Question

link

Analysis

This is one of the 2 most popular questions of DP. This is a sequences Dp. The equation isn’t difficult. Time complexity of DP solution is O(n²).

There’s also a binary search solution which the time complexity is O(nlgn). It’s very complex, and very hard to explain, but I’ll try:

Maintain an array called ‘array’. A[i] denotes the tail of sequence for LIS = i. Initially A[0] = first element of the input, then keep inserting elements into this array. It’s explained in this post.

I will give an example for the input: 0, 8, 4, 12, 2

Our strategy determined by the following conditions:

1. If A[i] is smallest among all end candidates of active lists, we will start new active list of length 1.

2. If A[i] is largest among all end candidates of active lists, we will clone the largest active list, and extend it by A[i].

3. If A[i] is in between, we will find a list with largest end element that is smaller than A[i]. Clone and extend this list by A[i]. We will discard all other lists of same length as that of this modified list.

A[0] = 0. Case 1. There are no active lists, create one.

array = 0

A[1] = 8. Case 2. Clone and extend.

array = 0, 8

A[2] = 4. Case 3. Clone, extend and discard.

array = 0, 4

A[3] = 12. Case 2. Clone and extend.

array = 0, 4, 12

A[4] = 2. Case 3. Clone, extend and discard.

array = 0, 2, 12

So the LIS is (0, 2, 12) of length 3.

Code

DP solution, by me

public int longestIncreasingSubsequence(int[] nums) {
    // write your code here
    if (nums == null || nums.length == 0) {
        return 0;
    }
    int len = nums.length;
    int[] dp = new int[len];
    dp[0] = 1;
    for (int i = 1; i < len; i++) {
        dp[i] = 1;
        for (int j = 0; j < i; j++) {
            if (nums[j] <= nums[i]) {
                dp[i] = Math.max(dp[i], dp[j] + 1);
            }
        }
    }
    int lis = 1;
    for (Integer seq : dp) {
        lis = Math.max(lis, seq);
    }
    return lis;
}

Binary search solution, C++ code from this post. I don’t think I am able code this solution.

#include <iostream>
#include <string.h>
#include <stdio.h>

using namespace std;

#define ARRAY_SIZE(A) sizeof(A)/sizeof(A[0])
// Binary search (note boundaries in the caller)
// A[] is ceilIndex in the caller
int CeilIndex(int A[], int l, int r, int key) {
    int m;

    while( r - l > 1 ) {
        m = l + (r - l)/2;
        (A[m] >= key ? r : l) = m; // ternary expression returns an l-value
    }

    return r;
}

int LongestIncreasingSubsequenceLength(int A[], int size) {
    // Add boundary case, when array size is one

    int *tailTable   = new int[size];
    int len; // always points empty slot

    memset(tailTable, 0, sizeof(tailTable[0])*size);

    tailTable[0] = A[0];
    len = 1;
    for( int i = 1; i < size; i++ ) {
        if( A[i] < tailTable[0] )
            // new smallest value
            tailTable[0] = A[i];
        else if( A[i] > tailTable[len-1] )
            // A[i] wants to extend largest subsequence
            tailTable[len++] = A[i];
        else
            // A[i] wants to be current end candidate of an existing subsequence
            // It will replace ceil value in tailTable
            tailTable[CeilIndex(tailTable, -1, len-1, A[i])] = A[i];
    }

    delete[] tailTable;

    return len;
}

int main() {
    int A[] = { 2, 5, 3, 7, 11, 8, 10, 13, 6 };
    int n = ARRAY_SIZE(A);

    printf("Length of Longest Increasing Subsequence is %d\n",
            LongestIncreasingSubsequenceLength(A, n));

    return 0;
}