LeetCode in Kotlin
2460. Apply Operations to an Array
Easy
You are given a 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 ith operation (0-indexed), you will apply the following on the ith element of nums:
- If
nums[i] == nums[i + 1], then multiplynums[i]by2and setnums[i + 1]to0. Otherwise, you skip this operation.
After performing all the operations, shift all the 0’s to the end of the array.
- For example, the array
[1,0,2,0,0,1]after shifting all its0’s to the end, is[1,2,1,0,0,0].
Return the resulting array.
Note that the operations are applied sequentially, not all at once.
Example 1:
Input: nums = [1,2,2,1,1,0]
Output: [1,4,2,0,0,0]
Explanation: We do the following operations:
-
i = 0: nums[0] and nums[1] are not equal, so we skip this operation.
-
i = 1: nums[1] and nums[2] are equal, we multiply nums[1] by 2 and change nums[2] to 0. The array becomes [1,4,0,1,1,0].
-
i = 2: nums[2] and nums[3] are not equal, so we skip this operation.
-
i = 3: nums[3] and nums[4] are equal, we multiply nums[3] by 2 and change nums[4] to 0. The array becomes [1,4,0,2,0,0].
-
i = 4: nums[4] and nums[5] are equal, we multiply nums[4] by 2 and change nums[5] to 0. The array becomes [1,4,0,2,0,0].
After that, we shift the 0’s to the end, which gives the array [1,4,2,0,0,0].
Example 2:
Input: nums = [0,1]
Output: [1,0]
Explanation: No operation can be applied, we just shift the 0 to the end.
Constraints:
2 <= nums.length <= 20000 <= nums[i] <= 1000
Solution
class Solution {
fun applyOperations(nums: IntArray): IntArray {
for (i in 0 until nums.size - 1) {
if (nums[i] == nums[i + 1]) {
nums[i] *= 2
nums[i + 1] = 0
}
}
var index = 0
for (i in nums.indices) {
if (nums[i] != 0) {
nums[index] = nums[i]
index++
}
}
for (i in index until nums.size) {
nums[i] = 0
}
return nums
}
}