Search In Rotated Sorted Array Problem Solution

Search In Rotated Sorted Array Problem: There is an integer array nums sorted in ascending order (with distinct values).

Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].

Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.

You must write an algorithm with O(log n) runtime complexity.

Example :

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

Output: 4

Search In Rotated Sorted Array Problem Solution

Problem Solution In Python

class Solution:
    def search(self, nums, target):    
        def search(nums, l, r):
            m = (l + r) // 2
            if nums[m] == target:
                return m
            if l == r:
                return -1
            if nums[l] < nums[m]:
                if target >= nums[l] and target < nums[m]:
                    return search(nums, l, m - 1)
                else:
                    return search(nums, m + 1, r)
            elif r - m == 1:
                if nums[r] == target:
                    return r
                else:
                    return search(nums, l, m)
            elif nums[m + 1] < nums[r]:
                if target >= nums[m + 1] and target <= nums[r]:
                    return search(nums, m + 1, r)
                else:
                    return search(nums, l, m - 1)
            return -1

        return -1 if not nums else search(nums, 0, len(nums) - 1)

Problem Solution In Java

class Solution {
    public int search(int[] nums, int target) {
        
        if(nums.length == 1){
            return nums[0] == target ? 0 : -1;    
        }
        int i=0;
        int j = nums.length-1;
        while(i<j){
            if(nums[i] == target){
                return i;
            }
            else if(nums[j] == target){
                return j;
            }
            else if(target > nums[i]){
                i++;
            }
            else{
                j--;
            }
        }
        return -1;
    }
}

Problem Solution In C++

class Solution {
public:
    int b_Search(vector<int>& nums, int target, int low, int high) {
        while(low <= high) {
            int mid = low + (high - low) / 2;
            if(nums[mid] == target) return mid;
            else if(nums[mid] > target) high = mid - 1;
            else low = mid + 1;
        }
        return -1;
    }
    
    int search(vector<int>& nums, int target) {
        int i = 0;
        for(; i < nums.size() - 1; i++) {
            if(nums[i] > nums[i + 1]) break;
        }
        int ans = b_Search(nums, target, 0, i);
        if(ans != -1 || i == nums.size() - 1) return ans; 
        ans = b_Search(nums, target, i+1, nums.size() - 1);
        return ans;
    }
};

Problem Solution In C

int search(int* nums, int numsSize, int target){

    int i = 0;
for(i = 0;i<numsSize;i++)
{
    if((nums[i] == target) ) 
        return i;
    if((nums[numsSize - i-1] == target ))
        return (numsSize-i-1);
}

return -1;
}


If You Like This Page Then Make Sure To Follow Us on Facebook, G News and Subscribe Our YouTube Channel. We will provide you updates daily.
Shares
facebook sharing button Share
twitter sharing button Tweet
whatsapp sharing button Share
telegram sharing button Share
pinterest sharing button Pin

Leave a Comment