From 9f64e82838dbdbf3884ffd6ef7ce35fd346e7326 Mon Sep 17 00:00:00 2001 From: Jack Sangdahl Date: Fri, 15 Nov 2024 21:56:59 -0700 Subject: 3254. find the power of k-size subarrays 1 --- find_power_k_subarr1.c | 44 ++++++++++++++++++++++++++++++++++++++++++++ find_power_k_subarr1.py | 41 +++++++++++++++++++++++++++++++++++++++++ 2 files changed, 85 insertions(+) create mode 100644 find_power_k_subarr1.c create mode 100644 find_power_k_subarr1.py diff --git a/find_power_k_subarr1.c b/find_power_k_subarr1.c new file mode 100644 index 0000000..286b54a --- /dev/null +++ b/find_power_k_subarr1.c @@ -0,0 +1,44 @@ +// 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); +} diff --git a/find_power_k_subarr1.py b/find_power_k_subarr1.py new file mode 100644 index 0000000..b1588ee --- /dev/null +++ b/find_power_k_subarr1.py @@ -0,0 +1,41 @@ +# 3254. Find the Power of K-Size Subarrays I +from collections import deque + +""" +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)]'. +""" + + +class Solution(object): + def resultsArray(self, nums, k): + """ + :type nums: List[int] + :type k: int + :rtype: List[int] + """ + if k == 1: + return nums + n, ans, window = len(nums), [], deque() + for i in range(n): + while window and i - window[0] >= k: + window.popleft() + if not window or nums[i] - nums[i - 1] == 1: + window.append(i) + else: + window.clear() + window.append(i) + if i >= k - 1: + ans.append(nums[i] if len(window) == k else -1) + return ans + + +if __name__ == "__main__": + obj = Solution() + print(obj.resultsArray(nums=[1, 2, 3, 4, 3, 2, 5], k=3)) + print(obj.resultsArray(nums=[2, 2, 2, 2, 2], k=4)) + print(obj.resultsArray(nums=[3, 2, 3, 2, 3, 2], k=2)) -- cgit v1.2.3