Minimum One Bit Operations to Make Integers Zero

Minimum One Bit Operations to Make Integers Zero - Problem

You're given an integer n and need to transform it to 0 using the minimum number of bit operations. This is a fascinating problem that combines bit manipulation with dynamic programming.

You have exactly two types of operations:
1. Toggle the rightmost bit (0th bit) - You can always flip the least significant bit
2. Toggle the i-th bit - But only if bit (i-1) is set to 1 AND all bits from (i-2) down to 0 are set to 0

The challenge is finding the minimum number of operations to reach 0. This problem has deep connections to the Gray Code sequence and requires understanding recursive patterns in binary representations.

Input & Output

example_1.py — Basic Case

$ Input: n = 3

› Output: 2

💡 Note: 3 in binary is 11. We can: 1) Flip bit 1 (11 → 01), 2) Flip bit 0 (01 → 00). Total: 2 operations.

example_2.py — Single Bit

$ Input: n = 6

› Output: 4

💡 Note: 6 in binary is 110. Using the Gray Code pattern: result = 6 ⊕ 3 ⊕ 1 = 4 operations needed.

example_3.py — Edge Case

$ Input: n = 0

› Output: 0

💡 Note: Already at 0, so no operations needed.

Constraints

0 ≤ n ≤ 10⁹
The answer will fit in a 32-bit signed integer
Only two types of operations are allowed

Visualization

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Understanding the Visualization

Analyze Constraints

Understand which bit positions can be flipped based on the rules

Recognize Pattern

Identify that this follows inverse Gray Code sequence properties

Apply Formula

Use iterative XOR to calculate the minimum operations directly

Key Takeaway

🎯 Key Insight: This problem maps to the inverse Gray Code sequence, allowing us to calculate the answer directly using iterative XOR operations in O(log n) time instead of exploring all possible operation paths.

Asked in

G Google 15 f Meta 8 ⊞ Microsoft 6 a Amazon 4

The optimal solution uses Gray Code inverse properties to calculate the answer directly in O(log n) time. The key insight is recognizing that this problem follows the inverse Gray Code sequence pattern, where we can compute the result using iterative XOR operations instead of exploring all possible bit operation sequences.

Common Approaches

Approach	Time	Space	Notes
✓ Recursive Exploration (Brute Force)	O(2^n)	O(2^n)	Try all possible operations recursively with memoization
Gray Code Pattern Recognition (Optimal)	O(log n)	O(log n)	Recognize the mathematical pattern and calculate directly using Gray Code inverse

Recursive Exploration (Brute Force) — Algorithm Steps

Start with the given number n
For each possible operation, apply it and get a new number
Recursively calculate minimum operations for the new number
Take the minimum across all possibilities and add 1
Use memoization to cache results for efficiency

Visualization

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Step-by-Step Walkthrough

Initial State

Start with number n and find highest set bit

Apply Formula

Use recursive formula: 2^(k+1) - 1 - f(remaining)

Memoize

Cache result to avoid recalculation

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

#define MAX_MEMO 100000
int memo[MAX_MEMO];
int memo_initialized = 0;

int minimumOneBitOperations(int n) {
    if (!memo_initialized) {
        for (int i = 0; i < MAX_MEMO; i++) {
            memo[i] = -1;
        }
        memo_initialized = 1;
    }
    
    if (n == 0) return 0;
    if (n < MAX_MEMO && memo[n] != -1) return memo[n];
    
    // Find the highest set bit
    int highestBit = 0;
    int temp = n;
    while (temp > 1) {
        temp >>= 1;
        highestBit++;
    }
    
    // Calculate result using the recursive formula
    int remaining = n ^ (1 << highestBit);
    int result = (1 << (highestBit + 1)) - 1 - minimumOneBitOperations(remaining);
    
    if (n < MAX_MEMO) {
        memo[n] = result;
    }
    return result;
}

int main() {
    int n;
    scanf("%d", &n);
    printf("%d\n", minimumOneBitOperations(n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n)

Without memoization, we explore all possible operation sequences. With memoization, it's much better in practice

⚠ Quadratic Growth

Space Complexity

O(2^n)

Memoization table can store up to 2^n different states in worst case

⚠ Quadratic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Recursive Exploration (Brute Force) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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