numberic pattern 232 and 233 solved

Author: Vipinchaudhary31122002

Numeric Patterns/numericpattern232.py

@@ -0,0 +1,12 @@
+for i in range(1,6):
+    for j in range(1,6):
+        if i==int(5/2)+1 or j==int(5/2)+1:
+            if abs(i-j)==2:
+                print(1,end=' ')
+            elif abs(i-j)==1:
+                print(2,end=' ')
+            else:
+                print(3,end=' ')
+        else:
+            print(" ",end=' ')
+    print(end='\n')

Numeric Patterns/numericpattern233.py

@@ -0,0 +1,7 @@
+for i in range(1,6):
+    for j in range(1,6):
+        if i==int(5/2)+1 or j==int(5/2)+1:
+            print(abs(i-j),end=' ')
+        else:
+            print(" ",end=' ')
+    print(end='\n')