LeetCode Entry
2460. Apply Operations to an Array
If a[i] == a[i + 1], a[i] = 2, a[i + 1] = 0
2460. Apply Operations to an Array easy
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Problem TLDR
If a[i] == a[i + 1], a[i] *= 2, a[i + 1] = 0 #easy
Intuition
The operations should be applied left-to-right. I can’t find a way to do this with iterators.
Approach
- careful with zeroing if i == j
Complexity
-
Time complexity: \(O(n)\)
-
Space complexity: \(O(1)\)
Code
fun applyOperations(nums: IntArray) = nums.apply {
for (i in 0..<size - 1) if (nums[i] == nums[i + 1])
nums[i + 1] = 0.also { nums[i] *= 2 }
}.sortedBy { it == 0 }
pub fn apply_operations(mut nums: Vec<i32>) -> Vec<i32> {
let mut j = 0;
for i in 0..nums.len() {
if i < nums.len() - 1 && nums[i] == nums[i + 1]
{ nums[i + 1] = 0; nums[i] *= 2 }
if nums[i] > 0 { nums.swap(i, j); j += 1 }
}; nums
}
vector<int> applyOperations(vector<int>& a) {
for (int i = 0; i < size(a) - 1; ++i)
if (a[i] == a[i + 1]) a[i] *= 2, a[i + 1] = 0;
stable_partition(begin(a), end(a), [](int n){return n;});
return a;
}