From 2db8bfda73886fa98ec1e3875f58ec3767b6672f Mon Sep 17 00:00:00 2001
From: Krish Modi <krishmodi3030@gmail.com>
Date: Thu, 30 Oct 2025 15:38:14 +0530
Subject: [PATCH 1/2] Create number_of_longest_increasing_subsequence.py

Implement function to count number of longest increasing subsequence
---
 ...umber_of_longest_increasing_subsequence.py | 52 +++++++++++++++++++
 1 file changed, 52 insertions(+)
 create mode 100644 dynamic_programming/number_of_longest_increasing_subsequence.py

diff --git a/dynamic_programming/number_of_longest_increasing_subsequence.py b/dynamic_programming/number_of_longest_increasing_subsequence.py
new file mode 100644
index 000000000000..6ddd1827da39
--- /dev/null
+++ b/dynamic_programming/number_of_longest_increasing_subsequence.py
@@ -0,0 +1,52 @@
+'''
+Leetcode Problem:673. Number of Longest Increasing Subsequence
+Link: https://leetcode.com/problems/number-of-longest-increasing-subsequence/description/
+
+Given an integer array nums, return the number of longest increasing subsequences.
+Notice that the sequence has to be strictly increasing.
+
+Example 1:
+
+Input: nums = [1,3,5,4,7]
+Output: 2
+Explanation: The two longest increasing subsequences are [1, 3, 4, 7] and [1, 3, 5, 7].
+Example 2:
+
+Input: nums = [2,2,2,2,2]
+Output: 5
+Explanation: The length of the longest increasing subsequence is 1, and there are 5 increasing subsequences of length 1, so output 5.
+ 
+
+Constraints:
+
+1 <= nums.length <= 2000
+-10**6 <= nums[i] <= 10**6
+The answer is guaranteed to fit inside a 32-bit integer.
+'''
+
+
+def findNumberOfLIS(nums):
+    n = len(nums)
+    if n == 0:
+        return 0
+
+    length = [1] * n
+    count = [1] * n
+
+    for i in range(n):
+        for j in range(i):
+            if nums[j] < nums[i]:
+                if length[j] + 1 > length[i]:
+                    length[i] = length[j] + 1
+                    count[i] = count[j]
+                elif length[j] + 1 == length[i]:
+                    count[i] += count[j]
+
+    max_len = max(length)
+    return sum(c for l, c in zip(length, count) if l == max_len)
+
+
+# For testing...
+n=int(input())
+nums=list(map(int,input().split()))
+print(findNumberOfLIS(nums))

From 7cbae762c9330d1697455eeb7de4172456e4063e Mon Sep 17 00:00:00 2001
From: "pre-commit-ci[bot]"
 <66853113+pre-commit-ci[bot]@users.noreply.github.com>
Date: Thu, 30 Oct 2025 10:11:22 +0000
Subject: [PATCH 2/2] [pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci
---
 .../number_of_longest_increasing_subsequence.py        | 10 +++++-----
 1 file changed, 5 insertions(+), 5 deletions(-)

diff --git a/dynamic_programming/number_of_longest_increasing_subsequence.py b/dynamic_programming/number_of_longest_increasing_subsequence.py
index 6ddd1827da39..c9f018dbc3f1 100644
--- a/dynamic_programming/number_of_longest_increasing_subsequence.py
+++ b/dynamic_programming/number_of_longest_increasing_subsequence.py
@@ -1,4 +1,4 @@
-'''
+"""
 Leetcode Problem:673. Number of Longest Increasing Subsequence
 Link: https://leetcode.com/problems/number-of-longest-increasing-subsequence/description/
 
@@ -15,14 +15,14 @@
 Input: nums = [2,2,2,2,2]
 Output: 5
 Explanation: The length of the longest increasing subsequence is 1, and there are 5 increasing subsequences of length 1, so output 5.
- 
+
 
 Constraints:
 
 1 <= nums.length <= 2000
 -10**6 <= nums[i] <= 10**6
 The answer is guaranteed to fit inside a 32-bit integer.
-'''
+"""
 
 
 def findNumberOfLIS(nums):
@@ -47,6 +47,6 @@ def findNumberOfLIS(nums):
 
 
 # For testing...
-n=int(input())
-nums=list(map(int,input().split()))
+n = int(input())
+nums = list(map(int, input().split()))
 print(findNumberOfLIS(nums))