Computer Science, asked by anaskhan14301, 3 months ago

C++ program


1. Write down generic code + dry run of given pattern
1 2 3 4 5
4 4 4 4 4
9 9 9 9 9
16 6 16 16 16
17 25 25 25 25 25

Answers

Answered by maansatveer101
6

Answer:

12345................

Answered by deeptigvns
0

Answer:

Input : n = 4

Output :

1

3*2

4*5*6

10*9*8*7

Input : n = 5

Output :

1

3*2

4*5*6

10*9*8*7

11*12*13*14*15

Source: Amdocs Interview Experience | Set 1

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: For odd number row, values are being displayed in increasing order and for even number row, values are being displayed in decreasing order. The only other trick is to how to iterate the loops.

Algorithm:

printPattern(int n)

Declare j, k

Initialize k = 0

for i = 1 to n

if i%2 != 0

for j = k+1, j < k+i, j++

print j and "*"

print j and new line

k = ++j

else

k = k+i-1

for j = k, j > k-i+1, j--

print j and "*";

print j and new line

C++

// C++ implementation to print the following

// pyramid pattern

#include <bits/stdc++.h>

using namespace std;

// function to print the following pyramid pattern

void printPattern(int n)

{

int j, k = 0;

// loop to decide the row number

for (int i=1; i<=n; i++)

{

// if row number is odd

if (i%2 != 0)

{

// print numbers with the '*' sign in

// increasing order

for (j=k+1; j<k+i; j++)

cout << j << "*";

cout << j++ << endl;

// update value of 'k'

k = j;

}

// if row number is even

else

{

// update value of 'k'

k = k+i-1;

// print numbers with the '*' in

// decreasing order

for (j=k; j>k-i+1; j--)

cout << j << "*";

cout << j << endl;

}

}

}

// Driver program to test above

int main()

{

int n = 5;

printPattern(n);

return 0;

}

Java

Python 3

C#

PHP

Output:

1

3*2

4*5*

10*9*8*7

11*12*13*14*15

Time Complexity: O((n * (n + 1)) / 2)

Explanation:

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