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Diffstat (limited to 'arr_elements_neq_neighbour.c')
-rw-r--r-- | arr_elements_neq_neighbour.c | 44 |
1 files changed, 44 insertions, 0 deletions
diff --git a/arr_elements_neq_neighbour.c b/arr_elements_neq_neighbour.c new file mode 100644 index 0000000..59a7b8f --- /dev/null +++ b/arr_elements_neq_neighbour.c @@ -0,0 +1,44 @@ +// 1968. Array With Elements Not Equal to Average of Neighbors +#include "leetcode.h" + +/* + * you are given a 0-indexed array 'nums' of distinct integers. you want to + * rearrange the elements in the array such that every element in the rearranged + * array is not equal to the average of its neighbors. more formally the + * rearranged array should have the property such that for every 'i' in the + * range '1 <= nums.len - 1', '(nums[i - 1] + nums[i + 1]) / 2' is not equal to + * 'nums[i]'. return any rearrangement of 'nums' that meets the requirements. + */ + +void swap(int *x, int *y) { + if (x == y) + return; + *x ^= *y; + *y ^= *x; + *x ^= *y; +} + +int *rearrangeArray(int *nums, int numsSize, int *returnSize) { + int *ans = (int *)malloc(numsSize * sizeof(int)), i, j = 0; + *returnSize = numsSize; + for (int i = 1; i < numsSize - 1; i++) + if (nums[i] * 2 == nums[i - 1] + nums[i + 1]) + swap(&nums[i], &nums[i + 1]); + for (int i = numsSize - 2; i; i--) + if (nums[i] * 2 == nums[i - 1] + nums[i + 1]) + swap(&nums[i], &nums[i - 1]); + return ans; +} + +int main() { + int n1[] = {1, 2, 3, 4, 5}, n2[] = {6, 2, 0, 9, 7}, rs1, rs2; + int *ra1 = rearrangeArray(n1, ARRAY_SIZE(n1), &rs1); + int *ra2 = rearrangeArray(n2, ARRAY_SIZE(n2), &rs2); + for (int i = 0; i < rs1; i++) + printf("%d ", ra1[i]); // expect: 1,2,4,5,3 + printf("\n"); + for (int i = 0; i < rs2; i++) + printf("%d ", ra2[i]); // expect: 9,7,6,2,0 + printf("\n"); + free(ra1), free(ra2); +} |