aboutsummaryrefslogtreecommitdiff
diff options
context:
space:
mode:
authorjack <[email protected]>2024-06-20 19:54:24 -0600
committerjack <[email protected]>2024-06-20 19:54:24 -0600
commitbd945a7d02858be19b5fa2a80cc4782dda112d50 (patch)
tree48ca87275d59e90ad7e2922fefea99587e75834c
parentab7c69280c3dca736e00e6b41372167045af27cc (diff)
downloadleetcode-bd945a7d02858be19b5fa2a80cc4782dda112d50.tar.gz
leetcode-bd945a7d02858be19b5fa2a80cc4782dda112d50.zip
2654. mininimum operations array elements 1
-rw-r--r--min_num_opp_arr_equal_1.c48
-rw-r--r--min_num_opp_arr_equal_1.py36
2 files changed, 84 insertions, 0 deletions
diff --git a/min_num_opp_arr_equal_1.c b/min_num_opp_arr_equal_1.c
new file mode 100644
index 0000000..c6b5dc3
--- /dev/null
+++ b/min_num_opp_arr_equal_1.c
@@ -0,0 +1,48 @@
+// 2654. Minimum Number of Operations to Make All Array Elements Equal to 1
+#include "leetcode.h"
+
+/*
+ * given a 0-indexed array 'nums' consisting of positive integers. you can do
+ * the following operation on the array any number of time: select an index 'i'
+ * such that '0 <= i < n - 1' and replace either of 'nums[i]' or 'nums[i + 1]'
+ * with their gcd values. return the minimum number of operations to make all
+ * elements of 'nums' equal to 1. if this is impossible, return -1. the gcd of
+ * two integers is the greatest common divisor of the two integers.
+ */
+
+int gcd(int a, int b) {
+ int rem = a % b;
+ if (!rem)
+ return b;
+ return gcd(b, rem);
+}
+
+int minOperations(int *nums, int numsSize) {
+ int one_cnt = 0;
+ for (int i = 0; i < numsSize; i++)
+ if (nums[i] == 1)
+ one_cnt++;
+ if (one_cnt >= 1)
+ return numsSize - one_cnt;
+ int min_subarr_len = numsSize + 1;
+ for (int i = 0; i < numsSize; i++) {
+ int curr = nums[i];
+ for (int j = i + 1; j < numsSize; j++) {
+ curr = gcd(curr, nums[j]);
+ if (curr == 1) {
+ int curr_n = (j + 1) - i;
+ if (curr_n < min_subarr_len)
+ min_subarr_len = curr_n;
+ }
+ break;
+ }
+ }
+ return min_subarr_len <= numsSize ? (min_subarr_len - 1) + (numsSize - 1)
+ : -1;
+}
+
+int main() {
+ int n1[] = {2, 6, 3, 4}, n2[] = {2, 10, 6, 14};
+ printf("%d\n", minOperations(n1, ARRAY_SIZE(n1))); // expect: 4
+ printf("%d\n", minOperations(n2, ARRAY_SIZE(n2))); // expect: -1
+}
diff --git a/min_num_opp_arr_equal_1.py b/min_num_opp_arr_equal_1.py
new file mode 100644
index 0000000..75551dc
--- /dev/null
+++ b/min_num_opp_arr_equal_1.py
@@ -0,0 +1,36 @@
+# 2654. Minimum Number of Operations to Make All Array Elements Equal to 1
+from math import gcd
+
+"""
+given a 0-indexed array 'nums' consisting of positive integers. you can do
+the following operation on the array any number of time: select an index 'i'
+such that '0 <= i < n - 1' and replace either of 'nums[i]' or 'nums[i + 1]'
+with their gcd values. return the minimum number of operations to make all
+elements of 'nums' equal to 1. if this is impossible, return -1. the gcd of
+two integers is the greatest common divisor of the two integers.
+"""
+
+
+class Solution(object):
+ def minOperations(self, nums):
+ """
+ :type nums: List[int]
+ :rtype: int
+ """
+ ones_cnt = nums.count(1)
+ if ones_cnt:
+ return len(nums) - ones_cnt
+ diff = float("inf")
+ for i in range(len(nums)):
+ g = nums[i]
+ for j in range(i + 1, len(nums)):
+ g = gcd(g, nums[j])
+ if g == 1:
+ diff = min(diff, j - i)
+ return -1 if diff == float("inf") else diff + len(nums) - 1
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.minOperations([2, 6, 3, 4]))
+ print(obj.minOperations([2, 10, 6, 14]))