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findMinRotateedSortedArrray.cpp
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findMinRotateedSortedArrray.cpp
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153. Find Minimum in Rotated Sorted Array
Medium
9.4K
450
Companies
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 rotated 4 times.
[0,1,2,4,5,6,7] if it was rotated 7 times.
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.
//CODE
class Solution {
public:
int findMin(vector<int>& nums) {
int low = 0;
int high = nums.size();
int middle;
while(low < high)
{
middle = (high-low)/2 + low;
//1st part
if(nums[middle] <= nums[low])
high = middle;
else
low = middle;
}
// 2nd part
if(low >= nums.size()-1)
return nums.at(0);
return nums.at(low+1);
}
};
/*class Solution {
public:
int findMin(vector<int>& nums) {
//first index return
if(nums[0] ==1) return nums[0];
if(nums[0]<nums.back()) return nums[0];
int l= 0 , r=nums.size()-1;
int ans = INT_MAX;
while(l<=r){
int mid = (l+r)/2;
//one part sorted
if(nums[mid]>=nums[0]){
l = mid+1;
}
// 2nd part sorted
else{
ans = min(ans,nums[mid]);
r= mid-1;
}
}
return ans;
}
}; */