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Greedy Algorithm - Question 2

300. Longest Increasing Subsequence

Given an integer array nums, return the length of the longest strictly increasing subsequence.

A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, [3,6,2,7] is a subsequence of the array [0,3,1,6,2,2,7].

Constraints:

1 <= nums.length <= 2500

-10^4 <= nums[i] <= 10^4


Follow up: Can you come up with an algorithm that runs in O(n log(n)) time complexity?


Analysis:

This questions can be solved by dynamic programming (dp) with a time complexity of O(N^2), where N is the length of the array. The key point is to define dp[i] as the longest length of the increasing sequence ending with element arr[i].

The pseudo code is:

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

The time complexity is O(N^2), which is not the optimal solution.

The optimal solution is using the so-called patient algorithm. The basic idea to maintain some decreasing sequences. And for each new element arr[i], we need to search for the position of it. The rule is to find the first sequences with an ending elements larger or equal to arr[i].

If the ending element is larger than arr[i], then we just place arr[i] after it as the new ending element;

If the ending element is equal to arr[i], nothing needs to do;

If arr[i] is larger than all the ending elements, then we need to create a new sequence with arr[i] as the first element (which is the ending element for the new sequence as well).

In this way, the ending elements is monotonically increasing. And the number of the sequences is the length of the longest increasing sub-sequence.

So we can use binary search to find the position to place the new element, or create a new sequence.

Now let us focus on the reason why this greedy method work.

Maybe it is relative easier to understand this with an example. Let say we have one array: [9, 10, 1, 2, 4, 11, 23, 20, 3].

The first element is 9, so we need to create a sequence [[9]];

The next element is 10, so we need to create a second one [[9], [10]].

When i = 2, the sequences becomes [[9, 1], [10]];

                3,                                       [[9, 1], [10, 2]]

                4,                                        [[9, 1], [10, 2], [4]]

                5,                                        [[9, 1], [10, 2], [4], [11]]

                6,                                        [[9, 1], [10, 2], [4], [11], [23]] 

                7,                                        [[9, 1], [10, 2], [4], [11], [23, 20]]

                8,                                        [[9, 1], [10, 2], [4,3], [11], [23, 20]]

The number of the sequences is 5, which is correct (1, 2, 4, 11, 23, or 1, 2, 4, 11, 20).

Then why the element of 1 can be greedily placed after 9? The reason is that element 1 appears later than 9, and also smaller than 9. So if there is any element after element 1 larger than 9, it would larger than 1 as well! Thus, keep 1 as the ending element rather than 9 would not miss any potentially longer increasing subsequence. 

This argument applies to every new element to be placed into the sorted sub-sequences.

One observation is that: the ending elements are sorted, for example in the above case, it is [1, 2, 3, 11, 20]. But the ending-element-sequence is not necessary as an valid solution (the solution to the above case is [1, 2, 4, 11, 23] or [1, 2, 4, 11, 20]). But the length is the same!

A different way to understand this method is that: the first element in an valid solution must be in the first sequence; the second is in the second sequence; ..., the nth is in the nth sequence, even though the sequences of all the ending elements of each sequence is not necessarily an valid solution. That is why the length is always the same, which is what we need for this question.

Since we just need the ending elements, we can maintain one single array called tails, which are the collections of the ending elements from the above sequences. When a new element arr[i] arrive, we just need to find a position for arr[i], using the same rules as the above: find the first element that is larger or equal to arr[i]. If exist, use arr[i] to replace it; if not, add arr[i] in the back.

After handling all the elements, the length of the array is the answer.


See the code below:


class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        vector<int> tails;
        for(auto &a : nums) {
            auto it = lower_bound(tails.begin(), tails.end(), a);
            if(it == tails.end()) tails.push_back(a);
            else *it = a;
        }
        return tails.size();
    }
};


Follow up 1: how about the length of the longest non-decreasing sub-sequence? For example, [2, 2, 2, 2]. The above method is for increasing sub-sequence, which generates 1. But if asks for non-decreasing, the answer should be 4.

We just need to change one place in the above code: from lower_bound to upper_bound.

See the code below:

class Solution {
public:
    int lengthOfLNDS(vector<int>& nums) {
        vector<int> tails;
        for(auto &a : nums) {
            auto it = upper_bound(tails.begin(), tails.end(), a);
            if(it == tails.end()) tails.push_back(a);
            else *it = a;
        }
        return tails.size();
    }
};


Follow up 2: Say we have one array, and we want to make the array sorted by decreasing some elements. What is the minimum number of elements to be deleted, to make the remained array sorted?

This question is essentially the same as the LIS question!




Upper Layer

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