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Programming Articles
Page 2288 of 2547
Upper bound and Lower bound for non increasing vector in C++
In this article we are going to discuss the vector::upper_bound() and vector::lower_bound() for an array sorted in non-increasing order in C++ STL.Vectors are similar to the dynamic arrays; they have the ability to modify its size itself whenever a value is inserted into or removed from the container where we are storing the value.In a Vector, lower bound returns an iterator pointing to the first element in the range that does not compare the given value. Upper Bound returns an iterator pointing element in the range that smaller than given value.Input 30 30 30 20 20 20 10 10Output Lower bound of ...
Read MoreSplit the sentence into words in C++
Given is the task to split the sentence into words. In this, we will separate all the words present in sentence.Input I am a good boyOutput I am a good boyIn the above example we will print the single word in single line.Example#include #include #include Using namespace std; void split( string st){ String word = “ “; for ( char s : st){ If (s== ‘ ‘){ Cout
Read MoreMaximum and Minimum element of a linked list which is divisible by a given number k in C++
A linked list is a linear data structure in which elements are linked via pointers. Each element or node of a linked list has a data part and link, or we can say pointer to the next element in sequence. The elements can take noncontiguous locations in memory.We are given a singly linked list in which there is a data part and link to the next element. The other input is a number K. Task is to find the Maximum and Minimum element of a linked list which is divisible by the number K. The linear linked list can only ...
Read MoreMaximum adjacent difference in an array in its sorted form in C++
We are given with an array. The array need not be sorted. The task is to find the maximum difference between adjacent elements of that array in its sorted form. So the first thing is to sort the array in increasing or decreasing order. Then we will iterate the array and calculate the adjacent difference of Arr[i+1]-Arr[i]. Then for each iteration we will compare this difference with the one which is found maximum so far.Input − Arr[] = [ 1, 5, 10, 2, 7 ]Output − Maximum adjacent difference in array in its sorted form is 3.Explanation − Sorted Arr[] ...
Read MoreMaximize the sum of X+Y elements by picking X and Y elements from 1st and 2nd array in C++
For the given two arrays each of size N, the task is to find the maximum sum by choosing X elements from array 1 and Y elements from array 2.Let’s now understand what we have to do using an example −Input arr1 = {1, 2, 3, 4, 5} ; X=2 arr2 = {1, 3, 5, 2, 7}; Y=3Output Maximum sum here is : 24Explanation − We are selecting 2 number(s) from arr1 and 3 from arr2. Largest 2 of arr1 are 4, 5 and largest 3 of arr2 are 3, 5, 7. Total sum of these 5 elements is 24 which is ...
Read MoreMaximize the given number by replacing a segment of digits with the alternate digits given in C++
Given the task is to maximize a given number with ‘N’ number of digits by replacing its digit using another array that contains 10 digits as an alternative for all single-digit numbers 0 to 9, The given condition is that only a consecutive segment of numbers can be replaced and only once.Input N=1234, arr[]={3 ,0 ,1 ,5 ,7 ,7 ,8 ,2 ,9 ,4}Output 1257Explanation The number 3 can be replaced with its alternative 5= arr[3]The number 4 can be replaced with its alternative 7= arr[4]Input N=5183, arr[]={3 ,0 ,1 ,5 ,7 ,7 ,8 ,2 ,9 ,4}Output 7183Approach used in the below program as followsIn function ...
Read MoreMaximize the profit by selling at-most M products in C++
Given the task is to calculate the maximum profit that can be made by selling at-most ‘M’ products.The total number of products are ‘N’ and the cost price and the selling price of each product is given in the lists CP[] and SP[] respectively.Input N=6, M=4 CP[]={1, 9, 5, 8, 2, 11} SP[]={1, 15, 10, 16, 5, 20}Output 28Explanation − The profit obtained from selling all the products are 0, 6, 5, 8, 3, 9 respectively.So, in order to make maximum profit by selling only 4 products, the products with the highest profit need to be chosen, that is, product number 2, ...
Read MoreFunctions that cannot be overloaded in C++
Function overloading is also known as method overloading. Function overloading is the feature provided by the concept of polymorphism which is widely used in object-oriented programming.To achieve function overloading, functions should satisfy these conditions −Return type of functions should be sameName of the functions should be sameParameters can be different in type but should be same in numberExampleint display(int a); int display(float a); // both the functions can be overloaded int display(int a); float display(float b); //both the functions can’t be overloaded as the return type of one function is different from anotherlet’s discuss the functions that can’t overloaded in ...
Read MoreHow to merge two unequal data frames and replace missing values to 0 in R?
Often, we get data frames that are not of same size that means some of the rows or columns are missing in any of the data frames. Therefore, to merge these types of data frames we can merge them with all of their values and convert the missing values to zero if necessary. This can be done by using merge function and replacement of missing values NA to zeros should be done by using single square brackets.ExampleConsider the below data frames −> C1 df1 df1 C1 1 A 2 B 3 C 4 D 5 E 6 F 7 G ...
Read MoreHow to find the date after a number of days in R?
In our daily life, often we want to know what will be date after some number of days. This is also required in professional life, especially in those professions where we work on projects and have tight deadlines. To find the date after a number a certain number of days we can just use plus sign after reading the date with as.Date.Example> as.Date("2001-01-01")+30 [1] "2001-01-31" > as.Date("2020-06-30")+30 [1] "2020-07-30" > as.Date("2020-06-30")+50 [1] "2020-08-19" > as.Date("2020-06-30")+100 [1] "2020-10-08" > as.Date("2020-06-30")+120 [1] "2020-10-28" > as.Date("2020-06-30")+15 [1] "2020-07-15" > as.Date("2020-06-30")+45 [1] "2020-08-14" > as.Date("2020-06-30")+40 [1] "2020-08-09" > as.Date("2020-12-25")+20 [1] "2021-01-14" > as.Date("2020-12-25")+300 [1] ...
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