// 3254. Find the Power of K-Size Subarrays I #include "leetcode.h" /* * you are given an array of integers 'nums' of length 'n' and a positive * integer 'k'. the power of an array is defined as its maximum element if all * its elements are consecutive and sorted in ascending order, -1 otherwise. you * need to find the power of all subarrays of 'nums' of size 'k'. return an * integer array 'results' of size 'n - k + 1' where 'results[i]' is the power * of 'nums[i ... (i + k - 1)]'. */ int *resultsArray(int *nums, int numsSize, int k, int *returnSize) { int *ans = (int *)malloc((numsSize - k + 1) * sizeof(int)), n = 0; for (int i = 0; i <= numsSize - k; i++) { bool consec = true; for (int j = 0; j < k - 1; j++) if (nums[i + j] + 1 != nums[i + j + 1]) { consec = false; break; } ans[n++] = consec ? nums[i + k - 1] : -1; } *returnSize = n; return ans; } int main() { int n1[] = {1, 2, 3, 4, 3, 2, 5}, n2[] = {2, 2, 2, 2, 2}, n3[] = {3, 2, 3, 2, 3, 2}, rs1, rs2, rs3; int *ra1 = resultsArray(n1, ARRAY_SIZE(n1), 3, &rs1); int *ra2 = resultsArray(n2, ARRAY_SIZE(n2), 4, &rs2); int *ra3 = resultsArray(n3, ARRAY_SIZE(n3), 2, &rs3); for (int i = 0; i < rs1; i++) printf("%d ", ra1[i]); // expect: 3 4 -1 -1 -1 printf("\n"); for (int i = 0; i < rs2; i++) printf("%d ", ra2[i]); // expect: -1 -1 printf("\n"); for (int i = 0; i < rs3; i++) printf("%d ", ra3[i]); // expect: -1 3 -1 3 -1 printf("\n"); free(ra1), free(ra2), free(ra3); }