Overflow in intermediate arithmetic
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
Given an integer n, you must transform it into 0 using the following operations any number of times:
0th) bit in the binary representation of n.ith bit in the binary representation of n if the (i-1)th bit is set to 1 and the (i-2)th through 0th bits are set to 0.Return the minimum number of operations to transform n into 0.
Example 1:
Input: n = 3 Output: 2 Explanation: The binary representation of 3 is "11". "11" -> "01" with the 2nd operation since the 0th bit is 1. "01" -> "00" with the 1st operation.
Example 2:
Input: n = 6 Output: 4 Explanation: The binary representation of 6 is "110". "110" -> "010" with the 2nd operation since the 1st bit is 1 and 0th through 0th bits are 0. "010" -> "011" with the 1st operation. "011" -> "001" with the 2nd operation since the 0th bit is 1. "001" -> "000" with the 1st operation.
Constraints:
0 <= n <= 109Problem summary: Given an integer n, you must transform it into 0 using the following operations any number of times: Change the rightmost (0th) bit in the binary representation of n. Change the ith bit in the binary representation of n if the (i-1)th bit is set to 1 and the (i-2)th through 0th bits are set to 0. Return the minimum number of operations to transform n into 0.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math · Dynamic Programming · Bit Manipulation
3
6
minimum-number-of-operations-to-make-array-continuous)apply-bitwise-operations-to-make-strings-equal)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1611: Minimum One Bit Operations to Make Integers Zero
class Solution {
public int minimumOneBitOperations(int n) {
int ans = 0;
for (; n > 0; n >>= 1) {
ans ^= n;
}
return ans;
}
}
// Accepted solution for LeetCode #1611: Minimum One Bit Operations to Make Integers Zero
func minimumOneBitOperations(n int) (ans int) {
for ; n > 0; n >>= 1 {
ans ^= n
}
return
}
# Accepted solution for LeetCode #1611: Minimum One Bit Operations to Make Integers Zero
class Solution:
def minimumOneBitOperations(self, n: int) -> int:
ans = 0
while n:
ans ^= n
n >>= 1
return ans
// Accepted solution for LeetCode #1611: Minimum One Bit Operations to Make Integers Zero
impl Solution {
pub fn minimum_one_bit_operations(mut n: i32) -> i32 {
let mut ans = 0;
while n > 0 {
ans ^= n;
n >>= 1;
}
ans
}
}
// Accepted solution for LeetCode #1611: Minimum One Bit Operations to Make Integers Zero
function minimumOneBitOperations(n: number): number {
let ans = 0;
for (; n > 0; n >>= 1) {
ans ^= n;
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.