nums contains distinct values sorted in ascending order.

-10⁴ <= target <= 10⁴

23:T56a,\n# Approach\n- `binary seaerch` approach is damn easy for this question and also easy to think of it.\n- the array is sorted so we just have to fugure out where should it be placed.\n- so we just have to figure out the position at which the `prevoius value` is less that our element and `next` value is more than our element.\n- Thats how its figured out that it\'s a binary seach problem.\n\n\n# Complexity\n- Time complexity: O(log(n))\n\n\n- Space complexity: O(1)\n\n\n## `Upvote! It only takes 1 click :)`\n\n# Code\n```\nclass Solution {\npublic:\n int searchInsert(vector& nums, int target) {\n int low=0;\n int high=nums.size();\n int mid;\n if(target>nums[high-1]){\n return high;\n }\n while(low<=high){\n mid=(low+high)/2;\n if(nums[mid]==target){ \n return mid;\n }\n \n if(target

Search Insert Position

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Problem Description

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