1404. Number of Steps to Reduce a Number in Binary Representation to One - Detailed Explanation

Problem Statement

Given a binary string s representing a positive integer, the goal is to reduce the integer to 1 by performing the following operations:

  • If the current number is even, divide it by 2.
  • If the current number is odd, add 1.

Return the number of steps required to reduce the number to 1.

Example Inputs, Outputs, and Explanations

Example 1

  • Input: s = "1101"
  • Output: 6
  • Explanation:
    1. "1101" (13 in decimal) is odd, so add 1 → "1110"
    2. "1110" (14) is even, divide by 2 → "111"
    3. "111" (7) is odd, add 1 → "1000"
    4. "1000" (8) is even, divide by 2 → "100"
    5. "100" (4) is even, divide by 2 → "10"
    6. "10" (2) is even, divide by 2 → "1"

Example 2

  • Input: s = "10"
  • Output: 1
  • Explanation:
    1. "10" (2) is even, divide by 2 → "1"

Example 3

  • Input: s = "1"
  • Output: 0
  • Explanation:
    The number is already 1, so no steps are required.

Constraints

  • 1 <= s.length <= 500
  • s consists only of the characters '0' and '1'.
  • s[0] == '1' (i.e., there are no leading zeros).

Hints

  1. Even Number Shortcut: Notice that when the binary number is even (ends with '0'), dividing by 2 is equivalent to removing the trailing zero.
  2. Odd Number Handling: When the binary number is odd (ends with '1'), adding 1 may involve a cascade of changes due to carry propagation. Consider how adding 1 affects a sequence of trailing ones.

Approach 1: Brute Force Simulation

Explanation

  • Conversion: Convert the binary string to an integer.
  • Simulation:
    • While the number is greater than 1, check if it is even or odd.
    • If it is even, divide it by 2.
    • If it is odd, add 1.
    • Count each operation as a step.
  • Drawbacks:
    • Although Python can handle large integers, continuously updating a potentially huge number might be less efficient when the binary string is long and the operations are many.
    • The worst-case time complexity may be higher due to the repeated arithmetic operations.

Step-by-Step Walkthrough (Brute Force)

Consider the input "1101":

  1. Convert "1101" to integer → 13.
  2. 13 is odd → 13 + 1 = 14. (Step count = 1)
  3. 14 is even → 14 / 2 = 7. (Step count = 2)
  4. 7 is odd → 7 + 1 = 8. (Step count = 3)
  5. 8 is even → 8 / 2 = 4. (Step count = 4)
  6. 4 is even → 4 / 2 = 2. (Step count = 5)
  7. 2 is even → 2 / 2 = 1. (Step count = 6)

Approach 2: Optimal Simulation Using the Binary String Directly

Explanation

  • Observation:
    • Dividing by 2 for an even binary number means simply removing the last digit (which is '0').
    • For an odd number, adding 1 can change a sequence of trailing '1's to '0's and propagate a carry to the left.
  • Method:
    • Process the binary string from right to left while maintaining a carry flag.
    • For each bit (ignoring the most significant bit initially):
      • If the current bit plus any carry is even, it corresponds to a division by 2 (a single step).
      • If it is odd, perform an "add 1" operation (which in our simulation counts as 2 steps: one for addition and one for the subsequent division), and set the carry for future bits.
    • Finally, if a carry remains at the most significant bit, account for an extra step.
  • Benefits:
    • This approach works in O(n) time (where n is the length of the binary string) and avoids costly arithmetic on potentially huge numbers.

Step-by-Step Walkthrough (Optimal Approach)

Consider the input "1101":

  1. Start with a carry = 0 and a step count = 0.
  2. Process from the least significant digit to the second digit:
    • Digit: '1' (plus carry 0) → odd, so add 2 steps (for add and division) and set carry = 1.
    • Next Digit: '0' (plus carry 1 gives 1) → odd, add 2 steps, carry remains 1.
    • Next Digit: '1' (plus carry 1 gives 2) → even, add 1 step.
  3. Finally, process the most significant digit with the carry: add 1 extra step if a carry is present.
  4. Total steps equal the accumulated count.

Code in Python

Python3
Python3

. . . .

Code in Java

Java
Java

. . . .

Complexity Analysis

  • Time Complexity:

    • Brute Force Approach: O(m) per step, where m is the number of bits in the integer. In the worst case (with many carry operations), the overall complexity can be higher.

    • Optimal Approach: O(n), where n is the length of the binary string. We traverse the string only once.

  • Space Complexity:

    • Both approaches use O(1) additional space (apart from the input string).

Step-by-Step Walkthrough and Visual Examples

Consider the input "1101" using the optimal approach:

  1. Initial String: "1101", carry = 0, steps = 0.

  2. Iteration 1 (rightmost digit '1'):

    • Current = 1 + 0 = 1 (odd) → Add 2 steps (steps become 2) and set carry = 1.
  3. Iteration 2 (next digit '0'):

    • Current = 0 + 1 = 1 (odd) → Add 2 steps (steps become 4) and carry remains 1.
  4. Iteration 3 (digit '1'):

    • Current = 1 + 1 = 2 (even) → Add 1 step (steps become 5).
  5. Final Step:

    • Process the most significant digit with carry (carry = 1) → Add 1 step (steps become 6).

Common Mistakes and Edge Cases

  • Mistake 1: Not handling the carry properly when adding 1 to an odd number.

  • Mistake 2: Forgetting that the most significant bit might require an extra step if there is a remaining carry.

  • Edge Case:

    • Input "1" should return 0 as no operations are needed.
    • Handling long strings where multiple carry propagations occur.

Alternative Variations

  • Variation 1: Changing the operations allowed (for example, subtracting 1 instead of adding 1 when the number is odd).

  • Variation 2: Minimizing operations with different cost weights for addition and division, which may require a modified dynamic programming approach.

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