Algorithm

LintCode/Longest Increasing Subsequence

Problem Summary

Given a sequence of integers, find the length of the longest increasing subsequence (LIS).

Solution 1: DP, O(N^2) time

Let us denote the given array with A[1], A[2], …, A[N].

Define f[i] as the length of the LIS ending at A[i]. So we have

f[i] = max{f[j]} + 1 , where j < i and A[j] < A[i]

The answer is max{f[i]}, i = 1,2,…,N. This method takes O(N^2) time and O(N) space.

Solution 2: DP + Binary Search, O(NlogN) time

Define g[i] as the minimal ending number of all increasing subsequences whose length are i.

Naturally, we realize that g[i] is non-decreasing, which means we can use binary search on it.

Plus, if g[l] has a legal value, the answer L is equal to or greater than l.That is, L = max{l}, where g[l] has a legal value.

These thoughts lead to the following algorithm:

First, we initialize g[1] with A[1] and L with 1.

Then we iterate through the array. For each i,

  1. If A[i] > g[L], it is a great thing because we can now increase L by 1. For new L, we let g[L] = A[i].
  2. If A[i] == g[L], A[i] is not helpful so we continue.
  3. If A[i] < g[L], we use binary search to find an index k, such that g[k-1] < A[i] ≤ g[k]. (We suppose g[0] is INT_MIN, so there will always be a legal k.) Then we use A[i] to update g[k].

This method has a lower time complexity, but also requires more thinking.

Code for Soluton 2

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class Solution {
public:
    /*
     * @param nums: An integer array
     * @return: The length of LIS (longest increasing subsequence)
     */
    int longestIncreasingSubsequence(vector<int> &nums) {
    	int n = nums.size();
    	if (n<2)
    		return n;
    	int f[n+2];
    	f[1] = nums[0];
    	int len = 1;
    	for (int i = 1; i < n; ++i)
    	{
    		if (nums[i] > f[len])
    			f[++len] = nums[i];
    		else
    		if (nums[i] == f[len])
    			continue;
    		else
    		{
    			int pos = binary_search(nums[i],len,f);
    			f[pos] = nums[i];
    		}
    	}
    	return len;
    }
    int binary_search(int cur,int size,int *f) /* return k, s.t. f[k-1] < cur <= f[k] */
    {
    	int l = 1, r = size;
    	while (l<r)
    	{
    		int mid = l + ((r-l)>>1);
    		if (f[mid] == cur)
    			return mid;
    		else
    		if (f[mid] < cur)
    			l = mid+1;
    		else
    			r = mid;
    	}
    	return l;
    }
};