[LintCode] Longest Increasing Continuous subsequence II
Give you an integer matrix (with row size n, column size m)，find the longest increasing continuous subsequence in this matrix. (The definition of the longest increasing continuous subsequence here can start at any row or column and go up/down/right/left any direction).
Given a matrix:
[
[1 ,2 ,3 ,4 ,5],
[16,17,24,23,6],
[15,18,25,22,7],
[14,19,20,21,8],
[13,12,11,10,9]
]
return 25
O(nm) time and memory.
Solution 1. Brute force search
多重循环DP遇到了困难: 从上到下循环不能解决问题, 初始状态找不到. 我们可以用暴力搜索从大往小的搜索.
暴力搜索的解法很简单， 两层for loop 遍历每一个 A[i][j], 搜索以 A[i][j] 作为结尾元素的最长子序列. 返回所有 A.length * A[0].length 个搜索结果的最大值.
1 public class Solution { 2 private int[] dx = {-1, 0, 1, 0}; 3 private int[] dy = {0, 1, 0, -1}; 4 public int longestIncreasingContinuousSubsequenceII(int[][] A) { 5 if(A == null || A.length == 0){ 6 return 0; 7 } 8 int max = 0; 9 for(int i = 0; i < A.length; i++){ 10 for(int j = 0; j < A[0].length; j++){ 11 max = Math.max(max, search(A, i, j)); 12 } 13 } 14 return max; 15 } 16 private int search(int[][] A, int x, int y){ 17 int max = 1; 18 for(int i = 0; i < 4; i++){ 19 int nx = x + dx[i]; 20 int ny = y + dy[i]; 21 if(nx >= 0 && nx < A.length && ny >= 0 && ny < A[0].length && A[x][y] > A[nx][ny]){ 22 max = Math.max(max, search(A, nx, ny) + 1); 23 } 24 } 25 return max; 26 } 27 }
Solution 2. Dynamic Programming(Search with Memoization)
Solution 1 does a lot of redundant work computing the same subproblem over and over again. This is very inefficient.
To optimize this to get a better running time, we apply the time vs space tradeoff. Scarifice O(n * m) memory usage to
get a O(n * m) run time.
State:
dp[x][y]: the length of the longest ICS that ends at A[x][y];
computed[x][y]: whether dp[x][y] has been computed or not;
Function:
dp[x][y] = dp[nx][ny] + 1 (if A[x][y] > A[nx][ny])
Initialization:
dp[x][y] = 1;
Answer: Max of dp[x][y] for x in [0.....n - 1] y in [0.....m - 1]
Run time / Space complexity
Space: O(n * m)
Run time: since each position of (x, y) is only calculated once, so run time is O(n * m)
1 public class Solution { 2 private int[] dx = {-1, 0, 1, 0}; 3 private int[] dy = {0, 1, 0, -1}; 4 private int[][] dp; 5 private boolean[][] computed; 6 public int longestIncreasingContinuousSubsequenceII(int[][] A) { 7 if(A == null || A.length == 0){ 8 return 0; 9 } 10 int n = A.length, m = A[0].length; 11 dp = new int[n][m]; 12 computed = new boolean[n][m]; 13 int max = 0; 14 for(int i = 0; i < n; i++){ 15 for(int j = 0; j < m; j++){ 16 dp[i][j] = 1; 17 } 18 } 19 for(int i = 0; i < n; i++){ 20 for(int j = 0; j < m; j++){ 21 max = Math.max(max, search(A, i, j)); 22 } 23 } 24 return max; 25 } 26 private int search(int[][] A, int x, int y){ 27 if(computed[x][y]){ 28 return dp[x][y]; 29 } 30 for(int i = 0; i < 4; i++){ 31 int nx = x + dx[i]; 32 int ny = y + dy[i]; 33 if(nx >= 0 && nx < A.length && ny >= 0 && ny < A[0].length && A[x][y] > A[nx][ny]){ 34 dp[x][y] = Math.max(dp[x][y], search(A, nx, ny) + 1); 35 } 36 } 37 computed[x][y] = true; 38 return dp[x][y]; 39 } 40 }
Follow up questions
Q: Can you optimize the memory usage like some other dynamic programming problems?
A: No, since to calculate dp[x][y], we need all its 4 neighbors' result, so we can't use less memory. Also, search memoization is
hard to optimize memory usage even if we can. This is the downside of using top down memoization search compared to
bottom up typical dynamic programming.
Q: Can you reconstruct one optimal solution?
A: Easy. Pick one largest dp[x][y] if there are more than one. Then starting from this dp[x][y] to backtrack its neighbors of value dp[x][y] - 1,
do this until we find a dp[i][j] = 1.
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