Length of the largest subarray with contiguous elements | Set 2 - GeeksforGeeks

Length of the largest subarray with contiguous elements | Set 2 - GeeksforGeeks
Given an array of integers, find length of the longest subarray which contains numbers that can be arranged in a continuous sequence.
In the previous post, we have discussed a solution that assumes that elements in given array are distinct. Here we discuss a solution that works even if the input array has duplicates.

we checked whether maximum value minus minimum value is equal to ending index minus starting index or not. Since duplicate elements are allowed, we also need to check if the subarray contains duplicate elements or not. For example, the array {12, 14, 12} follows the first property, but numbers in it are not contiguous elements.
To check duplicate elements in a subarray, we create a hash set for every subarray and if we find an element already in hash, we don’t consider the current subarray.

    static int findLength(int arr[])

    {

        int n = arr.length;

        int max_len = 1; // Inialize result

 

        // One by one fix the starting points

        for (int i=0; i<n-1; i++)

        {

            // Create an empty hash set and add i'th element

            // to it.

            HashSet<Integer> set = new HashSet<>();

            set.add(arr[i]);

 

            // Initialize max and min in current subarray

            int mn = arr[i], mx = arr[i];

 

            // One by one fix ending points

            for (int j=i+1; j<n; j++)

            {

                // If current element is already in hash set, then

                // this subarray cannot contain contiguous elements

                if (set.contains(arr[j]))

                    break;

 

                // Else add curremt element to hash set and update

                // min, max if required.

                set.add(arr[j]);

                mn = Math.min(mn, arr[j]);

                mx = Math.max(mx, arr[j]);

 

                // We have already cheched for duplicates, now check

                // for other property and update max_len if needed

                if (mx-mn == j-i) // optimize here, if(max-min > n-i, then no point to continue to search

                    max_len = Math.max(max_len, mx-mn+1);

            }

        }

        return max_len; // Return result

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