jeremymanning.md

Problem 2220: Minimum Bit Flips to Convert Number

Initial thoughts (stream-of-consciousness)

• First we need to convert each number to its binary representation. We can do that using bin(x)[2:]
• Next, we need to append leading 0s as needed. We need to append abs(len(start) - len(goal)) leading 0s to whichever (of start or goal) has a shorter binary representation
• Finally, now that the two representations have the same length, we can just loop through bit by bit and count up the number of mismatches

Refining the problem, round 2 thoughts

• Nothing too complex here; let's implement it!

Attempted solution(s)

classSolution: defminBitFlips(self, start: int, goal: int) ->int: start=bin(start)[2:] goal=bin(goal)[2:] leading_zeros=''.join('0'*abs(len(start) -len(goal))) iflen(start) <len(goal): start=leading_zeros+startelse: goal=leading_zeros+goalflips=0fori, jinzip(start, goal): ifi!=j: flips+=1returnflips

• Given test cases pass
• Submitting...

Solved! Easy peasy lemon squeezy! 🍋