LeetCode #3226 — EASY

Number of Bit Changes to Make Two Integers Equal

Build confidence with an intuition-first walkthrough focused on bit manipulation fundamentals.

Solve on LeetCode

The Problem

Problem Statement

You are given two positive integers n and k.

You can choose any bit in the binary representation of n that is equal to 1 and change it to 0.

Return the number of changes needed to make n equal to k. If it is impossible, return -1.

Example 1:

Input: n = 13, k = 4

Output: 2

Explanation:
Initially, the binary representations of n and k are n = (1101)₂ and k = (0100)₂.
We can change the first and fourth bits of n. The resulting integer is n = (0100)₂ = k.

Example 2:

Input: n = 21, k = 21

Output: 0

Explanation:
n and k are already equal, so no changes are needed.

Example 3:

Input: n = 14, k = 13

Output: -1

Explanation:
It is not possible to make n equal to k.

Constraints:

1 <= n, k <= 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given two positive integers n and k. You can choose any bit in the binary representation of n that is equal to 1 and change it to 0. Return the number of changes needed to make n equal to k. If it is impossible, return -1.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Bit Manipulation

Example 1

13
4

Example 2

21
21

Example 3

14
13

Step 02

Core Insight

What unlocks the optimal approach

Find the binary representations of <code>n</code> and <code>k</code>.
Any bit that is equal to 1 in <code>n</code> and equal to 0 in <code>k</code> needs to be changed.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3226: Number of Bit Changes to Make Two Integers Equal
class Solution {
    public int minChanges(int n, int k) {
        return (n & k) != k ? -1 : Integer.bitCount(n ^ k);
    }
}

// Accepted solution for LeetCode #3226: Number of Bit Changes to Make Two Integers Equal
func minChanges(n int, k int) int {
	if n&k != k {
		return -1
	}
	return bits.OnesCount(uint(n ^ k))
}

# Accepted solution for LeetCode #3226: Number of Bit Changes to Make Two Integers Equal
class Solution:
    def minChanges(self, n: int, k: int) -> int:
        return -1 if n & k != k else (n ^ k).bit_count()

// Accepted solution for LeetCode #3226: Number of Bit Changes to Make Two Integers Equal
impl Solution {
    pub fn min_changes(n: i32, k: i32) -> i32 {
        if (n & k) != k {
            -1
        } else {
            (n ^ k).count_ones() as i32
        }
    }
}

// Accepted solution for LeetCode #3226: Number of Bit Changes to Make Two Integers Equal
function minChanges(n: number, k: number): number {
    return (n & k) !== k ? -1 : bitCount(n ^ k);
}

function bitCount(i: number): number {
    i = i - ((i >>> 1) & 0x55555555);
    i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
    i = (i + (i >>> 4)) & 0x0f0f0f0f;
    i = i + (i >>> 8);
    i = i + (i >>> 16);
    return i & 0x3f;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.