[35 -Search Insert Position]

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Problem Description

Given a sorted array of distinct integers 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.

Idea

Since the array is sorted, binary search is the most efficient choice (O(log n)).

Logic

1. Define your search range
2. Use a loop (while (left <= right) or while (left < right))
3. Calculate mid, check if target is found
4. change to the boundary accordingly

Code (Java - Closed Interval)

class Solution {
    public int searchInsert(int[] nums, int target) {

        int left = 0; int right = nums.length - 1;

        while(left <= right){
            int mid = left + ((right - left)/2);
            
            if(nums[mid] < target){
                left = mid + 1;
            }
            else if(nums[mid] > target){
                right = mid - 1;
            }
            else{
                return mid;
            }
        }
        return right+1;
        
    }
}

Code (Java - Half-Open Interval)

class Solution {
    public int searchInsert(int[] nums, int target) {

        int left = 0; int right = nums.length;

        while(left < right){
            int mid = left + ((right - left)/2);
            
            if(nums[mid] < target){
                left = mid + 1;
            }
            else if(nums[mid] > target){
                right = mid;
            }
            else{
                return mid;
            }
        }
        return right;
    }
}