Move from brute-force thinking to an efficient approach using dynamic programming strategy.
You are given a positive integer n, you can do the following operation any number of times:
2 from n.Return the minimum number of operations to make n equal to 0.
A number x is power of 2 if x == 2i where i >= 0.
Example 1:
Input: n = 39 Output: 3 Explanation: We can do the following operations: - Add 20 = 1 to n, so now n = 40. - Subtract 23 = 8 from n, so now n = 32. - Subtract 25 = 32 from n, so now n = 0. It can be shown that 3 is the minimum number of operations we need to make n equal to 0.
Example 2:
Input: n = 54 Output: 3 Explanation: We can do the following operations: - Add 21 = 2 to n, so now n = 56. - Add 23 = 8 to n, so now n = 64. - Subtract 26 = 64 from n, so now n = 0. So the minimum number of operations is 3.
Constraints:
1 <= n <= 105Problem summary: You are given a positive integer n, you can do the following operation any number of times: Add or subtract a power of 2 from n. Return the minimum number of operations to make n equal to 0. A number x is power of 2 if x == 2i where i >= 0.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming · Greedy · Bit Manipulation
39
54
plus-one)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2571: Minimum Operations to Reduce an Integer to 0
class Solution {
public int minOperations(int n) {
int ans = 0, cnt = 0;
for (; n > 0; n >>= 1) {
if ((n & 1) == 1) {
++cnt;
} else if (cnt > 0) {
++ans;
cnt = cnt == 1 ? 0 : 1;
}
}
ans += cnt == 1 ? 1 : 0;
ans += cnt > 1 ? 2 : 0;
return ans;
}
}
// Accepted solution for LeetCode #2571: Minimum Operations to Reduce an Integer to 0
func minOperations(n int) (ans int) {
cnt := 0
for ; n > 0; n >>= 1 {
if n&1 == 1 {
cnt++
} else if cnt > 0 {
ans++
if cnt == 1 {
cnt = 0
} else {
cnt = 1
}
}
}
if cnt == 1 {
ans++
} else if cnt > 1 {
ans += 2
}
return
}
# Accepted solution for LeetCode #2571: Minimum Operations to Reduce an Integer to 0
class Solution:
def minOperations(self, n: int) -> int:
ans = cnt = 0
while n:
if n & 1:
cnt += 1
elif cnt:
ans += 1
cnt = 0 if cnt == 1 else 1
n >>= 1
if cnt == 1:
ans += 1
elif cnt > 1:
ans += 2
return ans
// Accepted solution for LeetCode #2571: Minimum Operations to Reduce an Integer to 0
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (java):
// // Accepted solution for LeetCode #2571: Minimum Operations to Reduce an Integer to 0
// class Solution {
// public int minOperations(int n) {
// int ans = 0, cnt = 0;
// for (; n > 0; n >>= 1) {
// if ((n & 1) == 1) {
// ++cnt;
// } else if (cnt > 0) {
// ++ans;
// cnt = cnt == 1 ? 0 : 1;
// }
// }
// ans += cnt == 1 ? 1 : 0;
// ans += cnt > 1 ? 2 : 0;
// return ans;
// }
// }
// Accepted solution for LeetCode #2571: Minimum Operations to Reduce an Integer to 0
function minOperations(n: number): number {
let [ans, cnt] = [0, 0];
for (; n; n >>= 1) {
if (n & 1) {
++cnt;
} else if (cnt) {
++ans;
cnt = cnt === 1 ? 0 : 1;
}
}
if (cnt === 1) {
++ans;
} else if (cnt > 1) {
ans += 2;
}
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: 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.
Wrong move: Locally optimal choices may fail globally.
Usually fails on: Counterexamples appear on crafted input orderings.
Fix: Verify with exchange argument or monotonic objective before committing.