DSA interview question

Minimum Bit Flips to Convert a Number

Medium · Algorithms · Topics: bit manipulation

The problem

A configuration register is currently set to start, and it must be changed to goal. The hardware can only flip one bit at a time, and each flip is slow, so the operator wants the minimum number of flips. Every bit where the two numbers differ must be flipped exactly once, and no other flips are needed. XOR marks exactly those differing positions with 1s, so the answer is the number of set bits in start ^ goal.

Example: start = 10 (1010), goal = 7 (0111) → 3.

Function signature

int minBitFlips(int start, int goal);  // 0 <= start, goal <= 10^9

Test cases

Input `(start, goal)`	Output	Explanation
`(10, 7)`	`3`	1010 ^ 0111 = 1101
`(3, 4)`	`3`	011 ^ 100 = 111
`(0, 0)`	`0`
`(0, 1023)`	`10`	Ten 1 bits to create
`(1000000000, 0)`	`13`	10⁹ has 13 set bits

(Confirm: a flip changes one bit; any bit position may be flipped, including leading zeros.)

How interviewers phrase it

"Return the minimum number of bit flips to convert start to goal" (LeetCode 2220).
It's the Hamming distance of two integers (LeetCode 461).

Clarifying questions to ask

Can the numbers be negative? (Not in LeetCode's version.)
Is a flip always one bit? (Yes.)
Should leading zeros be considered? (Yes: any position may be flipped.)

Follow-ups to expect

"Why is XOR the right tool?"
"How do you count the 1s?"
"Total Hamming distance over all pairs in an array?"
"Why is this the minimum?"
"Complexity?"

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