96. Unique Binary Search Trees && 95. Unique Binary Search Trees II && 241. Different Ways to Add Parentheses && 282. Expression Add Operators

96. Unique Binary Search Trees

Given n, how many structurally unique BST's (binary search trees) that store values 1...n?

For example,
Given n = 3, there are a total of 5 unique BST's.

1         3     3      2      1
\       /     /      / \      \
3     2     1      1   3      2
/     /       \                 \
2     1         2                 3

Tree Dynamic Programming

public class Solution {
public int numTrees(int n) {
//numTrees(n) = Sum(numTrees(n-1-i) * numTrees(n-1-i)), where i is [0,n-1]
if(n==0)
return 1;
int[] nums = new int[n+1];
nums[0] = 1; nums[1] = 1;

for(int i = 2; i<n+1; ++i)
{
int totalChildren = i - 1;
for(int left = 0; left<=totalChildren;++left)
{
nums[i]+=nums[left]*nums[totalChildren-left];
}
}
return nums
;

/*
public int numTrees(int n) {
if(n==0)
return 1;
if(n <= 2)
return n;

int total = 0;
for(int i = 1; i<=n;++i)
{
int left = numTrees(i-1);
int right = numTrees(n-i);
total += left * right;
}
return total;
}
*/
}
}

95. Unique Binary Search Trees II

Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1...n.

Given n = 3, your program should return all 5 unique BST's shown below.

1         3     3      2      1
\       /     /      / \      \
3     2     1      1   3      2
/     /       \                 \
2     1         2                 3

Tree Dynamic Programming

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<TreeNode> generateTrees(int n) {
return generateTrees(1,n);
}

private List<TreeNode> generateTrees(int from, int to) {
List<TreeNode> results = new ArrayList<TreeNode>();
if(from > to)
return results;
if(from == to){
results.add(new TreeNode(from));
return results;
}

for(int i = from; i<=to; ++i){
List<TreeNode> left = generateTrees(from, i-1);
List<TreeNode> right = generateTrees(i+1, to);
if(left.size() == 0) {
for(TreeNode r:right){
TreeNode root = new TreeNode(i);
root.right = r;
results.add(root);
}
}
else if(right.size() == 0) {
for(TreeNode l:left){
TreeNode root = new TreeNode(i);
root.left = l;
results.add(root);
}
}
else{
for(TreeNode l:left){
for(TreeNode r:right){
TreeNode root = new TreeNode(i);
root.left = l;
root.right = r;
results.add(root);
}
}
}
}
return results;
}
}

241. Different Ways to Add Parentheses

Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are

+

,

-

and

*

.

Example 1
Input:

"2-1-1"

.

((2-1)-1) = 0
(2-(1-1)) = 2

Output:

[0, 2]

Example 2
Input:

"2*3-4*5"

(2*(3-(4*5))) = -34
((2*3)-(4*5)) = -14
((2*(3-4))*5) = -10
(2*((3-4)*5)) = -10
(((2*3)-4)*5) = 10

Output:

[-34, -14, -10, -10, 10]

Divide and Conquer

class Solution {
public List<Integer> diffWaysToCompute(String input) {
List<Integer> ret = new LinkedList<>();
for (int i = 0; i < input.length(); ++i) {
char c = input.charAt(i);
if (c == '-' || c == '*' || c == '+') {
for (int left : diffWaysToCompute(input.substring(0, i))) {
for (int right : diffWaysToCompute(input.substring(i + 1))) {
if (c == '+')
ret.add(left + right);
else if (c == '-')
ret.add(left - right);
else if (c == '*')
ret.add(left * right);
}
}
}
}
if (ret.size() == 0)
ret.add(Integer.valueOf(input));
return ret;
}
}

282. Expression Add Operators

Given a string that contains only digits

0-9

and a target value, return all possibilities to add binary operators (not unary)

+

,

-

, or

*

between the digits so they evaluate to the target value.

Examples:

"123", 6 -> ["1+2+3", "1*2*3"]
"232", 8 -> ["2*3+2", "2+3*2"]
"105", 5 -> ["1*0+5","10-5"]
"00", 0 -> ["0+0", "0-0", "0*0"]
"3456237490", 9191 -> []

Divide and Conquer

public class Solution {
public List<String> addOperators(String num, int target) {
List<String> ret = new ArrayList<>();
if (num == null || num.length() == 0)
return ret;
helper(ret, "", 0, num, target, 0, 0);
return ret;
}

/**
* Iterate through all possibilities to add operators to str,
* and add to results if target is achieved.
*/
private void helper(List<String> ret, String currentString, long currentResult, String str, int target, int startInd, long previousNum) {
if (startInd == str.length()) {
if (target == currentResult)
ret.add(currentString);
return;
}
for (int endInd = startInd; endInd < str.length(); ++endInd) {
//0 cannot be a starting number, unless it is 0 by itself
if (str.charAt(startInd) == '0' && endInd > startInd)
//e.g. "105", "05" might be treated as 5
//so it will errorly return ["1*0+5","1*5","10-5"]
break;
Long num = Long.parseLong(str.substring(startInd, endInd + 1));
if (startInd == 0)
helper(ret, num.toString(), num, str, target, endInd + 1, num);
else {
helper(ret, currentString + "+" + num, currentResult + num, str, target, endInd + 1, num);
helper(ret, currentString + "-" + num, currentResult - num, str, target, endInd + 1, -num);
//if previous number is used as a multiplier, we need to remove it from earlier sum
long product = previousNum * num;
helper(ret, currentString + "*" + num, currentResult - previousNum + product, str, target, endInd + 1, product);
}
}
}
}