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+
+## Remarks
+## https://www.geeksforgeeks.org/string-with-additive-sequence-set-2/
+
+## python3 code to implement the approach
+ 
+## Variable to store the result
+bResult = False
+arrStr = []
+ 
+## Function to check the additive sequence
+def check_additive(strIn, subStrIn):
+    global bResult
+    global arrStr
+     
+    ## If the end is reached and vector consists of more than 2 numbers, then one of the possible solution is found
+    if (subStrIn == len(strIn) and len(arrStr) > 2):
+
+        ## Mark the bResult as true to indicate the solution is found and to avoid for trying the rest of the combinations
+        bResult = True
+        return
+
+    nInt_01, nInt_02, nInt_03 = 0, 0, 0
+
+    if (len(arrStr) >= 2):
+        ## Store the previous two numbers of the sequence to check the additive sequence property for the next number
+        nInt_02 = arrStr[len(arrStr) - 1]
+        nInt_01 = arrStr[len(arrStr) - 2]
+
+    for nIndxLoop in range(subStrIn, len(strIn)):
+ 
+        ## Generate the number
+        nInt_03 = nInt_03 * 10 + ord(strIn[nIndxLoop]) - ord("0")
+ 
+        ## Try all the possible ways to generate the first two numbers
+        ## i.e. if vector consists of less than two numbers and no solution is found yet
+        if (len(arrStr) < 2):
+            arrStr.append(nInt_03)
+            check_additive(strIn, nIndxLoop + 1)
+ 
+            ## Pop the value to try for the other combination
+            arrStr.pop()
+
+            ## If the number generated so far is not equal the sum of previous two numbers in the sequence
+            ## then it cannot be a part of additive sequence hence no need to proceed further
+
+            return
+
+        if (nInt_03 != nInt_01 + nInt_02):
+            return
+
+        ## If the number generated so far is equal to the sum of previous two numbers
+        ## then it can be a part of additive sequence
+        ## push it into vector and check for remaining string
+
+        ## Store it in the vector 
+        arrStr.append(nInt_03)
+
+        ## Recur for remaining string
+        check_additive(strIn, nIndxLoop + 1)
+ 
+        ## If unable to find solution
+        ## pop it and try for other combination
+        arrStr.pop()
+
+    return
+
+## Function to check if additive sequence
+def IsAdditiveSeq(strInputFunc):
+    global bResult
+     
+    ## In order to form additive sequence, the length of the string must be at least three
+    if (len(strInputFunc) <= 2):
+        return False
+ 
+    bResult = False
+    check_additive(strInputFunc, 0)
+    return bResult
+ 
+## Driver Code
+if __name__ == "__main__":
+
+    ## strInput = "112358"  ## Example 1
+    ## strInput = "12345"  ## Example 2
+    strInput = "199100199"  ## Example 3
+
+    print (IsAdditiveSeq(strInput))