2460. Apply Operations to an Array
Explanation
To solve this problem, we can iterate through the array and apply the specified operations. For each pair of adjacent elements where the elements are equal, we update the first element by multiplying it by 2 and set the second element to 0. After performing all operations, we shift all zeros to the end of the array.
Here are the steps for the algorithm:
- Iterate through the array and apply the given operations.
- Shift all zeros to the end of the array.
The time complexity of this approach is O(n) where n is the number of elements in the array. The space complexity is O(1) as we are modifying the input array in place.
class Solution {
public int[] applyOperations(int[] nums) {
for (int i = 0; i < nums.length - 1; i++) {
if (nums[i] == nums[i + 1]) {
nums[i] *= 2;
nums[i + 1] = 0;
}
}
int k = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] != 0) {
nums[k++] = nums[i];
}
}
while (k < nums.length) {
nums[k++] = 0;
}
return nums;
}
}Code Editor (Testing phase)
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