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author | jack <[email protected]> | 2024-06-14 13:13:15 -0600 |
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committer | jack <[email protected]> | 2024-06-14 13:13:15 -0600 |
commit | d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa (patch) | |
tree | 96df922dae2a5073d9a50ea70c7aae6aa37e3ebe /max_odd_binary_num.c | |
download | leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.tar.gz leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.zip |
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Diffstat (limited to 'max_odd_binary_num.c')
-rw-r--r-- | max_odd_binary_num.c | 32 |
1 files changed, 32 insertions, 0 deletions
diff --git a/max_odd_binary_num.c b/max_odd_binary_num.c new file mode 100644 index 0000000..86f6af1 --- /dev/null +++ b/max_odd_binary_num.c @@ -0,0 +1,32 @@ +// 2864. Maximum Odd Binary Number +#include "leetcode.h" + +/* + * given a binary string 's' that contains at least one '1'. you have to + * rearrange the bits in such a way that the resulting binary number is the + * maximum odd binary number that can be created from this combination. return a + * string representing the maximum odd binary number that can be created from + * the given combination. + */ + +char *maximumOddBinaryNumber(char *s) { + int n = strlen(s), one = 0; + for (int i = 0; i < n; ++i) + if (s[i] == '1') + one++; + for (int i = 0; i < n - 1; ++i) { + if (one > 1) { + s[i] = '1'; + --one; + } else + s[i] = '0'; + } + s[n - 1] = '1'; + return s; +} + +int main() { + char *s1 = "010", *s2 = "0101"; + printf("%s\n", maximumOddBinaryNumber(s1)); // expect: 001 + printf("%s\n", maximumOddBinaryNumber(s2)); // expect: 1001 +} |