Here, we will see how to solve Apply Operations to an Array Solution of leet code 2460 problem.
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.
Apply Operations to an Array Solution code in C++
Code 1:
class Solution {
public:
vector<int> applyOperations(vector<int>& nums) {
for(int i = 0; i < nums.size() - 1; i++) {
if(nums[i] == nums[i+1]) {
nums[i] *= 2;
nums[i+1] = 0;
}
}
int i = 0;
for(int j = 0; j < nums.size();) {
if(nums[j] == 0) {
j++;
} else {
nums[i] = nums[j];
i++;
j++;
}
}
while(i < nums.size()) {
nums[i++] = 0;
}
return nums;
}
};Code 2:
class Solution {
public:
vector<int> applyOperations(vector<int>& nums) {
int n = nums.size();
for (int i = 0; i < n - 1; ++i) {
if (nums[i] == nums[i+1]) {
nums[i] *= 2;
nums[i+1] = 0;
}
}
vector<int> ret;
for (int i: nums) {
if (i != 0) {
ret.push_back(i);
}
}
while (ret.size() < n) {
ret.push_back(0);
}
return ret;
}
};Code 3:
class Solution {
public:
vector<int> applyOperations(vector<int>& nums) {
for(int i=1;i<nums.size();i++)
{
if(nums[i]==nums[i-1])
{
nums[i] = 0;
nums[i - 1] *= 2;
}
}
int cnt=0;
vector<int> ans;
for(int x:nums)
{
if(x==0)
cnt++;
else
ans.push_back(x);
}
while(cnt>0)
{
ans.push_back(0);
cnt--;
}
return ans;
}
};Code in Go
Code 1:
func applyOperations(nums []int) []int {
var i, j = 0, 0
for ;i < len(nums) - 1; i++ {
if nums[i] == nums[i+1] {
nums[i] *= 2
nums[i+1] = 0
}
}
i = 0
for ;j < len(nums); j++ {
if nums[j] != 0 {
nums[i] = nums[j]
i += 1
}
}
for i < len(nums) {
nums[i] = 0
i += 1
}
return nums
}Output:
Input: nums = [1,2,2,1,1,0] Output: [1,4,2,0,0,0]
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