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Posts tagged as “pair”

花花酱 LeetCode 2563. Count the Number of Fair Pairs

Given a 0-indexed integer array nums of size n and two integers lower and upper, return the number of fair pairs.

A pair (i, j) is fair if:

  • 0 <= i < j < n, and
  • lower <= nums[i] + nums[j] <= upper

Example 1:

Input: nums = [0,1,7,4,4,5], lower = 3, upper = 6
Output: 6
Explanation: There are 6 fair pairs: (0,3), (0,4), (0,5), (1,3), (1,4), and (1,5).

Example 2:

Input: nums = [1,7,9,2,5], lower = 11, upper = 11
Output: 1
Explanation: There is a single fair pair: (2,3).

Constraints:

  • 1 <= nums.length <= 105
  • nums.length == n
  • -109 <= nums[i] <= 109
  • -109 <= lower <= upper <= 109

Solution: Two Pointers

Sort the array, use two pointers to find how # of pairs (i, j) s.t. nums[i] + nums[j] <= limit.
Ans = count(upper) – count(lower – 1)

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

C++

花花酱 LeetCode 1913. Maximum Product Difference Between Two Pairs

The product difference between two pairs (a, b) and (c, d) is defined as (a * b) - (c * d).

  • For example, the product difference between (5, 6) and (2, 7) is (5 * 6) - (2 * 7) = 16.

Given an integer array nums, choose four distinct indices wxy, and z such that the product difference between pairs (nums[w], nums[x]) and (nums[y], nums[z]) is maximized.

Return the maximum such product difference.

Example 1:

Input: nums = [5,6,2,7,4]
Output: 34
Explanation: We can choose indices 1 and 3 for the first pair (6, 7) and indices 2 and 4 for the second pair (2, 4).
The product difference is (6 * 7) - (2 * 4) = 34.

Example 2:

Input: nums = [4,2,5,9,7,4,8]
Output: 64
Explanation: We can choose indices 3 and 6 for the first pair (9, 8) and indices 1 and 5 for the second pair (2, 4).
The product difference is (9 * 8) - (2 * 4) = 64.

Constraints:

  • 4 <= nums.length <= 104
  • 1 <= nums[i] <= 104

Solution: Greedy

Since all the numbers are positive, we just need to find the largest two numbers as the first pair and smallest two numbers are the second pair.

Time complexity: O(nlogn) / sorting, O(n) / finding min/max elements.
Space complexity: O(1)

C++

Python3

花花酱 LeetCode 1877. Minimize Maximum Pair Sum in Array

The pair sum of a pair (a,b) is equal to a + b. The maximum pair sum is the largest pair sum in a list of pairs.

  • For example, if we have pairs (1,5)(2,3), and (4,4), the maximum pair sum would be max(1+5, 2+3, 4+4) = max(6, 5, 8) = 8.

Given an array nums of even length n, pair up the elements of nums into n / 2 pairs such that:

  • Each element of nums is in exactly one pair, and
  • The maximum pair sum is minimized.

Return the minimized maximum pair sum after optimally pairing up the elements.

Example 1:

Input: nums = [3,5,2,3]
Output: 7
Explanation: The elements can be paired up into pairs (3,3) and (5,2).
The maximum pair sum is max(3+3, 5+2) = max(6, 7) = 7.

Example 2:

Input: nums = [3,5,4,2,4,6]
Output: 8
Explanation: The elements can be paired up into pairs (3,5), (4,4), and (6,2).
The maximum pair sum is max(3+5, 4+4, 6+2) = max(8, 8, 8) = 8.

Constraints:

  • n == nums.length
  • 2 <= n <= 105
  • n is even.
  • 1 <= nums[i] <= 105

Solution: Greedy

Sort the elements, pair nums[i] with nums[n – i – 1] and find the max pair.

Time complexity: O(nlogn) -> O(n) counting sort.
Space complexity: O(1)

C++

花花酱 LeetCode 981. Time Based Key-Value Store

Create a timebased key-value store class TimeMap, that supports two operations.

1. set(string key, string value, int timestamp)

  • Stores the key and value, along with the given timestamp.

2. get(string key, int timestamp)

  • Returns a value such that set(key, value, timestamp_prev) was called previously, with timestamp_prev <= timestamp.
  • If there are multiple such values, it returns the one with the largest timestamp_prev.
  • If there are no values, it returns the empty string ("").

Example 1:

Input: inputs = ["TimeMap","set","get","get","set","get","get"], inputs = [[],["foo","bar",1],["foo",1],["foo",3],["foo","bar2",4],["foo",4],["foo",5]]
Output: [null,null,"bar","bar",null,"bar2","bar2"]
Explanation:   
TimeMap kv;   
kv.set("foo", "bar", 1); // store the key "foo" and value "bar" along with timestamp = 1   
kv.get("foo", 1);  // output "bar"   
kv.get("foo", 3); // output "bar" since there is no value corresponding to foo at timestamp 3 and timestamp 2, then the only value is at timestamp 1 ie "bar"   
kv.set("foo", "bar2", 4);   
kv.get("foo", 4); // output "bar2"   
kv.get("foo", 5); //output "bar2"   

Example 2:

Input: inputs = ["TimeMap","set","set","get","get","get","get","get"], inputs = [[],["love","high",10],["love","low",20],["love",5],["love",10],["love",15],["love",20],["love",25]]
Output: [null,null,null,"","high","high","low","low"]

Note:

  1. All key/value strings are lowercase.
  2. All key/value strings have length in the range [1, 100]
  3. The timestamps for all TimeMap.set operations are strictly increasing.
  4. 1 <= timestamp <= 10^7
  5. TimeMap.set and TimeMap.get functions will be called a total of 120000 times (combined) per test case.

Solution: HashTable + Map

C++

花花酱 LeetCode 956. Tallest Billboard

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

如果直接暴力搜索的话时间复杂度是O(3^N),铁定超时。对于每一根我们可以选择1、放到左边,2、放到右边,3、不使用。最后再看一下左边和右边是否相同。

题目的数据规模中的这句话非常重要:

The sum of rods is at most 5000.

这句话就是告诉你算法的时间复杂度和sum of rods有关系,通常需要使用DP。

由于每根柱子只能使用一次(让我们想到了 回复 01背包),但是我们怎么去描述放到左边还是放到右边呢?

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

和暴力搜索比起来DP已经对状态进行了压缩,因为我并不需要关心h1, h2是通过哪些(在我之前的)柱子构成了,我只关心它们的当前高度。

然后我可以选择

1、不用第i根柱子

2、放到低的那一堆

3、放到高的那一堆

状态转移的伪代码:

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

假设 rods=[1,1,2]

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)}

但是dp[i]这个集合的大小可能达到sum^2,所以还是会超时…

时间复杂度 O(n*sum^2)

空间复杂度 O(n*sum^2) 可降维至 O(sum^2)

革命尚未成功,同志仍需努力!

all pairs的cost太大,我们还需要继续压缩状态!

重点来了

通过观察发现,若有2个pairs:

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

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

如果 min(h1, h2) < min(h3, h4) 那么(h1, h2) 不可能产生最优解,直接舍弃。

因为如果后面的柱子可以构成 h4 – h3/h2 – h1 填补高度差,使得两根柱子一样高,那么答案就是 h2 和 h4。但h2 < h4,所以最优解只能来自后者。

举个例子:我有 (1, 3) 和 (2, 4) 两个pairs,它们的高度差都是2,假设我还有一个长度为2的柱子,那么我可以构成(1+2, 3) 以及 (2+2, 4),虽然这两个都是解。但是后者的高度要大于前者,所以前者无法构成最优解,也就没必要存下来。

所以,我们可以把状态压缩到高度差对于相同的高度差,我只存h1最大的

我们用 dp[i][j] 来表示使用前i个柱子,高度差为j的情况下最大的公共高度h1是多少。

状态转移(如下图)

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))

时间复杂度 O(nsum)

空间复杂度 O(nsum) 可降维至 O(sum)

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.

answer: dp[0]

Time complexity: O(n*Sum)

Space complexity: O(Sum)

C++ hashmap

C++ / array

C++/2D array