solved Minimum flips in circular array

Author: AdarshGupta-dev

Minimum Flips in Circular Array.py

@@ -38,4 +38,35 @@
 
 Sample Explanation
 One of the solutions is to flip the digit at index 2 resulting in our array as 1 0 1 1 and 1 0(index 1 and 2) will be our required subarray
-"""
+"""
+
+# Solution with explanation: 
+
+
+def solve(arr, length, ones, zeros):
+    ansO = ones - sum(arr[0:ones])
+    ansZ = sum(arr[ones : ones + zeros])
+    ans = ansO + ansZ
+    print(ansO, ansZ, ans)
+    for i in range(1, length):
+        j = i + ones
+        k = j + zeros
+
+        if arr[j - 1] and not arr[i - 1]:
+            ansO = max(0, ansO - 1)
+        if not arr[j - 1] and arr[i - 1]:
+            ansO += 1
+
+        if arr[k - 1] and not arr[j - 1]:
+            ansZ += 1
+        if not arr[k - 1] and arr[j - 1]:
+            ansZ = max(0, ansZ - 1)
+        print(arr[i:k])
+        print(ansO, ansZ)
+        ans = min(ans, ansO + ansZ)
+    return ans
+
+
+N, O, Z = [int(x) for x in input().split(" ")]
+A = [int(x) for x in input().split(" ")]
+print(solve(A, N, O, Z))