190. Reverse Bits

At a Glance

Topic: bit-manipulation
Pattern: Bit ops / shift accumulate (Bit Manipulation)
Difficulty: Easy
Companies: Amazon, Apple, Microsoft, Google, Adobe
Frequency: High
LeetCode: 190

Problem (one-liner)

Reverse the bit order of a 32-bit unsigned integer and return the value (LeetCode treats as unsigned for this problem).

Core Basics

Opening move: classify the object (interval, multiset, trie node, bitmask, overlapping sub-intervals…) and whether the answer lives in an enumeration, ordering, partition, graph walk, DP table, etc.
Contracts: spell what a valid partial solution looks like at each step—the thing you inductively preserve.
Complexity anchors: state the brute upper bound and the structural fact (sorting, monotonicity, optimal substructure, greedy exchange, hashability) that should beat it.

Understanding

Why brute hurts: name the repeated work or state explosion in one sentence.
Why optimal is safe: the invariant / exchange argument / optimal-subproblem story you would tell a staff engineer.

Memory Hooks

One chant / rule: a single memorable handle (acronym, operator trick, or shape) you can recall under time pressure.

Recognition Cues

"Reverse bits" of fixed width
Build result by shifting in bits from LSB to MSB of output

Study Pattern (SDE3+)

0–3 min: restate I/O and brittle edges aloud, then verbalize two approaches with time/space before you touch the keyboard.
Implementation pass: one-line invariant above the tight loop; state what progress you make each iteration.
5 min extension: pick one constraint twist (immutable input, stream, bounded memory, parallel readers) and explain what in your solution breaks without a refactor.

Diagram

At-a-glance flow (replace with problem-specific Mermaid as you refine this note). camelCase node IDs; no spaces in IDs.

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Approaches

Brute force — stringify bits — O(32) with extra space.
Optimal — iterative bit extraction and push — O(32) time / O(1) space.

Optimal Solution

Go

package main

func reverseBits(value uint32) uint32 {
	var result uint32 = 0
	for shift := 0; shift < 32; shift++ {
		result <<= 1
		result |= value & 1
		value >>= 1
	}
	return result
}

JavaScript

function reverseBits(value) {
	let result = 0;
	let remaining = value >>> 0;
	for (let shift = 0; shift < 32; shift++) {
		result <<= 1;
		result |= remaining & 1;
		remaining >>>= 1;
	}
	return result >>> 0;
}