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Print matrix in antispiral form
Given a 2d array of n*n and the task is to find the antispiral arrangement of the given matrix
Input : arr[4][4]={1,2,3,4,
5,6,7,8,
9,10,11,12
13,14,15,16}
Output: 10 11 7 6 5 9 13 14 15 16 12 8 4 3 2 1
For this, stack can be used where the transpose of a matrix can be pushed inside stack and poped reversely
Algorithm
START STEP 1 -> declare stack vector element as stk and variables as int r=4, c=4, i, j, rs=0 and cs=0 Step 2 -> store matrix elements in 2-3 array Step 3 -> Loop For i=0 and o<4 and i++ Loop For j=0 and j<4 and j++ Print arr[i][j] End Print
End Step 4 -> Loop While rs<c and cs<r Loop For i=rs and i<c and i++ Push arr[rs][i] End cs++ Loop For i=cs and i<r-1 and ++i Push arr[r-1][i] End c— IF(cs<r) Loop For i=r-1 and i>=rs and –i Push arr[r-1][i] End r- - End IF(rs<c) Loop For i=c-1 and i>=cs and i- - Push arr[i][rs] End Rs++ End End Step 5 -> Loop While !stk.empty() Print stk.top() Call pop() End STOP
Example
#include<iostream>
#include <bits/stdc++.h>
using namespace std;
int main(){
stack <int> stk;
int R=4,C=4,i,j,RS=0,CS=0;
int mat[R][C] = { {1,2,3, 4}, {5,6,7,8},{9,10,11,12},{13,14,15,16}};
for(i=0;i<4;i++){
for(j=0;j<4;j++)
cout<<mat[i][j]<<" ";
cout<<"
";
}
while(RS<C&&CS<R) {
for(i=RS;i<C;i++)
stk.push(mat[RS][i]);
CS++;
for(i=CS;i<R-1;++i)
stk.push(mat[i][C-1]);
C--;
if(CS<R){
for(i=R-1;i>=RS;--i)
stk.push(mat[R-1][i]);
R--;
}
if(RS<C){
for(i=C-1;i>=CS;i--)
stk.push(mat[i][RS]);
RS++;
}
}
while(!stk.empty()){
cout<<stk.top()<<" ";
stk.pop();
}Output
if we run the above program then it will generate the following output
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 10 11 7 6 5 9 13 14 15 16 12 8 4 3 2 1
Updated on: 30-Jul-2019
260 Views
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