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Python Articles
Page 162 of 852
First Bad Version in Python
Suppose in a company, one product manager is leading a team who develops a new product. Suppose latest version fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version will be bad. So we have an array A with n elements [1, 2, … n] and we have to find the first bad version from this array.Consider we have a function isBadVersion(version_id), this will return whether the version is bad or not. For an example, suppose n = 5, and version = 4 is first bad version. So if ...
Read MoreIntersection of Two Arrays II in Python
Suppose we have two arrays A and B, there are few elements in these array. We have to find the intersection of them. So if A = [1, 4, 5, 3, 6], and B = [2, 3, 5, 7, 9], then intersection will be [3, 5]To solve this, we will follow these steps −Take two arrays A and Bif length of A is smaller than length of B, then swap themcalculate the frequency of elements in the array and store them into mfor each element e in B, if e is present in m, and frequency is non-zero, decrease frequency ...
Read MoreDiameter of Binary Tree in Python
Suppose we have a binary tree; we have to compute the length of the diameter of the tree. The diameter of a binary tree is actually the length of the longest path between any two nodes in a tree. This path not necessarily pass through the root. So if the tree is like below, then the diameter will be 3.as the length of the path [4, 2, 1, 3] or [5, 2, 1, 3] is 3To solve this, we will follow these steps −We will use the dfs to find the diameter, set answer := 0call the dfs function with ...
Read MoreJewels and Stones in Python
Suppose we have a string J that indicates some letters that are considered as Jewel, and another string S, that represents some stones that we have. Our task is to find how many of stones in S is also jewel. The letters in J and S are case sensitive. So if the J = “aZc”, and S = “catTableZebraPicnic” then there are 7 jewels.To solve this we will convert the string into a list of characters. If the character in J present in S, then increase the count.ExampleLet us see the following implementation to get better understanding −class Solution(object): ...
Read MoreSum of Even Numbers After Queries in Python
Suppose we have an array of integers called A, and an array queries. For the i-th query value = queries[i][0] and index = queries[i][1], we will add value to A[index]. Then, the answer of the i-th query is the sum of the even values of A. We have to find the answer to all queries. We will find an array, that should have answer[i] as the answer to the i-th query. So if the array is like [1, 2, 3, 4], and the query array is like [[1, 0], [-3, 1], [-4, 0], [2, 3]], then the answer array will ...
Read MorePairs of Songs With Total Durations Divisible by 60 in Python
Suppose we have a list of songs, the i-th song has a duration of time[i] seconds. We have to find the number of pairs of songs for which their total time in seconds is divisible by 60.So if the time array is like [30, 20, 150, 100, 40], then the answer will be 3. Three pairs will be (3, 150), (20, 100), (20, 40) for all cases the total duration is divisible by 60.To solve this, we will follow these steps −Take a map rem to store remainders. Set ans := 0for all elements i in time −if i is ...
Read MorePartition Array Into Three Parts With Equal Sum in Python
Suppose we have an array A of integers, our output will be true if and only if we can partition the array into three non-empty parts whose sum is equal.Formally, we can partition the array if we can find the indexes i+1 < j with (A[0] + A[1] + ... + A[i] is same as A[i+1] + A[i+2] + ... + A[j-1] and A[j] + A[j-1] + ... + A[A.length - 1])So if the input is [0, 2, 1, -6, 6, -7, 9, 1, 2, 0, 1], then the output will be true. Three arrays will be [0, 2, 1], ...
Read MoreLast Stone Weight in Python
Suppose we have some rocks, each rock has a positive integer weight. In each turn, we will take two heaviest rocks and smash them together. consider the stones have weights x and y and x 1: stones.sort() s1,s2=stones[-1],stones[-2] if s1==s2: stones.pop(-1) stones.pop(-1) else: s1 = abs(s1-s2) stones.pop(-1) stones[-1] = s1 if len(stones): return stones[-1] return 0 ob1 = Solution() print(ob1.lastStoneWeight([2,7,4,1,6,1]))Input[2,7,4,1,6,1]Output1
Read MoreOccurrences After Bigram in Python
Suppose there are words given. These are first and second, consider occurrences in some text of the form "first second third", here second comes immediately after the first, and third comes immediately after the second.For each such cases, add "third" into the answer, and show the answer. So if the text is like “lina is a good girl she is a good singer”, first = “a”, second = “good”, the answer will be [girl, singer]To solve this, we will follow these steps −text := split the string by spacesres is an empty listfor i := 0 to size of text ...
Read MoreDistribute Candies to People in Python
Suppose we want to distribute some number of candies to a row of n people in the following way −We then give 1 candy to the first people, 2 candies to the second people, and so on until we give n candies to the last people.After that, we go back to the start of the row again, give n + 1 candies to the first people, n + 2 candies to the second people, and so on until we give 2 * n candies to the last people.We will repeat this process until we run out of candies. The last ...
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