//Binary Search implementation //Time Complexity: O (log n) //return int {-1 , mid index}
int binarySearch(int tar, vector<int>& arr){
int lo=0, md, hi=arr.size()-1;
while(hi>=lo){
md = lo + (hi - lo) / 2;
if(arr[md]==tar) return md;
else if(arr[md]<tar) lo=md+1;
   		else hi=md-1;
}
return -1;
}
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//Binary Search by using Stls //Time Complexity: O (log n)
// for binary search in containers like vector (let target element=6)
binary_search(v.begin(), v.end(), 6); 
// return 1 or 0 as present or not
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//binary search with lower and upper bound by using stls
int x, ans; //x is passing value, ans is a value you want, v --> vector or any container
ans = lower_bound(v.begin(), v.end(), x) - v.begin(); //ans >= x (equal number or first number after x)
ans = upper_bound(v.begin(), v.end(), x) - v.begin(); // ans > x (first number after x)

//implementation
int lower_bound(vector<int> nums, int target){
        int l = 0, r = nums.size()-1, m = 0;
        while(l < r) {
            m = (l+r)/2;
            if(nums[m] < target)
                l = m+1;
            else 
                r = m;
        }
        return r;
    }

    int upper_bound(vector<int> nums, int target){
        int l = 0, r = nums.size()-1, m = 0;
        while(l < r) {
            m = (l+r)/2;
            if(nums[m] <= target)
                l = m+1;
            else 
                r = m;
        }
        return r;
    }
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// Two Pointer implementation
// Two pointer technique based solution to find
// if there is a pair in A[0..N-1] with a given sum.
int isPairSum(int A[], int N, int X)
{
    // represents first pointer
    int i = 0;
  
    // represents second pointer
    int j = N - 1;
  
    while (i < j) {
      
        int curSum = A[i]+arr[j];
      
        // If we find a pair
        if curSum == X)
            return 1;
  
        // If sum of elements at current
        // pointers is less, we move towards
        // higher values by doing i++
        else if (curSum < X)
            i++;
  
        // If sum of elements at current
        // pointers is more, we move towards
        // lower values by doing j--
        else
            j--;
    }
    return 0;
}

int search(int* arr, int n, int key) {
int si = 0;
 int ei = n-1;
//     core concept : if array is rotated and we visited any index then one part
// of array is going to be sorted 
    while(si <= ei){
       int mid = (si + ei)/2;
        
        if(arr[mid] == key){
            return mid;
        }
        if(arr[mid]  >= arr[si]){ 
//             if mid == si this means left part has one element and that is sorted
//           left  array is sorted
            if(key < arr[mid] && key >= arr[si]){
                ei = mid - 1;
                continue;
            }else{
                si = mid + 1;
            }
        }
        if(arr[mid]  <= arr[ei]){
//             here arr[mid] can be equal to arr[ei] if there is only one 
//             element in whole array
//             righty array is sorted
//             
            if(key <= arr[ei]  && key > arr[mid]){
                si = mid + 1;
                
            }else{
                ei = mid - 1;
            }
            
        }
    }
   return -1;
}