2749. Minimum Operations to Make the Integer Zero LeetCode Solution

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Table of Contents

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

Problem Statement of Minimum Operations to Make the Integer Zero

You are given two integers num1 and num2.
In one operation, you can choose integer i in the range [0, 60] and subtract 2i + num2 from num1.
Return the integer denoting the minimum number of operations needed to make num1 equal to 0.
If it is impossible to make num1 equal to 0, return -1.

Example 1:

Input: num1 = 3, num2 = -2
Output: 3
Explanation: We can make 3 equal to 0 with the following operations:
– We choose i = 2 and subtract 22 + (-2) from 3, 3 – (4 + (-2)) = 1.
– We choose i = 2 and subtract 22 + (-2) from 1, 1 – (4 + (-2)) = -1.
– We choose i = 0 and subtract 20 + (-2) from -1, (-1) – (1 + (-2)) = 0.
It can be proven, that 3 is the minimum number of operations that we need to perform.

Example 2:

Input: num1 = 5, num2 = 7
Output: -1
Explanation: It can be proven, that it is impossible to make 5 equal to 0 with the given operation.

Constraints:

1 <= num1 <= 109
-109 <= num2 <= 109

Complexity Analysis

• Time Complexity: O(1)
• Space Complexity: O(1)

2749. Minimum Operations to Make the Integer Zero LeetCode Solution in C++

class Solution {
 public:
  int makeTheIntegerZero(int num1, int num2) {
    // If k operations are used, num1 - [(num2 + 2^{i_1}) + (num2 + 2^{i_2}) +
    // ... + (num2 + 2^{i_k})] = 0. So, num1 - k * num2 = (2^{i_1} + 2^{i_2} +
    // ... + 2^{i_k}), where i_1, i_2, ..., i_k are in the range [0, 60].
    // Note that for any number x, we can use "x's bit count" operations to make
    // x equal to 0. Additionally, we can also use x operations to deduct x by
    // 2^0 (x times), which also results in 0.

    for (long ops = 0; ops <= 60; ++ops) {
      const long target = num1 - ops * num2;
      if (__builtin_popcountl(target) <= ops && ops <= target)
        return ops;
    }

    return -1;
  }
};
/* code provided by PROGIEZ */

2749. Minimum Operations to Make the Integer Zero LeetCode Solution in Java

class Solution {
  public int makeTheIntegerZero(int num1, int num2) {
    // If k operations are used, num1 - [(num2 + 2^{i_1}) + (num2 + 2^{i_2}) +
    // ... + (num2 + 2^{i_k})] = 0. So, num1 - k * num2 = (2^{i_1} + 2^{i_2} +
    // ... + 2^{i_k}), where i_1, i_2, ..., i_k are in the range [0, 60].
    // Note that for any number x, we can use "x's bit count" operations to make
    // x equal to 0. Additionally, we can also use x operations to deduct x by
    // 2^0 (x times), which also results in 0.

    for (long ops = 0; ops <= 60; ++ops) {
      long target = num1 - ops * num2;
      if (Long.bitCount(target) <= ops && ops <= target)
        return (int) ops;
    }

    return -1;
  }
}
// code provided by PROGIEZ

2749. Minimum Operations to Make the Integer Zero LeetCode Solution in Python

class Solution:
  def makeTheIntegerZero(self, num1: int, num2: int) -> int:
    # If k operations are used, num1 - [(num2 + 2^{i_1}) + (num2 + 2^{i_2}) +
    # ... + (num2 + 2^{i_k})] = 0. So, num1 - k * num2 = (2^{i_1} + 2^{i_2} +
    # ... + 2^{i_k}), where i_1, i_2, ..., i_k are in the range [0, 60].
    # Note that for any number x, we can use "x's bit count" operations to make
    # x equal to 0. Additionally, we can also use x operations to deduct x by
    # 2^0 (x times), which also results in 0.

    for ops in range(61):
      target = num1 - ops * num2
      if target.bit_count() <= ops <= target:
        return ops

    return -1
 # code by PROGIEZ

Additional Resources

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

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