Balance Array

Author: anantkaushik

Python/Interviewbit/Arrays/balance-array.py

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+"""
+Problem Link: https://www.interviewbit.com/problems/balance-array/
+
+Problem Description
+Given an integer array A of size N. You need to count the number of special elements in the given array.
+A element is special if removal of that element make the array balanced.
+Array will be balanced if sum of even index element equal to sum of odd index element.
+
+Problem Constraints
+1 <= N <= 105
+1 <= A[i] <= 109
+
+Input Format
+First and only argument is an integer array A of size N.
+
+Output Format
+Return an integer denoting the count of special elements.
+
+Example Input
+Input 1:
+A = [2, 1, 6, 4]
+Input 2:
+A = [5, 5, 2, 5, 8]
+
+Example Output
+Output 1:
+1
+
+Output 2:
+2
+
+Example Explanation
+Explanation 1:
+After deleting 1 from array : {2,6,4}
+(2+4) = (6)
+Hence 1 is the only special element, so count is 1
+
+Explanation 2:
+If we delete A[0] or A[1] , array will be balanced
+(5+5) = (2+8)
+So A[0] and A[1] are special elements, so count is 2.
+"""
+class Solution:
+    # @param A : list of integers
+    # @return an integer
+    def solve(self, A):
+        right_even = right_odd = 0
+        for i in range(len(A)):
+            if i % 2 == 0:
+                right_odd += A[i]
+            else:
+                right_even += A[i]
+        
+        res = left_odd = left_even = 0
+        for i in range(len(A)):
+            if i % 2 == 0:
+                right_odd -= A[i]
+            else:
+                right_even -= A[i]
+                
+            if right_odd + left_even == left_odd + right_even:
+                res += 1
+                
+            if i % 2 == 0:
+                left_odd += A[i]
+            else:
+                left_even += A[i]
+
+        return res