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PremiumBinary SearchEasy

Search Insert Position

Asked at:GoogleAmazon

Given a sorted array of distinct integers and a target, return the index where the target is found, or the index where it would be inserted to keep the array sorted.

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Problem

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.

Input

A sorted (ascending) array of distinct integers `nums`, and an integer `target`.

Output

The index of `target` if it exists in `nums`, otherwise the index where it would be inserted.

Examples

Input: nums = [1, 3, 5, 6], target = 5

Output: 2

5 is already in the array at index 2.

Input: nums = [1, 3, 5, 6], target = 2

Output: 1

2 is not in the array. It would go between 1 and 3, at index 1.

Input: nums = [1, 3, 5, 6], target = 7

Output: 4

7 is larger than everything, it would be appended at index 4.

The brute-force approach

Scan left to right. The moment you find a value that's >= target, that's your answer, either target is there, or that's where it would be inserted.

for i in 0 .. len(nums) - 1:
    if nums[i] >= target:
        return i       # found it or found the insertion point

return len(nums)       # target is larger than everything

Linear scan is O(n). For a sorted array, that's leaving half the available information on the table, the problem is set up perfectly for binary search.

Time: O(n)Space: O(1)

Spotting the pattern

This is a Binary Search problem. The key question to ask yourself:

If the target isn't in the array, how do I know exactly where it would go?

Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.

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