Dynamic Programming
The fundamental of DP is ‘merorized search’. It’s easy to implement but bad for memory. And it’s generally useless in the industry.
When to use DP?
- Input cannot sort
- Find minimum/maximum result
- Check the feasibility
- Count all possible solutions
If question asks you to find all possible solutions, it’s gonna be DFS, not DP.
5 Types of DP
- Matrix DP (10%)
- Sequence/Two Sequences DP (80%)
- Interval DP (5%)
- Tree DP (5%)
- States Compressing DP (0%)
Type 3 to 5 are less important.
Question List
Type 1: Matrix DP
Type 2.1: Sequence Dp
Jump Game II - tricky, index handling
Decode Ways - tricky, initial state
Type 2.2: Two Sequences Dp
Distinct Subsequences - difficult, state transition formula
Type 3: Interval Dp
Type 4: Tree Dp
Type 5: States Compressing DP
Ignore.
Additional questions
Code
Type 1: Matrix DP
Triangle
public int minimumTotal(List<List<Integer>> triangle) {
if (triangle == null || triangle.size() == 0) {
return 0;
}
int len = triangle.size();
int[][] dp = new int[len][len];
for (int i = len - 1; i >= 0; i--) {
// bottom-up approach (row by row)
for (int j = 0; j <= i; j++) {
if (i == len - 1) {
dp[i][j] = triangle.get(i).get(j);
} else {
dp[i][j] = triangle.get(i).get(j)
+ Math.min(dp[i+1][j], dp[i+1][j+1]);
}
}
}
return dp[0][0];
}
Unique Paths
public int uniquePaths(int m, int n) {
if (m == 0 && n == 0) {
return 0;
}
int[][] dp = new int[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0) {
dp[i][j] = 1;
} else {
dp[i][j] = dp[i-1][j] + dp[i][j-1];
}
}
}
return dp[m-1][n-1];
}
Unique Paths II
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
if (obstacleGrid == null) {
return 0;
}
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
int[][] dp = new int[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (obstacleGrid[i][j] != 0) {
dp[i][j] = 0;
} else if (i == 0 && j == 0) {
dp[i][j] = 1;
} else if (i == 0) {
dp[i][j] = dp[i][j-1];
} else if (j == 0) {
dp[i][j] = dp[i-1][j];
} else {
dp[i][j] = dp[i-1][j] + dp[i][j-1];
}
}
}
return dp[m-1][n-1];
}
Minimum Path Sum
public int minPathSum(int[][] grid) {
if (grid == null) {
return 0;
}
int m = grid.length;
int n = grid[0].length;
int[][] dp = new int[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 && j == 0) {
dp[i][j] = grid[i][j];
} else if (i == 0) {
dp[i][j] = dp[i][j-1] + grid[i][j];
} else if (j == 0) {
dp[i][j] = dp[i-1][j] + grid[i][j];
} else {
dp[i][j] = grid[i][j] + Math.min(dp[i-1][j], dp[i][j-1]);
}
}
}
return dp[m-1][n-1];
}
Type 2.1: Sequence Dp
Climbing Stairs
public int climbStairs(int n) {
if (n <= 2) {
return n;
}
int[] dp = new int[n];
dp[0] = 1;
dp[1] = 2;
for (int i = 2; i < n; i++) {
dp[i] = dp[i - 2] + dp[i - 1];
}
return dp[n - 1];
}
Jump Game
public boolean canJump(int[] A) {
if (A == null || A.length == 0)
return false;
int reach = 0;
int len = A.length;
for (int i = 0; i < len; i++) {
if (i > reach)
break;
reach = Math.max(reach, i + A[i]);
if (reach >= len - 1)
return true;
}
return false;
}
Jump Game II
public int jump(int[] A) {
if (A == null || A.length <= 1)
return 0;
int[] dp = new int[A.length];
int cur = 1;
for (int i = 0; i < A.length - 1; i++) {
while (cur <= i + A[i] && cur < dp.length) {
dp[cur] = dp[i] + 1;
cur++;
}
if (cur == dp.length) {
break;
}
}
return dp[A.length - 1];
}
Palindrome Partitioning II
public int minCut(String s) {
if (s == null || s.length() <= 1) {
return 0;
}
boolean[][] map = palindromeMap(s);
int len = s.length();
int[] dp = new int[len + 1];
dp[0] = -1;
for (int i = 1; i <= len; i++) {
dp[i] = Integer.MAX_VALUE;
for (int j = 1; j <= i; j++) {
if (map[i-1][j-1]) {
dp[i] = Math.min(dp[i], dp[j-1] + 1);
}
}
}
return dp[len];
}
private boolean[][] palindromeMap(String s) {
int len = s.length();
boolean[][] map = new boolean[len][len];
char[] ss = s.toCharArray();
for (int i = 0; i < len; i++) {
for (int j = i; j >= 0; j--) {
if (i == j) {
map[i][j] = true;
} else if (i - j == 1) {
map[i][j] = ss[i] == ss[j];
} else {
map[i][j] = (ss[i] == ss[j]) & map[i - 1][j + 1];
}
}
}
return map;
}
Word Break
public boolean wordBreak(String s, Set<String> dict) {
if (s == null || s.length() == 0) {
return true;
}
int len = s.length();
boolean[] dp = new boolean[len + 1];
dp[0] = true;
for (int i = 1; i <= len; i++) {
for (int j = 0; j < i; j++) {
if (!dp[j]) {
continue;
}
if (dict.contains(s.substring(j, i))) {
dp[i] = true;
break;
}
}
}
return dp[len];
}
Word Break II
My code is definitely correct, although it got TLE.
See original post for more.
Decode Ways
public int numDecodings(String s) {
if (s == null || s.length() == 0) {
return 0;
}
int len = s.length();
int[] dp = new int[len + 1];
dp[0] = 1;
dp[1] = 1; // pay attention to the initial state
if (!isValidNumber(s.substring(0, 1))) {
return 0;
}
for (int i = 2; i <= len; i++) {
if (isValidNumber(s.substring(i - 1, i))) {
dp[i] += dp[i - 1];
}
if (isValidNumber(s.substring(i - 2, i))) {
dp[i] += dp[i - 2];
}
}
return dp[len];
}
private boolean isValidNumber(String input) {
if (input.length() == 0 || input.length() > 2 || input.charAt(0) == '0') {
return false;
}
int num = Integer.parseInt(input);
return (1 <= num && num <= 26);
}
Longest Increasing Subsequence
There’s a new post in ‘Lintcode’ category.
Type 2.2: Two Sequences Dp
Distinct Subsequences
public int numDistinct(String S, String T) {
int m = S.length(), n = T.length();
int[][] dp = new int[m + 1][n + 1];
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
if (j == 0) {
dp[i][j] = 1;
} else if (i == 0) {
dp[i][j] = 0;
} else {
dp[i][j] = dp[i - 1][j];
if (S.charAt(i - 1) == T.charAt(j - 1)) {
dp[i][j] += dp[i - 1][j - 1];
}
}
}
}
return dp[m][n];
}
Edit Distance
public int minDistance(String A, String B) {
if (A == null || B == null)
return 0;
int m = A.length(), n = B.length();
int[][] dp = new int[m + 1][n + 1];
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
if (i == 0) {
dp[i][j] = j;
} else if (j == 0) {
dp[i][j] = i;
} else {
if (A.charAt(i - 1) == B.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = Math.min(dp[i - 1][j - 1], dp[i - 1][j]);
dp[i][j] = Math.min(dp[i][j], dp[i][j - 1]);
dp[i][j]++;
}
}
}
}
return dp[m][n];
}
Interleaving String
public boolean isInterleave(String s1, String s2, String s3) {
if (s1 == null || s2 == null) {
return false;
}
int m = s1.length(), n = s2.length();
if (m + n != s3.length()) {
return false;
}
boolean[][] dp = new boolean[m + 1][n + 1];
for (int i = 0; i <= m; i++) {
for (int j = 0; j <= n; j++) {
if (i == 0) {
dp[i][j] = s2.substring(0, j).equals(s3.substring(0, j));
continue;
} else if (j == 0) {
dp[i][j] = s1.substring(0, i).equals(s3.substring(0, i));
continue;
}
if (i > 0 && dp[i - 1][j]
&& s1.charAt(i - 1) == s3.charAt(i + j - 1)) {
dp[i][j] = true;
}
if (j > 0 && dp[i][j - 1]
&& s2.charAt(j - 1) == s3.charAt(i + j - 1)) {
dp[i][j] = true;
}
}
}
return dp[m][n];
}
Longest Common Subsequence
There’s a new post in ‘Lintcode’ category.
Type 3: Interval Dp
Merge Stone
有n堆石子排成一列，每堆石子有一个重量w[i], 每次合并可以合并相邻的两堆石子，一次合并的代价为两堆石子的重量和w[i]+w[i+1]。问安排怎样的合并顺序，能够使得总合并代价达到最小。
Solution explained:
sum[i[用于记录从第1堆到第i堆（包含i）石子的总重量。
dp[i][j]表示从第i堆（包含i）到第j堆（包含j）石子的合并的最小代价。
状态转移方程为：dp[i][j] = minimize{dp[i][k] + dp[k+1][j] + sum[j] - sum[i-1]}, k从i到j（不包含j）。
len=2表示第一次合并的情况，此时合并的石子为2堆。此时，i从1到n-len+1，j=i+len-1。
Quoted from this blog and the solution is well explained on wikiio.
Type 4: Tree Dp
Binary Tree Maximum Path Sum
This is a difficult question, but I solved it. I feel happy.
private class ResultType {
int maxPath;
int depth;
ResultType(int a, int b) {
this.maxPath = a;
this.depth = b;
}
}
public int maxPathSum(TreeNode root) {
if (root == null) {
return 0;
}
return helper(root).maxPath;
}
private ResultType helper(TreeNode node) {
if (node == null) {
return new ResultType(Integer.MIN_VALUE, 0);
}
ResultType ll = helper(node.left);
ResultType rr = helper(node.right);
int maxPath = node.val + ll.depth + rr.depth;
maxPath = Math.max(maxPath, ll.maxPath);
maxPath = Math.max(maxPath, rr.maxPath);
int depth = 0;
depth = Math.max(depth, node.val + Math.max(ll.depth, rr.depth));
return new ResultType(maxPath, depth);
}
Additional questions
Maximum Subarray
public int maxSubArray(int[] A) {
if (A == null || A.length == 0) {
return 0;
}
int len = A.length;
int[] dp = new int[len];
dp[0] = A[0];
for (int i = 1; i < len; i++) {
dp[i] = A[i];
if (dp[i - 1] > 0)
dp[i] += dp[i - 1];
}
int max = Integer.MIN_VALUE;
for (Integer i : dp)
max = Math.max(max, i);
return max;
}