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Alternate flip_bit_challenge solution

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This solution is clearer

Algorithm description updated for new solution.
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andy.boot 2017-04-06 23:02:33 +01:00
parent 075b51cde3
commit 7dfee269f1
1 changed files with 44 additions and 76 deletions
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120
bit_manipulation/flip_bit/flip_bit_solution.ipynb
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@ -71,35 +71,25 @@
"source": [
"## Algorithm\n",
"\n",
"* seen = []\n",
"* Build a list of sequence counts\n",
" * Look for 0's\n",
" * This will be 0 length if the input has trailing ones\n",
" * Add sequence length to seen\n",
" * Look for 1's\n",
" * Add sequence length to seen\n",
"* Find the largest sequence of ones looking at seen\n",
" * Loop through seen\n",
" * On each iteration of the loop, flip what we are looking for from 0 to 1 and vice versa\n",
" * If seen[i] represents 1's, continue, we only want to process 0's\n",
" * If this is our first iteration:\n",
" * max_result = seen[i+1] + 1 if seen[i] > 0\n",
" * continue\n",
" * If we are looking at leading zeroes (i == len(seen)-1):\n",
" * result = seen[i-1] + 1\n",
" * If we are looking at one zero:\n",
" * result = seen[i+1] + seen[i-1] + 1\n",
" * If we are looking at multiple zeroes:\n",
" * result = max(seen[i+1], seen[i-1]) + 1\n",
" * Update max_result based on result\n",
"* Overview:\n",
" * Use 2 counters. \n",
" * Each counter will track the number of 1's it has seen. \n",
" * When we see a 0 this indicates the end of the run for the long_counter and indicates the bit to flip for the short_counter. We then throw the long counter away, promote the short_counter to the long_counter and reset the short_counter\n",
"\n",
"\n",
"* For each bit:\n",
" * If we see a '1' \n",
" * Increment both counters.\n",
" * If we see a '0' \n",
" * For the long_counter this indicates the end of the row of '1's. Compare our count to best (and update it if better).\n",
" * For the short_counter this indicates the bit we should flip from 0 to 1. Promote it to be the long_counter, it will continue to count bits as the long_counter\n",
" * Reset the short_counter to 0\n",
" \n",
"We should make a note that Python does not have a logical right shift operator built in. We can either use a positive number or implement one for a 32 bit number:\n",
" \n",
" num % 0x100000000 >> n\n",
"\n",
"Complexity:\n",
"* Time: O(b)\n",
"* Space: O(b)"
"* Time: O(lg(n))\n",
"* Space: O(1)\n"
]
},
{
@ -118,62 +108,37 @@
"outputs": [],
"source": [
"class Bits(object):\n",
"\n",
" MAX_BITS = 32\n",
" \n",
" def _build_seen_list(self, num):\n",
" seen = []\n",
" looking_for = 0\n",
" count = 0\n",
" for _ in range(self.MAX_BITS):\n",
" if num & 1 != looking_for:\n",
" seen.append(count)\n",
" looking_for = not looking_for\n",
" count = 0\n",
" count += 1\n",
" num >>= 1\n",
" seen.append(count)\n",
" return seen\n",
" \n",
" def flip_bit(self, num):\n",
" if num is None:\n",
" raise TypeError('num cannot be None')\n",
" if num == -1:\n",
" return self.MAX_BITS\n",
" if num == 0:\n",
" return 1\n",
" seen = self._build_seen_list(num)\n",
" max_result = 0\n",
" looking_for = 0\n",
" for index, count in enumerate(seen):\n",
" result = 0\n",
" # Only look for zeroes\n",
" if looking_for == 1:\n",
" looking_for = not looking_for\n",
" continue\n",
" # First iteration, take trailing zeroes\n",
" # or trailing ones into account\n",
" if index == 0:\n",
" if count != 0:\n",
" # Trailing zeroes\n",
" try:\n",
" result = seen[index + 1] + 1\n",
" except IndexError:\n",
" result = 1\n",
" # Last iteration\n",
" elif index == len(seen) - 1:\n",
" result = 1 + seen[index - 1]\n",
" raise TypeError\n",
" count_no_bit_flip = 0\n",
" count_after_bit_flip = 0\n",
" best = 0\n",
" \n",
" # This handles negative numbers.\n",
" if num < 0:\n",
" num += 2**self.MAX_BITS\n",
" \n",
" for i in range(0, 32):\n",
" c = (num >> i) & 1\n",
" if c == 1:\n",
" count_no_bit_flip += 1\n",
" count_after_bit_flip += 1\n",
" else:\n",
" # One zero\n",
" if count == 1:\n",
" result = seen[index + 1] + seen[index - 1] + 1\n",
" # Multiple zeroes\n",
" else:\n",
" result = max(seen[index + 1], seen[index - 1]) + 1\n",
" if result > max_result:\n",
" max_result = result\n",
" looking_for = not looking_for\n",
" return max_result"
" # If the count after the bit flip > best -> save it.\n",
" if count_after_bit_flip > best:\n",
" best = count_after_bit_flip\n",
" \n",
" # This counter now assumes we flipped this bit\n",
" count_after_bit_flip = count_no_bit_flip + 1\n",
" \n",
" # We start a new counter from this index.\n",
" count_no_bit_flip = 0\n",
" \n",
" return max(count_after_bit_flip, best)\n",
" "
]
},
{
@ -210,6 +175,9 @@
" assert_raises(TypeError, bits.flip_bit, None)\n",
" assert_equal(bits.flip_bit(0), 1)\n",
" assert_equal(bits.flip_bit(-1), bits.MAX_BITS)\n",
" assert_equal(bits.flip_bit(-2), bits.MAX_BITS)\n",
" assert_equal(bits.flip_bit(-3), bits.MAX_BITS)\n",
" assert_equal(bits.flip_bit(-4), bits.MAX_BITS - 1)\n",
" num = int('00001111110111011110001111110000', base=2)\n",
" expected = 10\n",
" assert_equal(bits.flip_bit(num), expected)\n",