diff options
Diffstat (limited to 'apply_operations_array.c')
-rw-r--r-- | apply_operations_array.c | 42 |
1 files changed, 42 insertions, 0 deletions
diff --git a/apply_operations_array.c b/apply_operations_array.c new file mode 100644 index 0000000..4cd8159 --- /dev/null +++ b/apply_operations_array.c @@ -0,0 +1,42 @@ +// 2460. Apply Operations to an Array +#include "leetcode.h" + +/* + * you are given 0-indexed array 'nums' of size 'n' consisting of non-negative + * integers. you need to apply 'n - 1' operations to this array where in the + * i'th operation (0-indexed), you will apply the following on the i'th element + * of 'nums': if 'nums[i] == nums[i + 1]', then multiply 'nums[i]' by 2 and set + * 'nums[i + 1]' to operate 0. otherwise, you skip this operation. after + * performing all the operations, shift all the 0's to the end of the array. + * return the resulting array. note that the operations are applied sequentially + * not all at once. + */ + +int *applyOperations(int *nums, int numsSize, int *returnSize) { + for (int i = 0; i < numsSize - 1; i++) + if (nums[i] == nums[i + 1] && nums[i]) { + nums[i] *= 2; + nums[i + 1] = 0; + } + int *ans = (int *)malloc(numsSize * sizeof(int)), j = 0; + *returnSize = numsSize; + for (int i = 0; i < numsSize; i++) + if (nums[i]) + ans[j++] = nums[i]; + while (j < numsSize) + ans[j++] = 0; + return ans; +} + +int main() { + int n1[] = {1, 2, 2, 1, 1, 0}, n2[] = {0, 1}, rs1, rs2; + int *ao1 = applyOperations(n1, ARRAY_SIZE(n1), &rs1); + int *ao2 = applyOperations(n2, ARRAY_SIZE(n2), &rs2); + for (int i = 0; i < rs1; i++) + printf("%d ", ao1[i]); // expect: 1 4 2 0 0 0 + printf("\n"); + for (int i = 0; i < rs2; i++) + printf("%d ", ao2[i]); // expect: 1 0 + printf("\n"); + free(ao1), free(ao2); +} |