Previous Line Expansion Pattern

-

        Previous Line Expansion Pattern       

PROBLEM STATEMENT :



   


SOLUTIONS IN (C,C++,JAVA,PYTHON)

1)  C  (Programming Language) 


2) C++ (CPP)



3) JAVA 



4) PYTHON





Comments

Post a Comment

Popular posts from this blog

DP (1) - Count number of ways to cover a distance

-
Given a distance n, count total number  of ways to cover the distance with 1, 2 and 3 steps. Input: n = 4 Output: 7 Explantion: Below are the four ways  1 step + 1 step + 1 step + 1 step  1 step + 2 step + 1 step  2 step + 1 step + 1 step   1 step + 1 step + 2 step  2 step + 2 step  3 step + 1 step  1 step + 3 step // C++ program to count number of ways in DP  to cover a distance with 1, 2 and 3 steps  #include<iostream>  using namespace std;  int main() {      int n;     std::cin>>n;     int count[n+1];         // Initialize base values. There is one way to cover 0 and 1      // distances and two ways to cover 2 distance.     count[0]  = 1,  count[1] = 1,  count[2] = 2;         // Fill the count array in from bottom to up manner      for (int i=3; i<=n; i++)  ...
Read more

Zero Insert After K Times One

-
 Zero Insert After K Times One  PROBLEM STATEMENT : Given a bitstream of length N consisting of 0s and 1s, insert 0 after 1 has appeared K times consecutively.  Input Format:  The first line contains N and K separated by a space.  The second line contains the bitstream with 1s and 0s with each value separated by a space.  Output Format:   In the first line, the bitstream with the 0 inserted after 1s has appeared K times with each value separated by a space.  Boundary Conditions:   1 <= K <= 1000  2 <= N <= 1000  Example Input/Output 1: Input: ( ) 12 2 1 0 1 1 0 1 1 0 1 1 1 1 Output: 1 0 1 1 0 0 1 1 0 0 1 1 0 1 1 0 SOLUTION : C (Programming Language) Copy Code #include<stdio.h> #include<stdlib.h> int main() { int n,k,cnt=0,i,j,a[2002]; scanf("%d%d",&n,&k); for(i=0;i<n;++i) { scanf("%d",&a[i]); } //Count of 1 in first step for(i=0;i<k;++i...
Read more

Left Number Twice Right

-
 Left Number Twice Right  PROBLEM STATEMENT : A set of N numbers (separated by one or more spaces) is passed as input to the program. The program must identify the count of numbers C where the number on the left is twice the number on the right.  Boundary Conditions:   3 <= N <= 50  The value of the numbers can be from -99999999 to 99999999  Input Format: The first line will contain the N numbers separated by one or more spaces.  Output Format: The first line contains the value of C.  Example Input/Output 1:   Input: 10 20 5 40 15 -2 -30 -1 60  Output: 2 Explanation: The numbers meeting the criteria are 20, -30  Example Input/Output 2:  Input: 5 90 10 2 5 -4 10 6 5 3  Output: 3 Explanation: The numbers meeting the criteria are 2, 6, 5 SOLUTION : C (Programming Language) Copy Code #include<stdio.h> #include<stdlib.h> int main() { int a[50],n=-1,i,cnt=0; while(scanf("%d",&a[++n...
Read more