153. Find Minimum in Rotated Sorted Array
Explanation
To find the minimum element in a rotated sorted array, we can utilize a modified binary search algorithm. The idea is to compare the middle element of the array with the first and last elements to determine which half of the array to discard. This way, we can focus on the unsorted half where the minimum element lies.
- Initialize two pointers,
leftandright, pointing to the start and end of the array respectively. - Perform binary search until
leftandrightconverge, indicating that the minimum element has been found. - At each step, compare the middle element with the first and last elements to determine which half is unsorted and update
leftorrightaccordingly. - Return the element at the
leftindex, which will be the minimum element.
Time complexity: O(log n) Space complexity: O(1)
class Solution {
public int findMin(int[] nums) {
int left = 0, right = nums.length - 1;
while (left < right) {
int mid = left + (right - left) / 2;
if (nums[mid] < nums[right]) {
right = mid;
} else {
left = mid + 1;
}
}
return nums[left];
}
}Code Editor (Testing phase)
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