Problem Description
You are given an array of n
pairs pairs
where pairs[i] = [lefti, righti]
and lefti < righti
.
A pair p2 = [c, d]
follows a pair p1 = [a, b]
if b < c
. A chain of pairs can be formed in this fashion.
Return the length longest chain which can be formed.
You do not need to use up all the given intervals. You can select pairs in any order.
Example 1:
Input: pairs = [[1,2],[2,3],[3,4]] Output: 2 Explanation: The longest chain is [1,2] -> [3,4].
Example 2:
Input: pairs = [[1,2],[7,8],[4,5]] Output: 3 Explanation: The longest chain is [1,2] -> [4,5] -> [7,8].
Constraints:
n == pairs.length
1 <= n <= 1000
-1000 <= lefti < righti <= 1000
Difficulty: Medium
Tags: array, dynamic programming, greedy, sorting
Rating: 97.23%
Solution
Here’s my Python solution to this problem:
class Solution:
def findLongestChain(self, pairs: List[List[int]]) -> int:
pairs.sort(key=lambda x:x[1])
print(pairs)
cur_end = -float('inf')
res = 0
for l, r in pairs:
if l > cur_end:
cur_end = r
res += 1
return res
Example Walkthrough
Let’s walk through an example to understand the solution better.
pairs = [[1,2],[7,8],[4,5]]
- Sort the pairs based on the second element of each pair:
pairs = [[1,2],[4,5],[7,8]]
- Initialize the current end of the chain and the result:
cur_end = -inf
res = 0
- Iterate through the pairs:
- For the first pair
[1,2]
, the left element is greater than the current end. So, we update the current end to2
and increment the result by1
.
cur_end = 2
res = 1
- For the second pair
[4,5]
, the left element is greater than the current end. So, we update the current end to5
and increment the result by1
.
cur_end = 5
res = 2
- For the third pair
[7,8]
, the left element is greater than the current end. So, we update the current end to8
and increment the result by1
.
cur_end = 8
res = 3
- Return the result
3
.
return 3
Complexity Analysis
The solution has the following complexity characteristics:
- Time Complexity:
- Space Complexity:
Note: This is an automated analysis and may not capture all edge cases or specific algorithmic optimizations.