Article Categories
- All Categories
-
Data Structure
-
Networking
-
RDBMS
-
Operating System
-
Java
-
MS Excel
-
iOS
-
HTML
-
CSS
-
Android
-
Python
-
C Programming
-
C++
-
C#
-
MongoDB
-
MySQL
-
Javascript
-
PHP
-
Economics & Finance
C++ Articles
Page 226 of 597
Program for K Most Recently Used (MRU) Apps in C++
Given a number k and an array arr[n], containing n number of integer elements which are storing the id’s of the opened apps in a system; the task is to show k number of most recently used apps, like when we press alt+tab shows all the recent apps and most recent before the least recent. The position of every id represents different apps in a system −They are as follows −Id in arr[0] is the id of an app which is currently in use.Id in arr[1] is the id of an app which was most recently used.Id in arr[n-1] is ...
Read MoreMaximize array sum after K negation in C++
Problem statementGiven an array of size n and a number k. We have to modify an array k number of times.Modify array means in each operation we can replace any array element arr[i] by negating it i.e. arr[i] = -arr[i]. The task is to perform this operation in such a way that after k operations, the sum of an array must be maximum.If input arr[] = {7, -3, 5, 4, -1} then maximum sum will be 20First negate -3. Now array becomes {7, 3, 5, 4, -1}Negate -1. Now array becomes {7, 3, 5, 4, 1}Algorithm1. Replace the minimum element ...
Read MoreProgram for power of a complex number in O(log n) in C++
Given a complex number in the form of x+yi and an integer n; the task is calculate and print the value of the complex number if we power the complex number by n.What is a complex number?A complex number is number which can be written in the form of a + bi, where a and b are the real numbers and i is the solution of the equation or we can say an imaginary number. So, simply putting it we can say that complex number is a combination of Real number and imaginary number.Raising power of a complex numberTo raise ...
Read MoreMaximize elements using another array in C++
Problem statementGiven two arrays with size n, maximize the first array by using the elements from the second array such that the new array formed contains n greatest but unique elements of both the arrays giving the second array priority i.e. all elements of the second array appear before the first array. The order of appearance of elements should be kept the same in output as in inputIf arr1[] = {12, 15, 10} and arr2[] = {16, 17, 5} then {16, 17, 15} are the maximum elements from both array by maintaining order.Algorithm1. Create temporary array of size 2 * ...
Read MoreMaximize number of 0s by flipping a subarray in C++
Problem statementGiven a binary array, find the maximum number of zeros in an array with one flip of a subarray allowed. A flip operation switches all 0s to 1s and 1s to 0sIf arr1= {1, 1, 0, 0, 0, 0, 0}If we flip first 2 1’s to 0’s, then we can get subarray of size 7 as follows −{0, 0, 0, 0, 0, 0, 0}Algorithm1. Consider all subarrays and find a subarray with maximum value of (count of 1s) – (count of 0s) 2. Considers this value be maxDiff. Finally return count of zeros in original array + maxDiff.Example#include ...
Read MoreMaximize the bitwise OR of an array in C++
Problem statementGiven an array of N integers. The bitwise OR of all the elements of the array has to be maximized by performing one task. The task is to multiply any element of the array at-most k times with a given integer xIf input array is {4, 3, 6, 1}, k = 2 and x = 3 then maximum value can be obtained is 55Algorithm1. multiply an array element with (x^k) and do bitwise OR it with the bitwise OR of all previous elements 2. Multiply an array element with bitwise OR of all next elements 3. Return the maximum ...
Read MoreMaximize the maximum subarray sum after removing at most one element in C++
Problem statementGiven an array arr[] of N integers. The task is to first find the maximum sub-array sum and then remove at most one element from the sub-array. Remove at most a single element such that the maximum sum after removal is maximized.If given input array is {1, 2, 3, -2, 3} then maximum sub-array sum is {2, 3, -2, 3}. Then we can remove -2. After removing the remaining array becomes−{1, 2, 3, 3} with sum 9 which is maximum.Algorithm1. Use Kadane’s algorithm to find the maximum subarray sum. 2. Once the sum has beens find, re-apply Kadane’s algorithm ...
Read MoreMaximize the median of an array in C++
Problem statementGiven an array arr[] of N elements and an integer K where K < N. The task is to insert K integer elements to the same array such that the median of the resultant array is maximizedIf input array is {1, 3, 2, 5} and k = 3 then−Sorted array becomes {1, 2, 3, 5}Insert 3 elements that are greater than 5. After this operation array becomes {1, 2, 3, 5, 6, 6, 6}Median of the new array is 5Algorithm1. In order to maximize the median of the resultant array, all the elements that need to be inserted must ...
Read MoreMaximize the median of the given array after adding K elements to the same array in C++
Problem statementGiven an array arr[] of N elements and an integer K where K < N. The task is to insert K integer elements to the same array such that the median of the resultant array is maximizedIf input array is {1, 3, 2, 5} and k = 3 then −Sorted array becomes {1, 2, 3, 5}Insert 3 elements that are greater than 5. After this operation array becomes {1, 2, 3, 5, 6, 6, 6}Median of the new array is 5Algorithm1. In order to maximize the median of the resultant array, all the elements that need to be inserted ...
Read MoreMaximize the number of segments of length p, q and r in C++
Problem statementGiven a rod of length L, the task is to cut the rod in such a way that the total number of segments of length p, q and r is maximized. The segments can only be of length p, q, and rIf l = 15, p = 2, q = 3 and r = 5 then we can make 7 segments as follows −{2, 2, 2, 2, 2, 2, 3}AlgorithmWe can solve this problem using dynamic programming1. Initialize dp[] array to 0 2. Iterate till the length of the rod. For every i, a cut of p, q and ...
Read More