Russian Doll Envelopes

This page explains Java solution to problem Russian Doll Envelopes using Binary Search algorithm.

Problem Statement

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)

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])

Solution

package com.vc.hard;

import java.util.*;

class RussianDollEnvelopes {
    public int maxEnvelopes(int[][] envelopes) {
        if(envelopes == null || envelopes.length == 0) return 0;

        int n = envelopes.length;
        Arrays.sort(envelopes, new Comparator<int[]>(){
            public int compare(int[] x, int[] y) {
                int cmp = Integer.compare(x[0], y[0]);
                if(cmp == 0) return Integer.compare(y[1], x[1]);
                else return cmp;
            }
        });

        int[] secondDimension = new int[n];
        for(int i = 0; i < n; i++) secondDimension[i] = envelopes[i][1];
        return LISBinarySearch(secondDimension);
    }

    private int LISDynamicProgramming(int[] arr) {
        int n = arr.length;
        int[] dp = new int[n];

        int max = 0;
        for(int i = 1; i < n; i++) {
            for(int j = 0; j < i; j++) {
                if(arr[j] < arr[i]) {
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                    max = Math.max(max, dp[i]);
                }
            }
        }
        return max + 1;
    }

    private int LISBinarySearch(int[] arr) {
        int n = arr.length;
        int[] dp = new int[n];

        int len = 0;
        for(int num: arr) {
            int index = Arrays.binarySearch(dp, 0, len, num);
            if(index < 0) {
                index = -(index + 1);
            }
            dp[index] = num;
            if(index == len) len++;
        }
        return len;
    }
}

Time Complexity

O(N log N) Where
N is total number of elements in an input array

Space Complexity

O(N) Where
N is total number of elements in an input array