[python 3] brutal force with explanation
First, special cases:\n \tIf there is no 1 in list, we can just return [0, len(n)-1].\n \tIf number of 1 cant be divied by 3, return [-1, -1].\n\nThen we find t
First, special cases:\n* \tIf there is no 1 in list, we can just return ```[0, len(n)-1]```.\n* \tIf number of 1 cant be divied by 3, return ```[-1, -1]```.\n\nThen we find the valid binary number, since the leading zeros dont matter here, we just need to find the binary start with "1", therefore we traverse the list until we find the ```2 * m / 3 + 1``` th "1", this "1" is the start of the 3rd binary number.\nTherefore we esaily find what this binary number is.\n\nNext, we just need to find the ```1st``` "1" and the ```m / 3 + 1``` th "1", standing for the begining of 1st and 2nd binary number seperatly. Once we find this "1",  comparing the array consisting this "1" with the 3rd number we have already found.\n\nReturn the index of ```i``` and ```j``` if both arrays are same with 3rd binary number, otherwise ```[-1, -1]```.\n\n```\ndef threeEqualParts(self, arr: List[int]) -> List[int]:\n        m = sum(arr)\n\t\tif m == 0:\n            return [0, n - 1]\n        if m % 3:\n            return [-1, -1]\n\t\t\t\n        k = m // 3\n        i = 0\n        n = len(arr)\n        \n        \n        #start with finding the third (last) binary value in the list\n        for idx in range(n):\n            if arr[idx] == 1:\n                i += 1\n            if i == 2 * k + 1:\n                res = arr[idx :]\n                break\n        reslen = len(res)\n        \n        ans = []\n        \n        #find the first binary value in list\n        for idx in range(n):\n            if arr[idx] == 1:\n                break\n        if arr[idx : idx + reslen] == res:\n            ans.append(idx + reslen - 1)\n        else:\n            return [-1, -1]\n\n        #find the second binary in the list\n        for idx in range(idx + reslen, n):\n            if arr[idx] == 1:\n                break\n        if arr[idx : idx + reslen] == res:\n            ans.append(idx + reslen)\n            return ans\n        else:\n            return [-1, -1]\n```
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