how to traverse the spiral traverse matrix to its original form.

Question

function spiralTraversalIV(N, arr){
let left=0;
let right=N-1;
let top=0;
let bottom=N-1;
let count=0;
let s="";

while(count < N*N){
for(let i=bottom;i>=top;i--){
s+=(arr[i][right])+" ";
count++;
}
right--;
for(let i=right;i>=left;i--){
s+=(arr[top][i])+" ";
count++;
}
top++;
for(let i=top;i<=bottom;i++){
s+=(arr[i][left])+" ";
count++;
}
left++;

for(let i=left;i<=right;i++){
s+=(arr[bottom][i])+" ";
count++;
}
bottom--;
}
console.log(s);
}

spiralTraversalIV(3,[ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ])

out put
9 6 3 2 1 4 7 8 5

https://code.sololearn.com/cNF0nyRhHQy6

how can make my original string from my out-put?

Answer 1

Steve yes ,but I already solved this

Answer 2

TINKLE DASH Glad to hear it!

Answer 3

Are you asking how to generate [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] from 9 6 3 2 1 4 7 8 5?