Minimum Operations to Reduce N to 0 Using Powers of 2

Problem Statement

Given a non-negative integer n, in each step you can add or subtract any power of 2 (i.e., 2^k for some k >= 0). Return the minimum number of operations needed to reduce n to 0.

Example 1:

Input: n = 39
Output: 3
Explanation: 39 = 32 + 4 + 2 + 1 (binary 100111). You can reach 0 in 3 ops:
  39 + 1 = 40       (binary 101000)
  40 - 8 = 32       (binary 100000)
  32 - 32 = 0

Example 2:

Input: n = 54
Output: 3
Explanation: 54 = 110110 in binary. One optimal path:
  54 - 2 = 52       (110100)
  52 + 4 = 56       (111000)
  56 - 8 = 48... no, better:
  54 + 2 = 56 (111000), 56 - 8 = 48 (110000), 48 - 16 = 32, 32 - 32 = 0 is 4 ops.
  Actual: 54 = 64 - 8 - 2 = 3 ops: 54 + 2 = 56, 56 + 8 = 64, 64 - 64 = 0. That's 3 ops.

Constraints:

• 0 <= n <= 10^9 (fits in 32-bit unsigned).

Difficulty: Medium Round: OA

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Pattern Recognition

• What is the input structure? A single integer.
• What are we optimizing? Minimum operations, where each operation is ±2^k.
• What pattern does this fit? Bit manipulation + greedy. Walk the binary representation from LSB to MSB and decide at each 1-bit whether to add or subtract.
• Why does this pattern fit? Adding/subtracting 2^k toggles the k-th bit (and possibly ripples a carry). The minimum number of operations equals the minimum number of bit-toggles needed to reach zero, which is a local decision at each bit.

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Approach

Brute Force — BFS on Integer States

Each state is an integer. From state x, you can reach x + 2^k or x - 2^k for any valid k. BFS from n until you hit 0.

Time: Exponential in the worst case, and state space unbounded (upward). Space: Same. Why insufficient: Intractable for n up to 10^9.

Optimal — Greedy on Binary Representation

Key Insight: Look at the lowest set bit. If it is an isolated 1 (...01), it is cheapest to subtract 1 (one op). If it is part of a run of 1s (...11), it is cheaper to add 1 and let the carry collapse the run upward.

Step-by-step (walk from LSB to MSB):

1. If n is even, shift right (free — bit was 0, nothing to do).
2. Else (LSB is 1):
   • If n == 1 or the next bit is 0 (...01 pattern), subtract 1 (op++).
   • Else (next bit is 1, meaning ...11 run), add 1 (op++), which carries up and may collapse multiple 1s into a single higher 1.
3. Repeat until n == 0.

Why it works:

• For an isolated 1, adding 1 would produce ...10, then you still need to deal with that higher 1 later — net cost >= 1 op for this bit plus downstream work. Subtracting 1 is better.
• For a run of 1s (...0111), adding 1 makes it ...1000 — one op collapses the whole run into a single higher bit.

Time: O(log n) — one pass over the bits. Space: O(1).

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Walkthrough

Trace on n = 39 (binary 100111):

step	n (binary)	n (dec)	LSB	next bit	action	ops
0	100111	39	1	1	run → n + 1 = 40	1
1	101000	40	0	—	shift right	1
2	10100	20	0	—	shift right	1
3	1010	10	0	—	shift right	1
4	101	5	1	0	isolated → n - 1 = 4	2
5	100	4	0	—	shift right	2
6	10	2	0	—	shift right	2
7	1	1	1	0	isolated → n - 1 = 0	3

Answer: 3.

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Implementation

TypeScript

function minOperationsToZero(n: number): number {
    let ops = 0;
    while (n > 0) {
        if ((n & 1) === 0) {
            // Even: shift right (free — bit was 0).
            n >>>= 1;
        } else if (n === 1 || (n & 2) === 0) {
            // Isolated 1 (pattern ...01): subtract 1.
            n -= 1;
            ops++;
        } else {
            // Run of 1s (pattern ...11): add 1 to carry-collapse.
            n += 1;
            ops++;
        }
    }
    return ops;
}

C++

int minOperationsToZero(long long n) {
    int ops = 0;
    while (n > 0) {
        if ((n & 1) == 0) {
            // Even: shift right.
            n >>= 1;
        } else if (n == 1 || (n & 2) == 0) {
            // Isolated 1: subtract 1.
            n -= 1;
            ops++;
        } else {
            // Run of 1s: add 1 (carry).
            n += 1;
            ops++;
        }
    }
    return ops;
}

Watch out:

1. Greedy rule is subtle. Add when you see ...11, subtract when you see ...01. Reversing this doubles the op count on runs.
2. n == 1 special case. There is no "next bit" above bit 0; always subtract 1 here.
3. Using signed right shift in TypeScript (>>) instead of unsigned (>>>) — fine here because n >= 0, but habit-forming bugs appear in related problems.

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Complexity Analysis

	Time	Space
BFS	Exponential	Exponential
Optimal (Bit Greedy)	O(log n)	O(1)

Bottleneck: The number of bits in n. For n <= 10^9, this is at most 30 iterations — essentially free.

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Edge Cases

1. n = 0 — return 0. Loop never runs.
2. n is a power of 2 — one op (subtract n itself). The greedy shifts down to 1, then subtracts.
3. n has all bits set up to some k (like 7 = 111) — add 1 once to get 1000, then shift down. 2 ops total.
4. Alternating bits (like 21 = 10101) — three isolated 1s, three ops.
5. Very large n near 2^30 — still O(30) steps. No overflow risk since adding 1 stays within 32-bit unsigned for n < 2^31.

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Related Problems

• LC 397 — Integer Replacement — Given n, either divide by 2 (if even) or add/subtract 1. Same greedy shape, same "look at next bit" rule.
• LC 191 — Number of 1 Bits — Popcount. The lower bound on ops here is popcount(n), but runs allow you to do better.
• LC 338 — Counting Bits — DP on bits. Same thinking patterns.

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OA Strategy Notes

This is a mid-difficulty OA problem. Expect 15-25 minutes. The trap is convincing yourself of the greedy rule — when in doubt, test on small cases like n=3 (11), n=7 (111), n=11 (1011) to verify your rule before coding the full solution. With 75 minutes total for two problems, this one is usually the "tricky" one where your first instinct (popcount) is wrong.

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What is Expected at Each Level

Junior: Likely struggles. Might propose popcount(n) as the answer, which is wrong for runs (e.g., 111 = 3 popcount but 2 ops). Needs the hint about runs.

Mid-level: Figures out the run-collapse insight with some thought. Writes the greedy loop correctly. Tests on n = 3, 7, 15 to verify.

Senior: Derives the rule rigorously: "The minimum ops equals the number of runs-of-1s plus the number of isolated 1s" or equivalently uses the formula popcount(n XOR (n >> 1)) framing. Discusses the connection to Integer Replacement (LC 397).