Search Insert Position

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You may assume NO duplicates in the array.

Example

[1,3,5,6], 5 → 2

[1,3,5,6], 2 → 1

[1,3,5,6], 7 → 4

[1,3,5,6], 0 → 0

Challenge

O(log(n)) time

Solution 1

 Naively, we can just iterate the array and compare target with ith and (i+1)th element. Time complexity is O(n)

public int searchInsert(int[] A, int target) 
    {
        // write your code here
     if(A.length==0) return 0;
 
        if(target <= A[0]) return 0;
 
        for(int i=0; i<A.length-1; i++){
            if(target > A[i] && target <= A[i+1]){
                return i+1;
            }
        }
 
        return A.length;
        
    }

Solution 2

 This also looks like a binary search problem. We should try to make the complexity to be O(log(n)).

    public int searchInsert(int[] A, int target) 
    {
        // write your code here
     if(A.length==0) return 0;
     return helper(A, target, 0, A.length-1);
        
    }
    
    
    public int helper(int[] A, int target, int start, int end)
    {
        int mid = (start+end)/2;
        
        if (target == A[mid]) return mid;
        
        if(target<A[mid])
        {
            if (mid <= start) return start;
            else
            {
                return helper(A,target,start,mid-1);
            }
        }
        
        else  //target > A[mid]
        {
            if(mid >= end) return end+1;
            else
            {
                return helper(A,target,mid+1,end);
            }
        }
        
        
    }

reference: http://www.programcreek.com/2013/01/leetcode-search-insert-position/