Algorithm
Problem Name: 153. Find Minimum in Rotated Sorted Array
Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:
[4,5,6,7,0,1,2]if it was rotated4times.[0,1,2,4,5,6,7]if it was rotated7times.
Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].
Given the sorted rotated array nums of unique elements, return the minimum element of this array.
You must write an algorithm that runs in O(log n) time.
Example 1:
Input: nums = [3,4,5,1,2] Output: 1 Explanation: The original array was [1,2,3,4,5] rotated 3 times.
Example 2:
Input: nums = [4,5,6,7,0,1,2] Output: 0 Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.
Example 3:
Input: nums = [11,13,15,17] Output: 11 Explanation: The original array was [11,13,15,17] and it was rotated 4 times.
Constraints:
n == nums.length1 <= n <= 5000-5000 <= nums[i] <= 5000- All the integers of
numsare unique. numsis sorted and rotated between1andntimes.
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
#include <assert.h>
int findMin(int *nums, int numsSize) {
int l = 0;
int r = numsSize - 1;
while (l < r) {
int m = l + (r - l) / 2;
if (nums[l] > nums[m]) { /* right side is sorted */
r = m;
}
else if (nums[r] < nums[m]) { /* left side is sorted */
l = m + 1;
}
else { /* the sub-array is not rotated */
r = m;
}
}
return nums[l];
}
int main() {
int nums[] = { 3, 1, 2 };
assert(findMin(nums, sizeof(nums) / sizeof(nums[0])) == 1);
printf("all tests passed!\n");
return 0;
}
Input
Output
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int findMin(int[] nums) {
if (nums.length == 1) {
return nums[0];
}
int left = 0;
int right = nums.length - 1;
// Sorted but not rotated
if (nums[right] > nums[left]) {
return nums[left];
}
while (left < = right) {
int mid = (left + right) / 2;
// mid + 1 is point of rotation
if (nums[mid] > nums[mid + 1]) {
return nums[mid + 1];
}
// mid is point of rotation
if (nums[mid - 1] > nums[mid]) {
return nums[mid];
}
if (nums[mid] > nums[0]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return Integer.MAX_VALUE;
}
}
Input
Output
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const findMin = function (nums) {
let low = 0,
high = nums.length - 1
while (low < high) {
const mid = low + ((high - low) >> 1)
if (nums[mid] <= nums[high]) high = mid
else if (nums[mid] > nums[high]) low = mid + 1
}
return nums[low]
}
Input
#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
findMin = min
Input
#5 Code Example with C# Programming
Code -
C# Programming
namespace LeetCode
{
public class _0153_FindMinimumInRotatedSortedArray
{
public int FindMin(int[] nums)
{
if (nums.Length == 1) return nums[0];
if (nums[nums.Length - 1] > nums[0]) return nums[0];
int left = 0, right = nums.Length - 1;
while (left < right)
{
var mid = left + (right - left) / 2;
if (nums[mid] < nums[right]) right = mid;
else left = mid + 1;
}
return nums[right];
}
}
}
Input
Output