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Suppose we have a rooted binary tree, we have to return the lowest common ancestor of its deepest leaves. We have to keep in mind that −The node of a binary tree is a leaf node if and only if it has no childrenThe depth of the root of the tree is 0, and when the depth of a node is d, the depth of each of its children is d+1.The lowest common ancestor of a set S of nodes in the node A with the largest depth such that every node in S is in the subtree with root ... Read More
Suppose we have the root of a binary tree; we have to find the maximum average value of any subtree of that tree. So if the tree is like −The output will be 6, this is because, for node 5, it will be (5 + 6 + 1)/ 3 = 4, then for node 6 it will be 6 / 1 = 6, and for node 1, it will be 1 / 1 = 1, so the max is 6.To solve this, we will follow these steps −res := 0Define a method called solve(), this will take rootif root is ... Read More
Suppose we have a binary tree, we have to find the sum of values of its deepest leaves. So if the tree is like −Then the output will be 15.To solve this, we will follow these steps −Define a map m, and maxDepthDefine a recursive method solve(), this will take node and level, initially level is 0if node is not present, then returnmaxDepth := max of level and maxDepthincrease m[level] by value of nodesolve(left of node, level + 1)solve(right of node, level + 1)In the main method, setup maxDepth := 0, then solve(root, 0)return m[maxDepth]Example(C++)Let us see the following implementation ... Read More
Suppose we have a string s, we have to find the maximum number of occurrences of any substring that satisfies the following rules −The number of distinct characters in the substring must be less than or equal to maxLetters.The substring size must be in range minSize and maxSize inclusive.So if the input is like − “aababcaab”, maxLetters = 2, minSize = 3 and maxSize = 4, then the output will be 2. The substring "aab" has 2 occurrences in the original string. This satisfies the conditions, 2 unique letters and size 3 (between minSize and maxSize).To solve this, we will ... Read More
Suppose we have to create an iterator that iterates through a run-length encoded sequence. Here the iterator is initialized by calling RLEIterator(int[] A), where A is a run-length encoding of a sequence. So we can say that for all even i, A[i] tells us the number of times that the non-negative integer value A[i+1] is repeated in the sequence. Here iterator supports one function −next(int n): This function exhausts the next n elements (n >= 1) and returns the last element exhausted in this way. So if there is no element left to exhaust, next returns -1 instead.Suppose we start ... Read More
Suppose in an infinite binary tree where every node has two children, the nodes are labelled in row order. Now in the odd numbered rows (the first, third, fifth, ...), the labelling is left to right, and in the even numbered rows (second, fourth, sixth, ...), the labelling is right to left. So the tree will be like −So we have given the label of a node in this tree, we have to find the labels in the path from the root of the tree to the node with that label. So if the input is label = 14, then ... Read More
Suppose we have a m x n matrix mat and an integer threshold. We have to the maximum side-length of a square with a sum less than or equal to the given threshold or return 0 if there is no such square. So if the input is like −113243211324321132432113243211324321132432And threshold is 4, then output will be 2, as there are two squares of side length 2, so max is 2To solve this, we will follow these steps −Define a method called ok, this will take x and matrix m and threshold thset curr := 0for i in range x – ... Read More
Suppose we have an array A of non-negative integers. For every (contiguous) subarray say B = [A[i], A[i+1], ..., A[j]] (with i
Suppose in a social group, there are N different people, with unique integer ids from 0 to N-1. Here we have a list of logs, where each logs[i] = [time, id_A, id_B] contains a non-negative integer timestamp, and the ids of two different people. Each log is showing the time in which two different people became friends. If A is friends with B, then B is friends with A. Let's say that person A is acquainted with person B if A is friends with B, or A is a friend of someone acquainted with B. We have to find the ... Read More
Suppose we have a set of N people (they are numbered 1, 2, ..., N), we would like to split everyone into two subgroups of any size. Now each person may dislike some other people, and they should not go into the same group. So, if dislikes[i] = [a, b], it indicates that it is not allowed to put the people numbered a and b into the same group. We have to find if it is possible to split everyone into two groups in this way.So if the input is like N = 4 and dislike = [[1, 2], [1, ... Read More