# Simple and easy understanding Java solution

• ``````public int search(int[] nums, int target) {
int pivot = findPivot(nums);
return binarySearch(nums, 0, pivot, target) + binarySearch(nums, pivot + 1, nums.length - 1, target) + 1;
}

public int findPivot(int nums[]){
int i = 0, j = nums.length - 1;
while(i < j - 1){
int mid = i + (j - i) / 2;
if(nums[i] < nums[mid] && nums[j] < nums[mid]){
i = mid;
}else {
j = mid;
}
}
return i;
}

public int binarySearch(int a[], int start, int end, int key){
int i = start, j = end;
while(i <= j){
int mid = i + (j - i) / 2;
if(a[mid] > key){
j = mid - 1;
}else if(a[mid] < key){
i = mid + 1;
}
else return mid;
}
return -1;
}``````

• ``````public int findp(int[] nums) {
for (int i = 1; i < nums.length; i++) {
if (nums[i] < nums[i-1]) {
return i-1;
}
}
return -1;
}``````

• Your suggestion will require O(n) time complexity whereas solution in question is requiring just O(Logn) complexity.

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