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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 /count_triplets_form_arr_xor.c | |
download | leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.tar.gz leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.zip |
initial
Diffstat (limited to 'count_triplets_form_arr_xor.c')
-rw-r--r-- | count_triplets_form_arr_xor.c | 31 |
1 files changed, 31 insertions, 0 deletions
diff --git a/count_triplets_form_arr_xor.c b/count_triplets_form_arr_xor.c new file mode 100644 index 0000000..cfb7493 --- /dev/null +++ b/count_triplets_form_arr_xor.c @@ -0,0 +1,31 @@ +// 1442. Count Triplets That Can Form Two Arrays of Equal XOR +#include "leetcode.h" + +/* + * given an array of integers 'arr'. we want to select three indices 'i', 'j', + * and 'k' where '(0 <= i < j <= k < arr.length)'. lets define 'a' and 'b' as + * follows. 'a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]', 'b = arr[j] ^ arr[j + + * 1] ^ ... ^ arr[k]'. note that '^' denotes the bitwise xor operation. return + * the number of triplets (i, j, and k) where 'a == b' + */ + +int countTriplets(int *arr, int arrSize) { + if (arrSize < 2) + return 0; + int ans = 0; + for (int i = 0; i < arrSize; i++) { + int tmp = arr[i]; + for (int j = i + 1; j < arrSize; j++) { + tmp ^= arr[j]; + if (!tmp) + ans += j - i; + } + } + return ans; +} + +int main() { + int a1[] = {2, 3, 1, 6, 7}, a2[] = {1, 1, 1, 1, 1}; + printf("%d\n", countTriplets(a1, ARRAY_SIZE(a1))); // expect: 4 + printf("%d\n", countTriplets(a2, ARRAY_SIZE(a2))); // expect: 10 +} |