Using Binary Search with sorted Array with duplicates [duplicate]

Answer 1

It is using Modified Binary Search. It will be O(LogN). Space complexity will be O(1). We are calling BinarySearchModified two times. One for finding start index of element and another for finding end index of element.

private static int BinarySearchModified(int[] input, double toSearch)
    {
        int start = 0;
        int end = input.Length - 1;

        while (start <= end)
        {
            int mid = start + (end - start)/2;
            if (toSearch < input[mid]) end = mid - 1;
            else start = mid + 1;
        }

        return start;
    }


    public static Result GetRange(int[] input, int toSearch)
    {
        if (input == null) return new Result(-1, -1);

        int low = BinarySearchModified(input, toSearch - 0.5);

        if ((low >= input.Length) || (input[low] != toSearch)) return new Result(-1, -1);

        int high = BinarySearchModified(input, toSearch + 0.5);

        return new Result(low, high - 1);
    } 

 public struct Result
    {
        public int LowIndex;
        public int HighIndex;

        public Result(int low, int high)
        {
            LowIndex = low;
            HighIndex = high;
        }
    }

Answer 2

Below is the java code which returns the range for which the search-key is spread in the given sorted array:

public static int doBinarySearchRec(int[] array, int start, int end, int n) {
    if (start > end) {
        return -1;
    }
    int mid = start + (end - start) / 2;

    if (n == array[mid]) {
        return mid;
    } else if (n < array[mid]) {
        return doBinarySearchRec(array, start, mid - 1, n);
    } else {
        return doBinarySearchRec(array, mid + 1, end, n);
    }
}

/**
 * Given a sorted array with duplicates and a number, find the range in the
 * form of (startIndex, endIndex) of that number. For example,
 * 
 * find_range({0 2 3 3 3 10 10}, 3) should return (2,4). find_range({0 2 3 3
 * 3 10 10}, 6) should return (-1,-1). The array and the number of
 * duplicates can be large.
 * 
 */
public static int[] binarySearchArrayWithDup(int[] array, int n) {

    if (null == array) {
        return null;
    }
    int firstMatch = doBinarySearchRec(array, 0, array.length - 1, n);
    int[] resultArray = { -1, -1 };
    if (firstMatch == -1) {
        return resultArray;
    }
    int leftMost = firstMatch;
    int rightMost = firstMatch;

    for (int result = doBinarySearchRec(array, 0, leftMost - 1, n); result != -1;) {
        leftMost = result;
        result = doBinarySearchRec(array, 0, leftMost - 1, n);
    }

    for (int result = doBinarySearchRec(array, rightMost + 1, array.length - 1, n); result != -1;) {
        rightMost = result;
        result = doBinarySearchRec(array, rightMost + 1, array.length - 1, n);
    }

    resultArray[0] = leftMost;
    resultArray[1] = rightMost;

    return resultArray;
}

Answer 3

public void printCopies(int[] array)
{
    HashMap<Integer, Integer> memberMap = new HashMap<Integer, Integer>();
    for(int i = 0; i < array.size; i++)
       if(!memberMap.contains(array[i]))
           memberMap.put(array[i], 1);
       else
       {
           int temp = memberMap.get(array[i]); //get the number of occurances
           memberMap.put(array[i], ++temp); //increment his occurance
       }

    //check keys which occured more than once
    //dump them in a ArrayList
    //return this ArrayList
 }

Alternatevely, instead of counting the number of occurances, you can put their indices in a arraylist and put that in the map instead of the count.

   HashMap<Integer, ArrayList<Integer>> 
   //the integer is the value, the arraylist a list of their indices

public void printCopies(int[] array)
{
    HashMap<Integer, ArrayList<Integer>> memberMap = new HashMap<Integer, ArrayList<Integer>>();
    for(int i = 0; i < array.size; i++)
       if(!memberMap.contains(array[i]))
       {
           ArrayList temp = new ArrayList();
           temp.add(i);
           memberMap.put(array[i], temp);
       }
       else
       {
           ArrayList temp = memberMap.get(array[i]); //get the lsit of indices
           temp.add(i);
           memberMap.put(array[i], temp); //update the index list
       }

    //check keys which return lists with length > 1
    //handle the result any way you want
 }

heh, i guess this will have to be posted.

 int predefinedDuplicate = //value here;
 int index = Arrays.binarySearch(array, predefinedDuplicate);
 int leftIndex, rightIndex;
 //search left
 for(leftIndex = index; array[leftIndex] == array[index]; leftIndex--); //let it run thru it
 //leftIndex is now the first different element to the left of this duplicate number string
 for(rightIndex = index; array[rightIndex] == array[index]; rightIndex++); //let it run thru it

 //right index contains the first different element to the right of the string
 //you can arraycopy this [leftIndex+1, rightIndex-1] string or just print it
 for(int i = leftIndex+1; i<rightIndex; i++)
 System.out.println(array[i] + "\t");

Answer 4

Another result for log(n) binary search for leftmost target and rightmost target. This is in C++, but I think it is quite readable.

The idea is that we always end up when left = right + 1. So, to find leftmost target, if we can move right to rightmost number which is less than target, left will be at the leftmost target.

For leftmost target:

int binary_search(vector<int>& nums, int target){
    int n = nums.size();
    int left = 0, right = n - 1;

    // carry right to the greatest number which is less than target.
    while(left <= right){
        int mid = (left + right) / 2;
        if(nums[mid] < target)
            left = mid + 1;
        else
            right = mid - 1;
    }
    // when we are here, right is at the index of greatest number
    // which is less than target and since left is at the next, 
    // it is at the first target's index
    return left;
}

For the rightmost target, the idea is very similar:

int binary_search(vector<int>& nums, int target){
    while(left <= right){
        int mid = (left + right) / 2;
        // carry left to the smallest number which is greater than target.
        if(nums[mid] <= target)
            left = mid + 1;
        else
            right = mid - 1;
    }
    // when we are here, left is at the index of smallest number
    // which is greater than target and since right is at the next, 
    // it is at the first target's index
    return right;
}