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Added Solution for 900-1000q/1064 900-1000q/1065 1000-1100q/1064 1000-1100q/1065 1000-1100q/1071
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1000-1100q/1071.py

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'''
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For strings S and T, we say "T divides S" if and only if S = T + ... + T (T concatenated with itself 1 or more times)
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Return the largest string X such that X divides str1 and X divides str2.
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Example 1:
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Input: str1 = "ABCABC", str2 = "ABC"
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Output: "ABC"
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Example 2:
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Input: str1 = "ABABAB", str2 = "ABAB"
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Output: "AB"
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Example 3:
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Input: str1 = "LEET", str2 = "CODE"
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Output: ""
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Note:
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1 <= str1.length <= 1000
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1 <= str2.length <= 1000
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str1[i] and str2[i] are English uppercase letters.
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'''
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class Solution(object):
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def gcdOfStrings(self, str1, str2):
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"""
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:type str1: str
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:type str2: str
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:rtype: str
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"""
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if len(str1) > len(str2):
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str1, str2 = str2, str1
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l_str1 = len(str1)
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for index in range(1, len(str1)+1):
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if l_str1%index != 0:
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continue
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size_to_take = int(l_str1/index)
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substr1 = str1[:size_to_take]
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substr2 = str2
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while substr1 == substr2[:size_to_take]:
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substr2 = substr2[size_to_take:]
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if substr2 == "":
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return substr1
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return ""

README.md

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##### [Problems 1000-1100](./1000-1100q/)
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| # | Title | Solution | Difficulty |
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|---| ----- | -------- | ---------- |
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|1071|[Greatest Common Divisor of Strings](https://leetcode.com/problems/greatest-common-divisor-of-strings)|[Python](./1000-1100q/1071.py)|Easy|
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|1065|[Index Pairs of a String](https://leetcode.com/problems/index-pairs-of-a-string)|[Python](./1000-1100q/1065.py)|Easy|
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|1064|[Fixed Point](https://leetcode.com/problems/fixed-point)|[Python](./1000-1100q/1064.py)|Easy|
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|1054|[Distant Barcodes](https://leetcode.com/problems/distant-barcodes)|[Python](./1000-1100q/1054.py)|Medium|

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