Ninja has been given a task to print the required pattern and he asked for your help with the same.
You must print a matrix corresponding to the given number pattern.
Example:
Input: ‘N’ = 4
Output:
4444444
4333334
4322234
4321234
4322234
4333334
4444444
The first line contains an integer ‘N’.
Print a 2D array of integers that represent the number pattern.
1 <= N <= 10^2
Time Limit: 1 sec
3
33333
32223
32123
32223
33333
5
555555555
544444445
543333345
543222325
543212345
543222325
543333345
544444445
555555555
4
4444444
4333334
4322234
4321234
4322234
4333334
4444444
Can you iterate over the cells?
Approach:
Here the pattern will be of size ‘K * K’ where ‘K = 2 * N - 1’ . First, we will find two integers ‘X = abs(i - N + 1)’ and ‘Y = abs(j - N + 1)’ it will tell us what number should be in the current cell of the pattern ‘(i, j)’.The cell should contain the number ‘max(X, Y) + 1’. Because we can see the mathematical pattern here every cell ‘(i, j)’ contains the number max of the above two values.
Algorithm:
O(N^2), Where N is the given input integer.
Two nested loops are running (2*N - 1) * (2*N - 1) times so that time complexity would be the order of quadratic or N^2.
Hence the time complexity is O(N^2).
O(1)
Since we are only printing the pattern and not storing it, no extra space is used.
All tags
Sort by
Interview problems
c++ code
void getNumberPattern(int n) {
for (int i = 0; i < 2*n-1; i++) {
for (int j = 0; j < 2*n-1; j++) {
int minDist = min({i, j, 2*n-2-i, 2*n-2-j});
cout << (n - minDist);
}
cout << endl;
}
}Interview problems
getNumberpattern
void getNumberPattern(int n) {
// Write your code here.
for(int i=0;i<2*n-1;i++){
for(int j=0;j<2*n-1;j++){
int top=i;
int left=j;
int right=(2*n-2)-j;
int down=(2*n-2)-i;
cout<<(n-min(min(top,down),min(left,right)));
}
cout<<endl;
}
}
Interview problems
simple java code
public class Solution {
public static void getNumberPattern(int n) {
// Write your code here.
int size = 2 * n - 1;
int start = 0;
int end = size - 1;
int matrix[][] = new int[size][size];
while (n != 0) {
for (int i = start; i <= end; i++) {
for (int j = start; j <= end; j++) {
if (i == start || i == end || j == start || j == end) {
matrix[i][j] = n;
}
}
}
n--;
start++;
end--;
}
for (int i = 0; i < size; i++) {
for (int j = 0; j < size; j++) {
System.out.print(matrix[i][j]);
}
System.out.println();
}
}
}
Interview problems
Ninja and the number pattern
public class Solution {
public static void getNumberPattern(int n) {
// Write your code here.
int X,Y;
int k = 2*n-1;
for(int i =0;i<k;i++)
{
for(int j=0;j<k;j++)
{
X=Math.abs(i-n+1);
Y=Math.abs(j-n+1);
System.out.print(Math.max(X,Y)+1);
}
System.out.println();
}
}
}
Interview problems
Sol using array
void getNumberPattern(int n) {
int size = 2*n-1;
int start =0;
int end = size-1;
int arr[size][size];
while(n!=0){
for(int i=start; i<=end; i++)
{
for(int j=start; j<=end; j++)
{
if(i==start || i==end || j==start || j==end)
{
arr[i][j]=n;
}
}
}
start++;
end--;
n--;
}
for(int i=0; i<size; i++)
{
for(int j=0; j<size; j++)
{
cout << arr[i][j];
}
cout << endl;
}
}
Interview problems
Ninja And The Number Pattern
void getNumberPattern(int n) {
for(int i = 0; i<2*n-1; i++){
for(int j = 0; j<2*n-1; j++){
int top = i;
int down = j;
int left = (2*n-2)-j;
int right = (2*n-2)-i;
cout<<(n-min(min(top,down),min(left,right)));
}
cout<<endl;
}
}
Interview problems
Answer:
void getNumberPattern(int n) {
for (int i=0;i<(n*2)-1;i++){
for (int j=0;j<(n*2)-1;j++){
int top=i;
int left=j;
int right=(2*n-2)-j;
int down=(2*n-2)-i;
cout<<n-min(min(top,down),min(right,left));
}
cout<<endl;
}
}
Interview problems
best and short solution
for(int i=0;i<2*n-1;i++)
{
for(int j=0;j<2*n-1;j++)
{
int top=(2*n-2)-(i);
int bottom=(2*n-2)-(j);
cout<<(n-min(min(top,bottom),min(i,j)));
}
cout<<endl;
}
}
Interview problems
Simple C++ Approach | Better than 98.5%
void getNumberPattern(int n) {
for(int i = 0; i < 2*n-1; i++){
cout << n;
}
cout << endl;
for(int i = 0; i < n-1; i++){
for(int j = 0; j <= i; j++)
cout << n - j;
for(int k = 0; k < 2*n-3 -2*i; k++)
cout << n-i-1;
for(int j = 0; j <= i; j++)
cout << n-i+j;
cout << endl;
}
for (int i = 0; i < n - 1; i++)
{
for(int j = 0; j < n-1-i; j++)
cout << n-j;
for(int k = 0; k < 2*i+1; k++)
cout << 2+i;
for(int j = 0; j < n-i-1; j++)
cout << 2+j+i;
cout << endl;
}
}
