feat: add python solution to lc problem: No.2902 (doocs#1854)

No.2902.Count of Sub-Multisets With Bounded Sum

Author: nrhitik

solution/2900-2999/2902.Count of Sub-Multisets With Bounded Sum/README.md

@@ -68,6 +68,27 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
+class Solution:
+    def countSubMultisets(self, nums: List[int], l: int, r: int) -> int:
+        kMod = 1_000_000_007
+        # dp[i] := # of submultisets of nums with sum i
+        dp = [1] + [0] * r
+        count = collections.Counter(nums)
+        zeros = count.pop(0, 0)
+
+        for num, freq in count.items():
+            # stride[i] := dp[i] + dp[i - num] + dp[i - 2 * num] + ...
+            stride = dp.copy()
+            for i in range(num, r + 1):
+                stride[i] += stride[i - num]
+            for i in range(r, 0, -1):
+                if i >= num * (freq + 1):
+                    # dp[i] + dp[i - num] + dp[i - freq * num]
+                    dp[i] = stride[i] - stride[i - num * (freq + 1)]
+                else:
+                    dp[i] = stride[i]
+
+        return (zeros + 1) * sum(dp[l : r + 1]) % kMod
 
 ```
 

solution/2900-2999/2902.Count of Sub-Multisets With Bounded Sum/README_EN.md

@@ -60,6 +60,27 @@
 ### **Python3**
 
 ```python
+class Solution:
+    def countSubMultisets(self, nums: List[int], l: int, r: int) -> int:
+        kMod = 1_000_000_007
+        # dp[i] := # of submultisets of nums with sum i
+        dp = [1] + [0] * r
+        count = collections.Counter(nums)
+        zeros = count.pop(0, 0)
+
+        for num, freq in count.items():
+            # stride[i] := dp[i] + dp[i - num] + dp[i - 2 * num] + ...
+            stride = dp.copy()
+            for i in range(num, r + 1):
+                stride[i] += stride[i - num]
+            for i in range(r, 0, -1):
+                if i >= num * (freq + 1):
+                    # dp[i] + dp[i - num] + dp[i - freq * num]
+                    dp[i] = stride[i] - stride[i - num * (freq + 1)]
+                else:
+                    dp[i] = stride[i]
+
+        return (zeros + 1) * sum(dp[l : r + 1]) % kMod
 
 ```
 

solution/2900-2999/2902.Count of Sub-Multisets With Bounded Sum/Solution.py

@@ -0,0 +1,21 @@
+class Solution:
+    def countSubMultisets(self, nums: List[int], l: int, r: int) -> int:
+        kMod = 1_000_000_007
+        # dp[i] := # of submultisets of nums with sum i
+        dp = [1] + [0] * r
+        count = collections.Counter(nums)
+        zeros = count.pop(0, 0)
+
+        for num, freq in count.items():
+            # stride[i] := dp[i] + dp[i - num] + dp[i - 2 * num] + ...
+            stride = dp.copy()
+            for i in range(num, r + 1):
+                stride[i] += stride[i - num]
+            for i in range(r, 0, -1):
+                if i >= num * (freq + 1):
+                    # dp[i] + dp[i - num] + dp[i - freq * num]
+                    dp[i] = stride[i] - stride[i - num * (freq + 1)]
+                else:
+                    dp[i] = stride[i]
+
+        return (zeros + 1) * sum(dp[l : r + 1]) % kMod