Adding Simple Solution for Problem - 1221 - Split String Balance

Author: Vanditg

Split_String_Balance/SimpleSolution.py

@@ -0,0 +1,76 @@
+##==================================
+## Leetcode
+## Student: Vandit Jyotindra Gajjar
+## Year: 2020
+## Problem: 1221
+## Problem Name: Split a String in Balanced Strings
+##===================================
+#
+#Balanced strings are those who have equal quantity of 'L' and 'R' characters.
+#
+#Given a balanced string s split it in the maximum amount of balanced strings.
+#
+#Return the maximum amount of splitted balanced strings.
+#
+#Example 1:
+#
+#Input: s = "RLRRLLRLRL"
+#Output: 4
+#Explanation: s can be split into "RL", "RRLL", "RL", "RL", each substring contains same number of 'L' and 'R'.
+#
+#Example 2:
+#
+#Input: s = "RLLLLRRRLR"
+#Output: 3
+#Explanation: s can be split into "RL", "LLLRRR", "LR", each substring contains same number of 'L' and 'R'.
+#
+#Example 3:
+#
+#Input: s = "LLLLRRRR"
+#Output: 1
+#Explanation: s can be split into "LLLLRRRR".
+#
+#Example 4:
+#
+#Input: s = "RLRRRLLRLL"
+#Output: 2
+#Explanation: s can be split into "RL", "RRRLLRLL", since each substring contains an equal number of 'L' and 'R'
+
+class Solution:
+    def balanceStringSplit(self, string):
+        counter = 0    #Intialize counter 
+        finalCounter = 0     #Intialize finalCounter which will return the max amount of splitted balanced strings. 
+        for s in string:    #Loop through the string for extracting a character. 
+            #print(s)
+            if s == 'R':    #Condition-check: If the character is R. 
+                counter += 1    #We pass the If Condition-check and increase the counter by 1. 
+            elif s == 'L':    #Condition-check: Elif the character is L. 
+                counter -= 1    #We pass the Elif Condition-check and decrease the counter by 1. 
+            
+			if counter == 0:    #Condition-check: If the counter will be zero.
+                finalCounter += 1     #We pass the If Condition-check and increase the finalCounter by 1 because now counter is zero that means we have all substring in matching R and L form. 
+        return finalCounter
+#
+#Example:
+#Input string = "LLLLRRRR"
+#counter = 0
+#finalCounter = 0
+#s = 'L'
+#If Condition-check failed enter elif so counter = -1
+#s = 'L'
+#If Condition-check failed enter elif so counter = -1
+#s = 'L'
+#If Condition-check failed enter elif so counter = -1
+#s = 'L'
+#If Condition-check failed enter elif so counter = -1
+#s = 'R'
+#If Condition-check failed enter elif so counter = +1
+#s = 'R'
+#If Condition-check failed enter elif so counter = +1
+#s = 'R'
+#If Condition-check failed enter elif so counter = +1
+#s = 'R'
+#If Condition-check failed enter elif so counter = +1
+#Now our finalCounter is -1 -1 -1 -1 +1 +1 +1 +1 = 0, so we increase this finalCounter by +1. 
+#As we have completed the string, we simply return finalCounter. 
+#So finalCounter = 1 for this input.