Given an integer N. Return a N x N matrix such that every aspect (Coordinates(i, j))is having most absolute distinction doable with the adjoining aspect (i.e., (i + 1, j), (i – 1, j), (i, j + 1), (i, j – 1)). Additionally, It ought to include distinct components starting from 1 to N2.
Examples:
Enter: N = 2
Output: 4 1
2 3Enter: N = 3
Output: 1 3 4
9 2 7
5 8 6
Strategy: The issue could be solved primarily based on the next thought:
The distinct numbers don’t exceed N2 – 1, as a result of the minimal distinction between two components is no less than 1, and the most distinction doesn’t exceed N2 – 1 (the distinction between the utmost aspect N2 and the minimal aspect 1).
Observe the under steps to implement the thought:
- Initialize an empty 2D matrix v of dimension N*N and bool examine = true.
- Begin including the weather from 1 to N2 i.e., l and r pointers respectively.
- If examine = true, begin including the weather from the r pointer in reducing order till the row is stuffed and alter the examine = false.
- If examine = false, begin including the weather from the l pointer in rising order till the row is stuffed and alter the examine = true.
- Repeat this till all matrix is stuffed.
Beneath is the implementation of the above strategy:
C++
// C++ Implementation of the above strategy
#embrace <bits/stdc++.h>
utilizing namespace std;
// Operate to create required matrix
void findMaximumDistinct(int n, vector<vector<int> >& v)
{
// l is left pointer and r is
// proper pointer
int l = 1, r = n * n, t = 0;
bool examine = true;
// Creating matrix by nested for loop
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
// Inserting one aspect from
// beginning and one aspect
// from ending (1 to n^2-1)
if (examine == false) {
// Inserting the worth
// from beginning
v[i][j] = l++;
examine = true;
}
else {
// Inserting the worth
// from ending
v[i][j] = r--;
// Checking the bool for
// alternatingly insertion
examine = false;
}
}
// Revrse the current row for
// odd row
if (i % 2)
reverse(v[i].start(), v[i].finish());
}
}
// Driver Code
int principal()
{
int n = 3;
vector<vector<int> > v(n, vector<int>(n));
// Operate name
findMaximumDistinct(n, v);
// Displaying the matrix
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
cout << v[i][j] << " ";
}
cout << endl;
}
return 0;
}
Time Complexity: O(n2)
Auxiliary Area: O(n2)
