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3432. Count Partitions with Even Sum Difference.py

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+# 3432. Count Partitions with Even Sum Difference
+# Solved
+# Easy
+# Topics
+# premium lock icon
+# Companies
+# Hint
+# You are given an integer array nums of length n.
+
+# A partition is defined as an index i where 0 <= i < n - 1, splitting the array into two non-empty subarrays such that:
+
+# Left subarray contains indices [0, i].
+# Right subarray contains indices [i + 1, n - 1].
+# Return the number of partitions where the difference between the sum of the left and right subarrays is even.
+
+ 
+
+# Example 1:
+
+# Input: nums = [10,10,3,7,6]
+
+# Output: 4
+
+# Explanation:
+
+# The 4 partitions are:
+
+# [10], [10, 3, 7, 6] with a sum difference of 10 - 26 = -16, which is even.
+# [10, 10], [3, 7, 6] with a sum difference of 20 - 16 = 4, which is even.
+# [10, 10, 3], [7, 6] with a sum difference of 23 - 13 = 10, which is even.
+# [10, 10, 3, 7], [6] with a sum difference of 30 - 6 = 24, which is even.
+# Example 2:
+
+# Input: nums = [1,2,2]
+
+# Output: 0
+
+# Explanation:
+
+# No partition results in an even sum difference.
+
+# Example 3:
+
+# Input: nums = [2,4,6,8]
+
+# Output: 3
+
+# Explanation:
+
+# All partitions result in an even sum difference.
+
+ 
+
+# Constraints:
+
+# 2 <= n == nums.length <= 100
+# 1 <= nums[i] <= 100
+
+class Solution:
+    def countPartitions(self, nums: List[int]) -> int:
+        x = 0
+        total = sum(nums)
+        prefix = 0
+        for i in range(len(nums)-1):
+            prefix += nums[i]
+            suffix = total - prefix
+            y = prefix - suffix
+            if y % 2 == 0:
+                x += 1
+        return x
+        
+class solution:
+    def countPartitions2(self,nums):
+        if sum(nums) % 2 : return 0 
+        return len(nums) - 1