Added Solution for 1000-1100q/1072 1000-1100q/1073 1000-1100q/1074 1000-1100q/1078 1000-1100q/1079 1000-1100q/1080 1000-1100q/1081 1000-1100q/1085 1000-1100q/1086 1000-1100q/1087 1000-1100q/1088 1000-1100q/1089

Author: Garvit244

1000-1100q/1072.py

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+'''
+Given a matrix consisting of 0s and 1s, we may choose any number of columns in the matrix and flip every cell in that column.  Flipping a cell changes the value of that cell from 0 to 1 or from 1 to 0.
+
+Return the maximum number of rows that have all values equal after some number of flips.
+
+ 
+
+Example 1:
+
+Input: [[0,1],[1,1]]
+Output: 1
+Explanation: After flipping no values, 1 row has all values equal.
+Example 2:
+
+Input: [[0,1],[1,0]]
+Output: 2
+Explanation: After flipping values in the first column, both rows have equal values.
+Example 3:
+
+Input: [[0,0,0],[0,0,1],[1,1,0]]
+Output: 2
+Explanation: After flipping values in the first two columns, the last two rows have equal values.
+ 
+
+Note:
+
+1 <= matrix.length <= 300
+1 <= matrix[i].length <= 300
+All matrix[i].length's are equal
+matrix[i][j] is 0 or 1
+'''

1000-1100q/1073.py

@@ -0,0 +1,24 @@
+'''
+Given two numbers arr1 and arr2 in base -2, return the result of adding them together.
+
+Each number is given in array format:  as an array of 0s and 1s, from most significant bit to least significant bit.  For example, arr = [1,1,0,1] represents the number (-2)^3 + (-2)^2 + (-2)^0 = -3.  A number arr in array format is also guaranteed to have no leading zeros: either arr == [0] or arr[0] == 1.
+
+Return the result of adding arr1 and arr2 in the same format: as an array of 0s and 1s with no leading zeros.
+
+ 
+
+Example 1:
+
+Input: arr1 = [1,1,1,1,1], arr2 = [1,0,1]
+Output: [1,0,0,0,0]
+Explanation: arr1 represents 11, arr2 represents 5, the output represents 16.
+ 
+
+Note:
+
+1 <= arr1.length <= 1000
+1 <= arr2.length <= 1000
+arr1 and arr2 have no leading zeros
+arr1[i] is 0 or 1
+arr2[i] is 0 or 1
+'''

1000-1100q/1074.py

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+'''
+Given a matrix, and a target, return the number of non-empty submatrices that sum to target.
+
+A submatrix x1, y1, x2, y2 is the set of all cells matrix[x][y] with x1 <= x <= x2 and y1 <= y <= y2.
+
+Two submatrices (x1, y1, x2, y2) and (x1', y1', x2', y2') are different if they have some coordinate that is different: for example, if x1 != x1'.
+
+ 
+
+Example 1:
+
+Input: matrix = [[0,1,0],[1,1,1],[0,1,0]], target = 0
+Output: 4
+Explanation: The four 1x1 submatrices that only contain 0.
+Example 2:
+
+Input: matrix = [[1,-1],[-1,1]], target = 0
+Output: 5
+Explanation: The two 1x2 submatrices, plus the two 2x1 submatrices, plus the 2x2 submatrix.
+ 
+
+Note:
+
+1 <= matrix.length <= 300
+1 <= matrix[0].length <= 300
+-1000 <= matrix[i] <= 1000
+-10^8 <= target <= 10^8
+'''

1000-1100q/1078.py

@@ -0,0 +1,44 @@
+'''
+Given words first and second, consider occurrences in some text of the form "first second third", where second comes immediately after first, and third comes immediately after second.
+
+For each such occurrence, add "third" to the answer, and return the answer.
+
+ 
+
+Example 1:
+
+Input: text = "alice is a good girl she is a good student", first = "a", second = "good"
+Output: ["girl","student"]
+Example 2:
+
+Input: text = "we will we will rock you", first = "we", second = "will"
+Output: ["we","rock"]
+ 
+
+Note:
+
+1 <= text.length <= 1000
+text consists of space separated words, where each word consists of lowercase English letters.
+1 <= first.length, second.length <= 10
+first and second consist of lowercase English letters.
+'''
+
+class Solution(object):
+    def findOcurrences(self, text, first, second):
+        """
+        :type text: str
+        :type first: str
+        :type second: str
+        :rtype: List[str]
+        """
+        result = []
+        if not text:
+            return []
+        splitted_text = text.split(' ')
+        indi = 0
+        for index in range(len(splitted_text)-1):
+            if splitted_text[index] == first and splitted_text[index+1] == second:
+                index = index+2
+                if index < len(splitted_text):
+                    result.append(splitted_text[index])
+        return result

1000-1100q/1079.py

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+'''
+You have a set of tiles, where each tile has one letter tiles[i] printed on it.  Return the number of possible non-empty sequences of letters you can make.
+
+ 
+
+Example 1:
+
+Input: "AAB"
+Output: 8
+Explanation: The possible sequences are "A", "B", "AA", "AB", "BA", "AAB", "ABA", "BAA".
+Example 2:
+
+Input: "AAABBC"
+Output: 188
+'''
+
+class Solution(object):
+    def numTilePossibilities(self, tiles):
+        """
+        :type tiles: str
+        :rtype: int
+        """
+        
+        if not tiles:
+            return 0
+        
+        import collections
+        unique = set(tiles)
+        freq_map =  collections.Counter(tiles)
+        total_len = 1
+        while total_len < len(tiles):
+            new = set()
+            for char in tiles:
+                for comb in unique:
+                    new_seq = comb+char
+                    up_freq = collections.Counter(new_seq)
+                    flag =True
+                    for key, val in up_freq.items():
+                        if val > freq_map[key]:
+                            flag = False
+                    if flag:
+                        new.add(new_seq)
+            # print new
+            unique.update(new)
+                    
+            total_len += 1
+        return len(unique)

1000-1100q/1080.py

@@ -0,0 +1,68 @@
+'''
+Given the root of a binary tree, consider all root to leaf paths: paths from the root to any leaf.  (A leaf is a node with no children.)
+
+A node is insufficient if every such root to leaf path intersecting this node has sum strictly less than limit.
+
+Delete all insufficient nodes simultaneously, and return the root of the resulting binary tree.
+
+ 
+
+Example 1:
+
+
+Input: root = [1,2,3,4,-99,-99,7,8,9,-99,-99,12,13,-99,14], limit = 1
+
+Output: [1,2,3,4,null,null,7,8,9,null,14]
+Example 2:
+
+
+Input: root = [5,4,8,11,null,17,4,7,1,null,null,5,3], limit = 22
+
+Output: [5,4,8,11,null,17,4,7,null,null,null,5]
+ 
+
+Example 3:
+
+
+Input: root = [1,2,-3,-5,null,4,null], limit = -1
+
+Output: [1,null,-3,4]
+ 
+
+Note:
+
+The given tree will have between 1 and 5000 nodes.
+-10^5 <= node.val <= 10^5
+-10^9 <= limit <= 10^9
+'''
+
+# Definition for a binary tree node.
+# class TreeNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution(object):
+    def sufficientSubset(self, root, limit):
+        """
+        :type root: TreeNode
+        :type limit: int
+        :rtype: TreeNode
+        """
+        def reduce_tree(root, limit, curr_sum):
+            if not root:
+                return None
+            
+            l_sum = [curr_sum[0] + root.val]
+            r_sum = [l_sum[0]]
+            
+            root.left = reduce_tree(root.left, limit, l_sum)
+            root.right = reduce_tree(root.right, limit, r_sum)
+            
+            curr_sum[0] = max(l_sum[0], r_sum[0])
+            if curr_sum[0] < limit:
+                root = None
+            return root
+        curr_sum = [0]
+        return reduce_tree(root, limit, curr_sum)

1000-1100q/1081.py

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+'''
+Return the lexicographically smallest subsequence of text that contains all the distinct characters of text exactly once.
+
+ 
+Example 1:
+
+Input: "cdadabcc"
+Output: "adbc"
+Example 2:
+
+Input: "abcd"
+Output: "abcd"
+Example 3:
+
+Input: "ecbacba"
+Output: "eacb"
+Example 4:
+
+Input: "leetcode"
+Output: "letcod"
+ 
+
+Note:
+
+1 <= text.length <= 1000
+text consists of lowercase English letters.
+'''
+class Solution(object):
+    def smallestSubsequence(self, text):
+        """
+        :type text: str
+        :rtype: str
+        """
+        if not text:
+            return ''
+        import collections
+        freq_map = collections.Counter(text)
+        used = [False]*26
+        result = ''
+        
+        for char in text:
+            freq_map[char] -= 1
+            if used[ord(char)-97]:
+                continue
+            while (result and result[-1] > char and freq_map[result[-1]] > 0):
+                used[ord(result[-1])-97] = False
+                result = result[:-1]
+                
+            used[ord(char)-97] = True
+            result += char
+        return result

1000-1100q/1085.py

@@ -0,0 +1,44 @@
+'''
+Given an array A of positive integers, let S be the sum of the digits of the minimal element of A.
+
+Return 0 if S is odd, otherwise return 1.
+
+ 
+
+Example 1:
+
+Input: [34,23,1,24,75,33,54,8]
+Output: 0
+Explanation: 
+The minimal element is 1, and the sum of those digits is S = 1 which is odd, so the answer is 0.
+Example 2:
+
+Input: [99,77,33,66,55]
+Output: 1
+Explanation: 
+The minimal element is 33, and the sum of those digits is S = 3 + 3 = 6 which is even, so the answer is 1.
+ 
+
+Note:
+
+1 <= A.length <= 100
+1 <= A[i].length <= 100
+'''
+class Solution(object):
+    def sumOfDigits(self, A):
+        """
+        :type A: List[int]
+        :rtype: int
+        """
+        if not A:
+            return 0
+        
+        mini = min(A)
+        result = 0
+        while mini > 0:
+            quo = mini%10
+            rem = mini/10
+            result += quo
+            mini = rem
+            
+        return 0 if result%2 else 1