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Server Side Programming Articles
Page 1366 of 2109
Video Stitching in C++
Suppose we have a series of video clips from a sporting event that lasted T seconds. Now these video clips can be overlapping with each other and have varied lengths. Here each video clip clips[i] is an interval − it starts at clips[i][0] time and ends at clips[i][1] time. We can cut these clips into segments freely − We have to find the minimum number of clips needed so that we can cut the clips into segments that cover the entire sporting event ([0, T]). If the task is impossible, return -1. So if the input is like [[0, 2], ...
Read MoreRobot Bounded In Circle C++
Suppose we have an infinite plane, a robot initially stands at position (0, 0) and faces north. The robot can receive one of three instructions −G − go straight 1 unit;L − turn 90 degrees to the left direction;R − turn 90 degrees to the right direction.The robot performs the instructions given in order, Instructions are repeated forever. We have to check whether there exists a circle in the plane such that the robot never leaves the circle. So if the input is like [GGLLGG], then the answer will be true. from (0, 0) to (0, 2), it will loop ...
Read MoreUncrossed Lines in C++
Suppose we have written the integers of A and B (in the order they are given) on two separate horizontal lines. Now, we may draw connecting lines: a straight line connecting two numbers A[i] and B[j] such that −A[i] == B[j];The line we draw that does not intersect any other connecting (non-horizontal) line.We have to keep in mind that connecting lines cannot intersect even at the endpoints − each number can only belong to one connecting line. Find the maximum number of connecting lines. So if the input is like [1, 4, 2] and [1, 2, 4], then the output ...
Read MoreLast Stone Weight II in C++
Suppose we have a collection of rocks, now each rock has a positive integer weight. In each turn, we choose any two rocks and smash them together. If the stones have weights x and y with x = 0, decrease j by 1dp[j] := false when (dp[j] and dp[j – stones[i]]) both are false, otherwise trueif dp[j] is true, then reach := max of reach and jreturn total – (2 * reach)Let us see the following implementation to get better understanding −Example#include using namespace std; class Solution { public: int lastStoneWeightII(vector& stones) { int ...
Read MoreLongest String Chain in C++
Suppose we have a list of words, here each word consists of lowercase letters. So one word word1 is a predecessor of another word word2 if and only if we can add exactly one letter anywhere in word1 to make it equal to word2. For the example of the predecessor is like, "abc" is a predecessor of "abac". Now a word chain is a sequence of words [word_1, word_2, ..., word_k] with k >= 1, where word_1 is a predecessor of word_2, word_2 is a predecessor of word_3, and so on. We have to find the longest possible length of ...
Read MoreLongest Repeating Substring in C++
Suppose we have a string S, we have to find the length of the longest repeating substring(s). We will return 0 if no repeating substring is present. So if the string is like “abbaba”, then the output will be 2. As the longest repeating substring is “ab” or “ba”.Return all words that can be formed in this manner, in lexicographical order.To solve this, we will follow these steps −n := size of Sset S := one blank space concatenated with Sset ret := 0create one matrix dp of size (n + 1) x (n + 1)for i in range 1 ...
Read MoreShortest Way to Form String in C++
Suppose we have a string, we can form a subsequence of that string by deleting some number of characters (possibly no deletions). So if there is two strings source and target, we have to find the minimum number of subsequences of source such that their concatenation equals target. If the task is impossible, then return -1. So if source is “abc” and target is “abcbc”, then the output will be 2.To solve this, we will follow these steps −Define a string called possible, this will take s and t as inputcreate a map mfor each character c in s mark ...
Read MoreMaximum of Absolute Value Expression in C++
Suppose we have two arrays of integers with equal lengths, we have to find the maximum value of: |arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - j|. Where the maximum value is taken over all 0
Read MoreMinimize Rounding Error to Meet Target in C++
Suppose we have an array of prices P [p1, p2..., pn] and a target value, we have to round each price Pi to Roundi(Pi) so that the rounded array [Round1(P1), Round2(P2)..., Roundn(Pn)] sums to the given target value. Here each operation Roundi(pi) could be either Floor(Pi) or Ceil(Pi).We have to return the string "-1" if the rounded array is impossible to sum to target. Otherwise, return the smallest rounding error, which will be (as a string with three places after the decimal) defined as −$\displaystyle\sum\limits_{i-1}^n |Round_{i} (???? ) - ????$So if the input is like [“0.700”, “2.800”, “4.900”], and the ...
Read MoreMerge operations using STL in C++ | merge(), includes(), set_union(), set_intersection(), set_difference(), inplace_merge
In this tutorial, we will be discussing a program to understand the various merge operations using STL in C++.The merge() function is used to merge two sorted containers in a way that the new container is also sorted. Further includes() is used to check if the elements from first container are present in the second one.Example#include #include #include using namespace std; int main(){ vector v1 = {1, 3, 4, 5, 20, 30}; vector v2 = {1, 5, 6, 7, 25, 30}; //initializing resultant vector vector v3(12); merge(v1.begin(), v1.end(), v2.begin(), v2.end(), v3.begin()); cout
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