2571. Minimum Operations to Reduce an Integer to 0 LeetCode Solution

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Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Minimum Operations to Reduce an Integer to solution in C++
4. Minimum Operations to Reduce an Integer to solution in Java
5. Minimum Operations to Reduce an Integer to solution in Python
6. Additional Resources

Problem Statement of Minimum Operations to Reduce an Integer to

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.

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 <= 105

Complexity Analysis

• Time Complexity: O(\log n)
• Space Complexity: O(1)

2571. Minimum Operations to Reduce an Integer to 0 LeetCode Solution in C++

class Solution {
 public:
  int minOperations(int n) {
    // The strategy is that when the end of n is
    //   1. consecutive 1s, add 1 (2^0).
    //   2. single 1, substract 1 (2^0).
    //   3. 0, substract 2^k to omit the last 1. Equivalently, n >> 1.
    //
    // e.g.
    //
    //         n = 0b101
    // n -= 2^0 -> 0b100
    // n -= 2^2 -> 0b0
    //         n = 0b1011
    // n += 2^0 -> 0b1100
    // n -= 2^2 -> 0b1000
    // n -= 2^3 -> 0b0
    int ans = 0;

    while (n > 0)
      if ((n & 3) == 3) {
        ++n;
        ++ans;
      } else if (n % 2 == 1) {
        --n;
        ++ans;
      } else {
        n >>= 1;
      }

    return ans;
  }
};
/* code provided by PROGIEZ */

2571. Minimum Operations to Reduce an Integer to 0 LeetCode Solution in Java

class Solution {
  public int minOperations(int n) {
    // The strategy is that when the end of n is
    //   1. consecutive 1s, add 1 (2^0).
    //   2. single 1, substract 1 (2^0).
    //   3. 0, substract 2^k to omit the last 1. Equivalently, n >> 1.
    //
    // e.g.
    //
    //         n = 0b101
    // n -= 2^0 -> 0b100
    // n -= 2^2 -> 0b0
    //         n = 0b1011
    // n += 2^0 -> 0b1100
    // n -= 2^2 -> 0b1000
    // n -= 2^3 -> 0b0
    int ans = 0;

    while (n > 0)
      if ((n & 3) == 3) {
        ++n;
        ++ans;
      } else if (n % 2 == 1) {
        --n;
        ++ans;
      } else {
        n >>= 1;
      }

    return ans;
  }
}
// code provided by PROGIEZ

2571. Minimum Operations to Reduce an Integer to 0 LeetCode Solution in Python

class Solution:
  def minOperations(self, n: int) -> int:
    # The strategy is that when the end of n is
    #   1. consecutive 1s, add 1 (2^0).
    #   2. single 1, substract 1 (2^0).
    #   3. 0, substract 2^k to omit the last 1. Equivalently, n >> 1.
    #
    # e.g.
    #
    #         n = 0b101
    # n -= 2^0 -> 0b100
    # n -= 2^2 -> 0b0
    #         n = 0b1011
    # n += 2^0 -> 0b1100
    # n -= 2^2 -> 0b1000
    # n -= 2^3 -> 0b0
    ans = 0

    while n > 0:
      if (n & 3) == 3:
        n += 1
        ans += 1
      elif n % 2 == 1:
        n -= 1
        ans += 1
      else:
        n >>= 1

    return ans
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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