# Posts tagged as “difference”

You are given an array of equal-length strings words. Assume that the length of each string is n.

Each string words[i] can be converted into a difference integer array difference[i] of length n - 1 where difference[i][j] = words[i][j+1] - words[i][j] where 0 <= j <= n - 2. Note that the difference between two letters is the difference between their positions in the alphabet i.e. the position of 'a' is 0'b' is 1, and 'z' is 25.

• For example, for the string "acb", the difference integer array is [2 - 0, 1 - 2] = [2, -1].

All the strings in words have the same difference integer array, except one. You should find that string.

Return the string in words that has different difference integer array.

Example 1:

Input: words = ["adc","wzy","abc"]
Output: "abc"
Explanation:
- The difference integer array of "adc" is [3 - 0, 2 - 3] = [3, -1].
- The difference integer array of "wzy" is [25 - 22, 24 - 25]= [3, -1].
- The difference integer array of "abc" is [1 - 0, 2 - 1] = [1, 1].
The odd array out is [1, 1], so we return the corresponding string, "abc".


Example 2:

Input: words = ["aaa","bob","ccc","ddd"]
Output: "bob"
Explanation: All the integer arrays are [0, 0] except for "bob", which corresponds to [13, -13].


Constraints:

• 3 <= words.length <= 100
• n == words[i].length
• 2 <= n <= 20
• words[i] consists of lowercase English letters.

## Solution: Comparing with first string.

Let us pick words[0] as a reference for comparison, assuming it’s valid. If we only found one instance say words[i], that is different than words[0], we know that words[i] is bad, otherwise we should see m – 1 different words which means words[0] itself is bad.

Time complexity: O(m*n)
Space complexity: O(1)

# Problem

You are installing a billboard and want it to have the largest height.  The billboard will have two steel supports, one on each side.  Each steel support must be an equal height.

You have a collection of rods which can be welded together.  For example, if you have rods of lengths 1, 2, and 3, you can weld them together to make a support of length 6.

Return the largest possible height of your billboard installation.  If you cannot support the billboard, return 0.

Example 1:

Input: [1,2,3,6]
Output: 6
Explanation: We have two disjoint subsets {1,2,3} and {6}, which have the same sum = 6.


Example 2:

Input: [1,2,3,4,5,6]
Output: 10
Explanation: We have two disjoint subsets {2,3,5} and {4,6}, which have the same sum = 10.


Example 3:

Input: [1,2]
Output: 0
Explanation: The billboard cannot be supported, so we return 0.


Note:

1. 0 <= rods.length <= 20
2. 1 <= rods[i] <= 1000
3. The sum of rods is at most 5000.

# Solution: DP

The sum of rods is at most 5000.

Naive的方法是用 dp[i] 表示使用前i个柱子能够构成的柱子高度的集合。

e.g. dp[i] = {(h1, h2)},  h1 <= h2

1、不用第i根柱子

2、放到低的那一堆

3、放到高的那一堆

for h1, h2 in dp[i – 1]:

dp[i] += (h1, h2)        # not used

dp[i] += (h1, h2 + h)  # put on higher

if h1 + h < h2:

dp[i] += (h1 + h, h2)  # put on lower

else:

dp[i] += (h2, h1 + h)  # put on lower

dp[0] = {(0,0)}

dp[1] = {(0,0), (0,1)}

dp[2] = {(0,0), (0,1), (0,2), (1,1)}

dp[3] = {(0,0), (0,1), (0,2), (0,3), (0,4), (1,1), (1,2), (1,3), (2,2)}

all pairs的cost太大，我们还需要继续压缩状态！

(h1, h2), (h3, h4),

h1 <= h2, h3 <= h4, h1 < h3, h2 – h1 = h4 – h3 即 高度差 相同

dp[i][j] = max(dp[i][j], dp[i – 1][j])

dp[i][j+h] = max(dp[i][j + h], dp[i – 1][j])

dp[i][|j-h|] = max(dp[i][|j-h|], dp[i – 1][j] + min(j, h))

dp[i] := max common height of two piles of height difference i.

e.g. y1 = 5, y2 = 9 => dp[9 – 5] = min(5, 9) => dp[4] = 5.