Russian Doll Envelopes

You have a number of envelopes with widths and heights given as a pair of integers (w, h). One envelope can fit into another if and only if both the width and height of one envelope is greater than the width and height of the other envelope.

What is the maximum number of envelopes can you Russian doll? (put one inside other)

Note: Rotation is not allowed.

Example:

Input: [[5,4],[6,4],[6,7],[2,3]]
Output: 3 
Explanation: The maximum number of envelopes you can Russian doll is 3 ([2,3] => [5,4] => [6,7]).

class Solution {
    // O(NlogN) Solution
    public int maxEnvelopes(int[][] envelopes) {
        // Sort on basis of Side
        // Descending sort on unsorted side
        Arrays.sort(envelopes, (a, b) -> a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]);
        // Now we can apply LIS on unsorted side
        int dp[] = new int[envelopes.length];
        int ans = 0;
        for (int[] envelope : envelopes) {
            // Searching for correct position of envelop in array
            int low = 0, high = ans;
            while (low < high) {
                int mid = low + (high - low) / 2;
                if (envelope[1] > dp[mid])
                    low = mid + 1;
                else
                    high = mid;
            }
            dp[low] = envelope[1];
            if (low == ans)
                ans++;
        }
        return ans;
    }

    // O(n^2) Solution
    public int maxEnvelopes(int[][] envelopes) {
        // Sort on basis of Side
        Arrays.sort(envelopes, (a, b) -> a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]);
        // Now we can apply LIS on unsorted side
        int[] dp = new int[envelopes.length];
        int max = 0;
        for (int i = 0; i < envelopes.length; i++) {
            // Base case
            dp[i] = 1;
            for (int j = 0; j < i; j++) {
                if (envelopes[j][0] < envelopes[i][0] && envelopes[j][1] < envelopes[i][1])
                    dp[i] = Math.max(dp[i], dp[j] + 1);
            }
            max = Math.max(max, dp[i]);
        }
        return max;
    }
}