LeetCode 2717: Semi-Ordered Permutation Solution
Master LeetCode problem 2717 (Semi-Ordered Permutation), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
2717. Semi-Ordered Permutation
Problem Explanation
Explanation
To solve this problem, we can iterate through the given permutation array nums and count the number of misplaced elements. A misplaced element is any element that is not at its correct position in a semi-ordered permutation. The number of operations needed would be equal to the number of misplaced elements. We can determine if an element is misplaced by checking if it should be at index i or i+1 based on the values 1 to n.
- Initialize a variable
misplacedto 0. - Iterate through the array
nums. - For each element at index
i, check if it is not equal toi+1(since the permutation should start from 1 to n). - If the element is misplaced, increment the
misplacedcount. - Finally, return the
misplacedcount as the minimum number of operations needed.
The time complexity of this solution is O(n) where n is the length of the input permutation array, and the space complexity is O(1) as we are using a constant amount of extra space.
Solution Code
class Solution {
public int minOperations(int[] nums) {
int misplaced = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] != i + 1) {
misplaced++;
}
}
return misplaced;
}
}Try It Yourself
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