Leetcode 154: Find Minimum in Rotated Sorted Array II

Problem:

Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?
Would this affect the run-time complexity? How and why?
Suppose a sorted array 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).
Find the minimum element.
The array may contain duplicates.

Analysis:

Based on the previous analysis of Leetcode 154, we just need to specify the action if mid equals
to the left.
Leetcode 154: Find Minimum in Rotated Sorted Array

Code:

public class Solution {

   public int findMin(int[] num) {
       int left=0;
       int right=num.length-1;
       int min=Integer.MAX_VALUE;
       while(left<=right){
           int mid=(left+right)/2;
           min=Math.min(min,num[mid]);
           if(num[left]<num[mid]){ //A[left..mid] is sorted
               min=Math.min(min,num[left]);
               left=mid+1;
           }else if(num[left]>num[mid]){ //A[mid..right] is sorted
               right=mid-1;
           }else{
               left++;
           }
       }
       return min;
   }
}

Leetcode 153: Find Minimum in Rotated Sorted Array

Problem:
Suppose a sorted array 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).
Find the minimum element.
You may assume no duplicate exists in the array.

Analysis:
Similarly, we can easily check which part of the array is sorted and we just compare the lowest
element. Then, we continue to search the other part.

Code:

public class Solution {
    public int findMin(int[] num) {//[1],[2,1], [3,1,2]
        int left=0;
        int right=num.length-1;
        int min=Integer.MAX_VALUE;
        while(left<=right){
            int mid=(left+right)/2;
            if(num[left]<=num[mid]){
                min=Math.min(min,num[left]);
                left=mid+1;
            }else{
                right=mid;
            }
        }
        return min;
    }
}

Leetcode 81: Search in Rotated Sorted Array II

Problem:

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

Analysis:

Based on the analysis of  "Search in Rotated Sorted Array", we are unable to make sure which part

is sorted only by comparing the mid with the left element. In this case, we need to increase the left

pointer until it is not equal to the mid.

Related questions:

Leetcode 33: Search in Rotated Sorted Array

Code:

public class Solution {
    public boolean search(int[] A, int target) {
        if(A==null || A.length==0 || A.length==1 && A[0]!=target){
            return false;
        }
        int left=0;
        int right=A.length-1;
        while(left<=right){
            int mid=(left+right)/2;
            if(target==A[mid]){
                return true;
            }else if(A[left]<A[mid]){
                if(A[left]<=target && target<A[mid]){
                    right=mid-1;
                }else{
                    left=mid+1;
                }
            }else if(A[left]>A[mid]){
                if(A[mid]<target && target <=A[right]){
                    left=mid+1;
                }else{
                    right=mid-1;
                }
            }else{
                left++;
            }
        }
        return false;
    }
}

Leetcode 33: Search in Rotated Sorted Array

Problem:

Suppose a sorted array 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.

Analysis:

Two conditions should be paid attention. One is that before rotated, the array is sorted and the other

is there is no duplicate. We can modify binary search to search rotated array. 

Suppose A is the rotated array.

left=0, right=A.length-1, mid=(right+left)/2

Case 1: if A[mid] >A[left], A[left .. mid] is in order. 

Case 2: if A[mid] < A[left], A[mid] is also smaller than A[right], therefore A[mid .. right] is in order.

We can easily verify whether the target is in the sorted part or not. 

Code:

public class Solution {
    public int search(int[] A, int target) {
        if(A==null || A.length==0 || A.length==1 && A[0]!=target){
            return -1;
        }
        int left=0;
        int right=A.length-1;
        while(left<=right){
            int mid=(left+right)/2;
            if(A[mid]==target){
                return mid;
            }else if(A[mid]>=A[left]){ // A[left..mid] is sorted.
                if(A[left]<=target && target<A[mid]){
                    right=mid-1;
                }else{
                    left=mid+1;
                }
            }else{//A[mid..right] is sorted.
               if(target<=A[right] && target>A[mid]){
                   left=mid+1;
               }else{
                   right=mid-1;
               }
            }
        }
        return -1;
    }
}

Leetcode 80: Remove Duplicates from Sorted Array II

Problem:

Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?

For example,
Given sorted array A = [1,1,1,2,2,3],

Your function should return length = 5, and A is now [1,1,2,2,3]. 

Analysis:

As duplicates are allowed at most twice, the first thought is to compare the current element with the

previous two elements. However, we must remember that the array is a sorted array, which implies

that if current element equals to the second previous element, it must equal to the first previous one.

If not equal, current element can be added to the list.

Code:

public class Solution {
    public int removeDuplicates(int[] A) {
       if(A.length<=2){
           return A.length;
       }
       int tail=2;
       for(int i=2;i<A.length;i++){
           if(A[i]!=A[tail-2]){
               A[tail++]=A[i];
           }
       }
       return tail;
    }
}

Leetcode 26: Remove Duplicates from Sorted Array

Problem:

Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

For example,
Given input array A = [1,1,2],

Your function should return length = 2, and A is now [1,2].

Analysis:
As it is a sorted array, duplicated elements are neighboring with each other.
We can easily using two pointers to copy the first met element to the end of non-duplicated list.

Code:

public class Solution {
    public int removeDuplicates(int[] A) {
        if(A==null){
            return -1;
        }else if(A.length<=1){
            return A.length;
        }
        int len=0;// the last element
        for(int i=1;i<A.length;i++){//i is another pointer
            if(A[i]!=A[len]){
                A[++len]=A[i];
            }
        }
        return len+1;
    }
}

Change the thought when cracking a special type of algorithm problem

Problem: (Gas station)

There are N gas stations along a circular route, where the amount of gas at station i is gas[i].

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations.

Return the starting gas station's index if you can travel around the circuit once, otherwise return -1.

Solutions:

The first thought of this problem is to check whether there exists an index that can meet the requirement and check each of them in serial until we find one. Well, you will get a O(n^2) algorithm if you do so.

My second thought is, this problem can use dynamic programming to record the path in order to minimize the computation. However, it is still a O(n^2) algorithm. 

After reading the solution of a guy from Leetcode.com, I realize that I went to the wrong way at the beginning. If we take a look at the problem, it is actually asking to "return ONE index that satisfy the requirement". Wow, I never notice that. So let's ask "What the most possible index?". If this index does not pass the constraints, there would no index left. As such, we only need to check the MOST POSSIBLE index and get a O(n) algorithm. 

http://discuss.leetcode.com/questions/3005/gas-station

class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        if (gas.size() == 0)
            return -1;

        int sum = 0;
        int index = -1;
        bool last = true;
        for (int i = 0; i < gas.size(); i++) {
            int diff = gas[i] - cost[i];
            if (last && diff >= 0 && sum <= 0)
                index = i;
            last = (diff < 0);
            sum += diff;
        }

        if (index == -1)
            return index;

        for (int i = 0, sum = 0; i < gas.size(); i++) {
            sum += gas[(index + i) % gas.size()] - cost[(index + i) % gas.size()];
            if (sum < 0)
                return -1;
        }

        return index;
    }
};