# 16.02.2026 [190. Reverse Bits]
Reverse bits #easy #bits
16.02.2026
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Problem TLDR
Reverse bits #easy #bits
Intuition
Manually reverse all 31 positions.
Approach
if called many times: use built-in functions (they should be optimized and cached) and use lookup tables if memory allows
Complexity
Time complexity: $$O(log(n))$$
Space complexity: $$O(1)$$
Code
// 103ms
fun reverseBits(n: Int) =
Integer.reverse(n)
/*
(0..31).sumOf { b -> n shr b and 1 shl (31-b) }
n.toString(2).reversed().padEnd(32,'0').toInt(2)
(0..31).fold(0) { r, b -> n shr b and 1 shl (31-b) or r }
*/
// 2ms
pub const reverse_bits: fn(i32) -> i32 = i32::reverse_bits;
/*
pub fn reverse_bits(n: i32) -> i32 {
(0..32).map(|b| (n >> b & 1) << (31-b)).sum()
}
*/