# An O(n) solution with short code and another solution using binary search

• ``````public int findMin(int[] num) {
for(int i = 0;i < num.length-1; i++)
{
if(num[i] > num[i+1])
return num[i+1];
}
return num[0];
}
``````

search for the first item that is smaller than its previous one, and that is the answer. If no item is found, it means the first item is the answer.

``````public int findMin(int[] num) {

return rf(0,num.length-1,num);
}

private int rf(int left, int right, int[] num)
{
if(right - left <= 1)
{
if(num[left] >= num[right])
return num[right];
else
return num[left];
}

int mid = (left+right)/2;
if(num[mid] == num[left] && num[mid] == num[right])
{
int a = rf(left,mid,num);
int b = rf(mid,right,num);
return a<b? a : b;
}
else if(num[mid] <= num[left] && num[mid] <= num[right])
return rf(left,mid,num);
else if(num[mid] >= num[left] && num[mid] <= num[right])
return rf(left,mid,num);
else //(num[mid] >= num[left] && num[mid] >= num[right])
return rf(mid,right,num);
}
``````

one point I think needs to explain: only in (num[mid] == num[left] && num[mid] == num[right]) condition, we don't know where the minimum stays, so calculate both and compare.
if all items in the array are the same, time complexity is O(n).

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