leetcode

commit cb847deddeb96a01211dd1e4c4ea9c7c16a29109
parent 1bb2da19c46df3878ed2993e0da4c961f94622e1
Author: Jack Sangdahl <jack@pngu.org>
Date:   Sat,  1 Mar 2025 00:10:01 -0700

2460. apply operations to an array

Diffstat:
A
apply_operations_array.c
 | 
42
++++++++++++++++++++++++++++++++++++++++++
A
apply_operations_array.py
 | 
35
+++++++++++++++++++++++++++++++++++

2 files changed, 77 insertions(+), 0 deletions(-)
diff --git a/apply_operations_array.c 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
@@ -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]))