Added Solution for 900-1000q/1064 900-1000q/1065

Author: Garvit244

900-1000q/1064.py

@@ -0,0 +1,37 @@
+'''
+Given an array A of distinct integers sorted in ascending order, return the smallest index i that satisfies A[i] == i.  Return -1 if no such i exists.
+
+ 
+
+Example 1:
+
+Input: [-10,-5,0,3,7]
+Output: 3
+Explanation: 
+For the given array, A[0] = -10, A[1] = -5, A[2] = 0, A[3] = 3, thus the output is 3.
+Example 2:
+
+Input: [0,2,5,8,17]
+Output: 0
+Explanation: 
+A[0] = 0, thus the output is 0.
+Example 3:
+
+Input: [-10,-5,3,4,7,9]
+Output: -1
+Explanation: 
+There is no such i that A[i] = i, thus the output is -1.
+'''
+
+class Solution(object):
+    def fixedPoint(self, A):
+        """
+        :type A: List[int]
+        :rtype: int
+        """
+        if not A:
+            return -1
+        for index, num in enumerate(A):
+            if num == index:
+                return index
+        return -1

900-1000q/1065.py

@@ -0,0 +1,45 @@
+'''
+Given a text string and words (a list of strings), return all index pairs [i, j] so that the substring text[i]...text[j] is in the list of words.
+
+ 
+
+Example 1:
+
+Input: text = "thestoryofleetcodeandme", words = ["story","fleet","leetcode"]
+Output: [[3,7],[9,13],[10,17]]
+Example 2:
+
+Input: text = "ababa", words = ["aba","ab"]
+Output: [[0,1],[0,2],[2,3],[2,4]]
+Explanation: 
+Notice that matches can overlap, see "aba" is found in [0,2] and [2,4].
+ 
+
+Note:
+
+All strings contains only lowercase English letters.
+It's guaranteed that all strings in words are different.
+1 <= text.length <= 100
+1 <= words.length <= 20
+1 <= words[i].length <= 50
+Return the pairs [i,j] in sorted order (i.e. sort them by their first coordinate in case of ties sort them by their second coordinate).
+'''
+
+class Solution(object):
+    def indexPairs(self, text, words):
+        """
+        :type text: str
+        :type words: List[str]
+        :rtype: List[List[int]]
+        """
+        if not words:
+            return []
+        result = []
+        for word in words:
+            starting = [index for index in range(len(text)) if text.startswith(word, index)]
+            for start in starting:
+                result.append([start, start+len(word)-1])
+            # print starting
+        result.sort()
+        return result
+            

README.md

@@ -15,6 +15,8 @@ Python solution of problems from [LeetCode](https://leetcode.com/).
 ##### [Problems 1000-1100](./1000-1100q/)
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|1065|[Index Pairs of a String](https://leetcode.com/problems/index-pairs-of-a-string)|[Python](./1000-1100q/1065.py)|Easy|
+|1064|[Fixed Point](https://leetcode.com/problems/fixed-point)|[Python](./1000-1100q/1064.py)|Easy|
 |1054|[Distant Barcodes](https://leetcode.com/problems/distant-barcodes)|[Python](./1000-1100q/1054.py)|Medium|	
 |1053|[Previous Permutation With One Swap](https://leetcode.com/problems/previous-permutation-with-one-swap)|[Python](./1000-1100q/1053.py)|Medium|
 |1052|[Grumpy Bookstore Owner](https://leetcode.com/problems/grumpy-bookstore-owner)|[Python](./1000-1100q/1052.py)|Medium|