Server Side Programming Articles - Page 2219 of 2646

Here we will see one problem. We will add two array elements and store them into another array. But we will follow some constraints. These constraints are like below −Addition should be stated from 0th index of both of the arraySplit the sum if it is more than one-digit number, and place each digit to the corresponding locationsRemaining digits of the larger input array will be stored at output arrayLet us see the algorithm to get the idea.AlgorithmaddArrayConstraints(arr1, arr2)Begin    define empty vector out    i := 0    while i is less than both arr1.length and arr2.length, do   ... Read More

In this section we will see one interesting problem. Suppose one number is given. We have to increase this number by 1. This is extremely simple task. But here we will place the number as an array. each digit of that number is placed as an element of the array. If the number is 512, then it will be stored as {5, 1, 2}. And also we have to increase the number using recursive approach. Let us see the algorithm to get the clear idea.Algorithmincrement(arr, n, index)Initially the default value of index is 0begin    if index < n, then ... Read More

Here we will see how to add 1 with a number stored into a linked list. In the linked list, each digit of the numbers is stored. If the number is 512, then it will be stored like below −512 = (5)-->(1)-->(2)-->NULLWe are providing the list into the increment function. That will return another list after adding 1 with it. Here we are using the C++ STL linked list. Let us see the algorithm to bet better idea.AlgorithmincrementList(l1)Begin    carry := 1    res := an empty list    for each node n from l1, scan from last to first, ... Read More

Here we will see one interesting problem. Suppose one binary array is given of size n. Here n > 3. A true value or 1 value indicates that the active state, and 0 or false indicates inactive. Another number k is also given. We have to find active or inactive cells after k days. After every day state of ith cell will be active if the left and right cells are not same, if they are same, then it will be inactive. The left most and right most cell has no cell before and after it. So left most and ... Read More

The rat in maze problem is one of the well-known problem of the backtracking. Here we will see that problem with little variation. Suppose one NxN maze M is given. The starting point is top left corner M[0, 0], and the destination is right bottom corner M[N – 1, N - 1]. One rat is placed at the starting point. Our goal is to find a path from starting point to ending point that can be by the rat to reach the destination. Here the rat can jump (The variation). Now there are some constraintsThe rat can move either towards ... Read More

Here we will see one puzzle. Suppose there is a program which is given as below, we have to tell what will be the output and why?Example#include using namespace std; int main() {    int x = 0xab;    ~x;    cout

Here we will see another sorting problem named Pancake sort. This problem is simple. We have one array. We have to sort this. But we can use only one operation called rev(arr, i). This will reverse the elements of arr from 0 to ith position.This idea is like the selection sort. We repeatedly place the max element at end reduce the size of the array. Let us see the algorithm to get the idea.AlgorithmpancakeSort(arr, n)Begin    size := n    while size > 1, do       index := index of max element in arr from [0 to size ... Read More

In this section we will see one interesting problem. We will see two code segments. Both are with two nested loops. We have to identify which one will run faster. (We will assume that the compiler is not optimizing the code).Segment 1for(int i = 0; i < 10; i++){    for(int j = 0; j

Here we will see one matrix probability problem. We have one rectangular matrix. We can move four directions from the current cell with equal probability. These four directions are left, right, up and down. We have to calculate the probability after N moves from position M[i, j].Here we will do something related to DFS. We will traverse recursively traverse in each of the four possible rooms from the current room. Then we will calculate the probability with one less move. As each of the four directions has equal probability, then each direction will contribute 0.25 of total probability. If we ... Read More

Here, as per the given problem pattern needs to be displayed using recursive approach.Recursive function is the one that calls itself n number of times. There can be ‘n’ number of recursive function in a program. The problem working with recursive function is their complexity.AlgorithmSTART Step 1 -> function int printpattern(int n)    If n>0       Printpattern(n-1)       Print *    End IF End Step 2 -> function int pattern(int n)    If n>0       pattern(n-1)    End IF    Printpattern(n)    Print End STOPExample#include int printpattern(int n) {    if(n>0) { ... Read More