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2749. Minimum Operations to Make the Integer Zero

Description

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 substract 22 + (-2) from 3, 3 - (4 + (-2)) = 1.
- We choose i = 2 and substract 22 + (-2) from 1, 1 - (4 + (-2)) = -1.
- We choose i = 0 and substract 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

Solutions

  • class Solution {
    public int makeTheIntegerZero(int num1, int num2) {
        for (long k = 1;; ++k) {
            long x = num1 - k * num2;
            if (x < 0) {
                break;
            }
            if (Long.bitCount(x) <= k && k <= x) {
                return (int) k;
            }
        }
        return -1;
    }
}

  • class Solution {
public:
    int makeTheIntegerZero(int num1, int num2) {
        using ll = long long;
        for (ll k = 1;; ++k) {
            ll x = num1 - k * num2;
            if (x < 0) {
                break;
            }
            if (__builtin_popcountll(x) <= k && k <= x) {
                return k;
            }
        }
        return -1;
    }
};

  • class Solution:
    def makeTheIntegerZero(self, num1: int, num2: int) -> int:
        for k in count(1):
            x = num1 - k * num2
            if x < 0:
                break
            if x.bit_count() <= k <= x:
                return k
        return -1


  • func makeTheIntegerZero(num1 int, num2 int) int {
	for k := 1; ; k++ {
		x := num1 - k*num2
		if x < 0 {
			break
		}
		if bits.OnesCount(uint(x)) <= k && k <= x {
			return k
		}
	}
	return -1
}

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