\'b\',\'a\',B, A))\n```\n\nSimple and Intuitive.\nPlease reply if any further explanation is needed.
['Greedy', 'Python3']