Interview-Increasing Sequence with Length 3.

Given an array, determine whether there are three elements A[i],A[j],A[k], such that A[i]<A[j]<A[k] & i<j<k.

Analysis:

It is a special case of the Longest Increasing Sequence problem. We can use the O(nlog(n)) algorithm, since we only need sequence with length three, we can do it in O(n).

Solution:

 public static boolean threeIncSeq(int[] A){
         if (A.length<3) return false;

         int oneLen = 0;
         int twoLen = -1;
         for (int i=1;i<A.length;i++){
             //check whether current element is larger then A[twoLen].
             if (twoLen!=-1 && A[i]>A[twoLen]) return true;
             if (twoLen!=-1 && A[i]>A[twoLen] && A[i]<A[oneLen]){
                 twoLen = i;
                 continue;
             }
             if (twoLen==-1 && A[i]>A[oneLen]){
                 twoLen = i;
                 continue;
             }
             if (A[i]<A[oneLen]){
                 oneLen = i;
                 continue;
             }
         }

         return false;
     }

Variant:

If we need to output the sequence, we then need to record the sequence of current length 1 seq and length 2 seq.

 public static List<Integer> threeIncSeq(int[] A){
         if (A.length<3) return (new ArrayList<Integer>());

         int oneLen = 0;
         int twoLen = -1;
         List<Integer> oneList = new ArrayList<Integer>();
         List<Integer> twoList = new ArrayList<Integer>();
         oneList.add(A[0]);
         for (int i=1;i<A.length;i++){
             //check whether current element is larger then A[twoLen].
             if (twoLen!=-1 && A[i]>A[twoLen]){
                 twoList.add(A[i]);
                 return twoList;
             }
             if (twoLen!=-1 && A[i]>A[twoLen] && A[i]<A[oneLen]){
                 twoLen = i;
                 twoList = new ArrayList<Integer>();
                 twoList.addAll(oneList);
                 twoList.add(A[i]);
                 continue;
             }
             if (twoLen==-1 && A[i]>A[oneLen]){
                 twoLen = i;
                 twoList = new ArrayList<Integer>();
                 twoList.addAll(oneList);
                 twoList.add(A[i]);
                 continue;
             }
             if (A[i]<A[oneLen]){
                 oneLen = i;
                 oneList = new ArrayList<Integer>();
                 oneList.add(A[i]);
                 continue;
             }
         }

         return (new ArrayList<Integer>());

     }

NOTE: This is more compliated then needed, when using List<> in this case, but this idea can be used to print the LIS.