2442. Count Number of Distinct Integers After Reverse Operations
Explanation
To solve this problem, we can iterate through the given array nums and for each number, reverse its digits and add it to a new set. We can use a set data structure to store the distinct integers in the final array. By iterating through each number in nums and its reversed version, we can count the number of distinct integers in the final array.
-
Algorithm:
- Initialize a set
distinctIntegersto store distinct integers. - Iterate through each number in
nums:- Reverse the digits of the number.
- Add the original number and its reversed version to the set
distinctIntegers.
- Return the size of the set
distinctIntegers.
- Initialize a set
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Time Complexity: O(N * M), where N is the number of elements in
numsand M is the maximum number of digits in any element. -
Space Complexity: O(N), where N is the number of elements in
nums.
class Solution {
public int countDistinct(int[] nums) {
Set<Integer> distinctIntegers = new HashSet<>();
for (int num : nums) {
int reverseNum = 0;
while (num > 0) {
reverseNum = reverseNum * 10 + num % 10;
num /= 10;
}
distinctIntegers.add(num);
distinctIntegers.add(reverseNum);
}
return distinctIntegers.size();
}
}Code Editor (Testing phase)
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