# LeetCode – Russian Doll Envelopes (Java)

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)

This problem is similar to Longeset Increasing Subsequence.

Java Solution 1 - Naive

```public int maxEnvelopes(int[][] envelopes) { if(envelopes==null||envelopes.length==0) return 0;   Arrays.sort(envelopes, new Comparator<int[]>(){ public int compare(int[] a, int[] b){ if(a[0]!=b[0]){ return a[0]-b[0]; }else{ return a[1]-b[1]; } } }); int max=1; int[] arr = new int[envelopes.length]; for(int i=0; i<envelopes.length; i++){ arr[i]=1; for(int j=i-1; j>=0; j--){ if(envelopes[i][0]>envelopes[j][0]&&envelopes[i][1]>envelopes[j][1]){ arr[i]=Math.max(arr[i], arr[j]+1); } } max = Math.max(max, arr[i]); }   return max; }```

Java Solution 2 - Binary Search

```public int maxEnvelopes(int[][] envelopes) { if(envelopes==null||envelopes.length==0) return 0;   Arrays.sort(envelopes, new Comparator<int[]>(){ public int compare(int[] a, int[] b){ if(a[0]!=b[0]){ return a[0]-b[0]; //ascending order }else{ return b[1]-a[1]; // descending order } } });   ArrayList<Integer> list = new ArrayList<Integer>();   for(int i=0; i<envelopes.length; i++){   if(list.size()==0 || list.get(list.size()-1)<envelopes[i][1]) list.add(envelopes[i][1]);   int l=0; int r=list.size()-1;   while(l<r){ int m=l+(r-l)/2; if(list.get(m)<envelopes[i][1]){ l=m+1; }else{ r=m; } }   list.set(r, envelopes[i][1]); }   return list.size(); }```
Category >> Algorithms
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