LeetCode 1752: Check if Array Is Sorted and Rotated
LeetCode 1752 Solution Explanation
Explanation:
To solve this problem, we can iterate through the given array and find the point where the array is not in non-decreasing order. If we find more than one such point, then the array cannot be sorted and rotated. If we find exactly one such point, then we check if rotating the array at that point gives us the original sorted array.
- Find the point where the array is not in non-decreasing order.
- If there is no such point, the array is already sorted, return true.
- If there is one such point, check if rotating the array at that point gives the original sorted array.
- If yes, return true. Otherwise, return false.
Time Complexity: O(n), where n is the number of elements in the given array. Space Complexity: O(1)
:
LeetCode 1752 Solutions in Java, C++, Python
class Solution {
public boolean check(int[] nums) {
int n = nums.length;
int count = 0;
for (int i = 0; i < n; i++) {
if (nums[i] > nums[(i + 1) % n]) {
count++;
}
if (count > 1) {
return false;
}
}
return count == 0 || nums[0] >= nums[n - 1];
}
}Interactive Code Editor for LeetCode 1752
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