search-in-rotated-sorted-array

Problem

Search in Rotated Sorted Array

Problem Description

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Your algorithm's runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

Solution

This problem asked to implement in O(logn). Binary search should come into your mind when you see O(logn), sorted and search keywords.

1. Intuitive solution is simple, iterate through array, find target in nums return pos, not, return -1, O(n)

2. Binary search, given array is rotated sorted array, meaning there are 2 sorted array. when do binary search, need to carefully compare

   target and nums[mid] and nums[lo] to chose whether go right (higher) or left (lower).

   • define lo, hi, and mid = (lo + hi)/2.

   • if target == nums[mid] found, return mid and exit.

   • if nums[mid] >= nums[lo] meaning mid is within sorted parts, now compare target,

     • if target >= nums[lo] && target <= nums[md], target is located in [lo, mid] sorted part, search left, hi = mid - 1

     • [mid, hi]lo = mid + 1

   • else check compare target with hi and mid value

     • if target >= nums[mid] && target <= nums[hi], target is located in [mid, hi] part, search right, lo = mid + 1

       else target located in [lo, mid] part, search left, hi = mid - 1

For example:

Search in Rotated Sorted Array

Complexity Analysis

Time Complexity: O(log(N))

Space Complexity: O(1)

• N - the length of array nums

Code - Binary search

class Solution {
    public int search(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        int lo = 0;
        int hi = nums.length - 1;
        while (lo < hi) {
            int mid = lo + (hi - lo) / 2;
            // found, return mid, exit
            if (nums[mid] == target) {
                return mid;
            }
            // mid is within sorted parts
            if (nums[lo] <= nums[mid]) {
                // target located at [lo, mid] part, search left
                if (target >= nums[lo] && target <= nums[mid]) {
                    hi = mid - 1;
                } else {
                    lo = mid + 1;
                }
            } else {
                // target located at [mid, hi] part, search right
                if (target >= nums[mid] && target <= nums[hi]) {
                    lo = mid + 1;
                } else {
                    hi = mid - 1;
                }
            }
        }
        return nums[lo] == target ? lo : -1;
    }
}

Naive solution - O(n)

class Solution {
    public int search(int[] nums, int target) {
        if (nums == null || nums.length == 0) return -1;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == target) return i;
        }
        return -1;
    }
}