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authorJack Sangdahl <[email protected]>2025-03-01 00:10:01 -0700
committerJack Sangdahl <[email protected]>2025-03-01 00:10:01 -0700
commitcb847deddeb96a01211dd1e4c4ea9c7c16a29109 (patch)
tree19fcc19b8d2b39ab1c9447893771f0c279f5dec1
parent1bb2da19c46df3878ed2993e0da4c961f94622e1 (diff)
downloadleetcode-cb847deddeb96a01211dd1e4c4ea9c7c16a29109.tar.gz
leetcode-cb847deddeb96a01211dd1e4c4ea9c7c16a29109.zip
2460. apply operations to an array
-rw-r--r--apply_operations_array.c42
-rw-r--r--apply_operations_array.py35
2 files changed, 77 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);
+}
diff --git a/apply_operations_array.py b/apply_operations_array.py
new file mode 100644
index 0000000..3dc8e7c
--- /dev/null
+++ b/apply_operations_array.py
@@ -0,0 +1,35 @@
+# 2460. Apply Operations to an Array
+
+"""
+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.
+"""
+
+
+class Solution(object):
+ def applyOperations(self, nums):
+ """
+ :type nums: List[int]
+ :rtype: List[int]
+ """
+ j = 0
+ for i in range(len(nums)):
+ if i + 1 < len(nums) and nums[i] == nums[i + 1]:
+ nums[i] *= 2
+ nums[i + 1] = 0
+ if nums[i]:
+ nums[j], nums[i] = nums[i], nums[j]
+ j += 1
+ return nums
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.applyOperations(nums=[1, 2, 2, 1, 1, 0]))
+ print(obj.applyOperations(nums=[0, 1]))