Aryan and his friends are very fond of the pattern. For a given integer ‘N’, they want to make the N-Binary Number Triangle.
You are required to print the pattern as shown in the examples below.
Example:
Input: ‘N’ = 3
Output:
1
0 1
1 0 1
The first line contains an integer ‘N’.
Print the pattern as specified in the problem statement.
1 <= N <= 20
Time Limit: 1 sec
3
1
0 1
1 0 1
4
1
0 1
1 0 1
0 1 0 1
6
1
0 1
1 0 1
0 1 0 1
1 0 1 0 1
0 1 0 1 0 1
Iterate to all the cells and fill the cells in an alternate manner.
Approach:
The solution is iterating on the pattern and for the cell in the ‘i’th row and the ‘j’ column, the value will be 1 if ‘i’ + ‘j’ is even else the value will be 0.
The steps are as follows :
Function void nBinaryTriangle(int ‘N’)
O(N^2), Where N is the given input integer.
Two nested loops are running N * N times so that time complexity would be the order of N * N.
Hence the time complexity is O(N^2).
O(1)
Since we are using constant extra space.
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Interview problems
Java easy solution
public class Solution {
public static void nBinaryTriangle(int n) {
// Write your code here.
for(int i =1;i<=n;i++){
int start = 0;
if(i%2!=0){
start =1;
}
for(int j =0;j<i;j++){
System.out.print(start+" ");
start =1-start;
}
System.out.println();
}
}
}
Interview problems
C++ Solution
void nBinaryTriangle(int n) {
for(int i=0;i<n;i++){
for(int j=0; j<=i; j++){
if((i+j)%2==0){
cout<<"1 ";
}
else{
cout<<"0 ";
}
}
cout<<endl;
}
}
Interview problems
cpp easyyy
void nBinaryTriangle(int n) {
// Write your code here.
int start = 1;
for(int i=1; i<=n; i++){
if(i%2 == 0){
start = 0;
}else{
start = 1;
}
for(int j=1; j<=i; j++){
cout<<start<<" ";
start = 1-start;
}
cout<<endl;
}
}
Interview problems
C++ Simple Bruteforce Solution
void nBinaryTriangle(int n) {
// Write your code here.
for(int i=0;i<n;i++){
for(int j=0;j<=i;j++){
if((i%2) == 0){
if((j%2) != 0){
cout<<"0 ";
}
else{
cout<<"1 ";
}
}
else{
if((j%2) != 0){
cout<<"1 ";
}
else{
cout<<"0 ";
}
}
}
cout<<endl;
}
}
Interview problems
c++ code
void nBinaryTriangle(int n) {
for(int i=1;i<=n;i++)
{
for(int j=1;j<=i;j++)
{
if((i+j)%2==0)
{
cout<<1<<" ";
}
else{
cout<<0<<" ";
}
}
cout<<endl;
}
}Interview problems
c++ code
void nBinaryTriangle(int n) {
int start=1;
for(int i=0;i<n;i++){
if(i%2==0){
start=1;
}
else{
start=0;
}
for(int j=0;j<=i;j++){
cout<<start<<" ";
start=1-start;
}
cout<<endl;
}
}Interview problems
Using bit manipulation for flipping number
def nBinaryTriangle(n: int) -> None:
for i in range(1,n+1):
digit = i % 2
for _ in range(i):
print(digit, end=" ")
digit ^= 1
print()Interview problems
simple python solution
def nBinaryTriangle(n: int) -> None:
# Write your solution here.
for i in range(1, n+1):
x = 1
if i%2 == 0:
x = 0
for j in range(i):
print(x, end=' ')
x=0 if x==1 else 1
print()Interview problems
Beginner Friendly Solution || C++
void nBinaryTriangle(int n) {
// Write your code here.
for(int i = 0;i<n;i++){
for(int j = 0;j<=i;j++){
if((i+j)%2 == 0){
cout<<"1 ";
}
else{
cout<<"0 ";
}
}
cout<<endl;
}
}
Interview problems
Binary Triangle Java Solution
public class Solution {
public static void nBinaryTriangle(int n) {
// Write your code here.
int start =0;
for (int i=0;i<n;i++){
if(i%2==0)
start=1;
else start=0;
for (int j=0;j<=i;j++){
System.out.print(start +" ");
start=1-start;
}
System.out.println();
}
}
}
