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author | Jack Sangdahl <[email protected]> | 2025-04-30 11:57:26 -0600 |
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committer | Jack Sangdahl <[email protected]> | 2025-04-30 11:57:26 -0600 |
commit | 8f773da1404a7c3f1e4a9ccde3d7b33c27d8c3be (patch) | |
tree | 736c5c779d8d4726c47faf9548403828823708ac | |
parent | 56b6d146c7f33b89c4ef573d0746dc0f306684f0 (diff) | |
download | leetcode-8f773da1404a7c3f1e4a9ccde3d7b33c27d8c3be.tar.gz leetcode-8f773da1404a7c3f1e4a9ccde3d7b33c27d8c3be.zip |
2749. minimum operations to make the integer zero
-rw-r--r-- | min_ops_make_int_zero.c | 36 | ||||
-rw-r--r-- | min_ops_make_int_zero.py | 39 |
2 files changed, 75 insertions, 0 deletions
diff --git a/min_ops_make_int_zero.c b/min_ops_make_int_zero.c new file mode 100644 index 0000000..95d7255 --- /dev/null +++ b/min_ops_make_int_zero.c @@ -0,0 +1,36 @@ +// 2749. Minimum Operations to Make the Integer Zero +#include "leetcode.h" + +/* + * you are given two integers 'nums1' and 'nums2'. in one operation you can + * choose integer 'i' in the range '[0, 60]' and subtract '2^i + num2' from + * 'num1'. return the integer denoting the minimum number of operations needed + * to make 'nums1' equal to 0. if it is impossible to make 'nums1' equal to 0, + * return -1. + */ + +int bit_cnt(long long num) { + int cnt = 0; + while (num > 0) { + cnt += num & 1; + num >>= 1; + } + return cnt; +} + +int makeTheIntegerZero(int num1, int num2) { + if (num1 < num2) + return -1; + for (int i = 0; i <= 100; i++) { + long long diff = num1 - 1LL * num2 * i; + int bits = bit_cnt(diff); + if (bits <= i && i <= diff) + return i; + } + return -1; +} + +int main() { + printf("%d\n", makeTheIntegerZero(3, -2)); // expect: 3 + printf("%d\n", makeTheIntegerZero(5, 7)); // expect: -1 +} diff --git a/min_ops_make_int_zero.py b/min_ops_make_int_zero.py new file mode 100644 index 0000000..096921a --- /dev/null +++ b/min_ops_make_int_zero.py @@ -0,0 +1,39 @@ +# 2749. Minimum Operations to Make the Integer Zero + +""" +you are given two integers 'nums1' and 'nums2'. in one operation you can +choose integer 'i' in the range '[0, 60]' and subtract '2^i + num2' from +'num1'. return the integer denoting the minimum number of operations needed +to make 'nums1' equal to 0. if it is impossible to make 'nums1' equal to 0, +return -1. +""" + + +class Solution(object): + def bit_cnt(self, num): + cnt = 0 + while num > 0: + cnt += num & 1 + num >>= 1 + return cnt + + def makeTheIntegerZero(self, num1, num2): + """ + :type num1: int + :type num2: int + :rtype: int + """ + if num1 < num2: + return -1 + for i in range(101): + diff = num1 - 1 * num2 * i + bits = self.bit_cnt(diff) + if bits <= i and i <= diff: + return i + return -1 + + +if __name__ == "__main__": + obj = Solution() + print(obj.makeTheIntegerZero(num1=3, num2=-2)) + print(obj.makeTheIntegerZero(num1=5, num2=7)) |