JS

const l = s.length;
    const words = Math.ceil(Math.sqrt(l));
    const letters = Math.floor(Math.sqrt(l));
    const d = Number.isInteger(Math.sqrt(l)) ? 0 : 1;
    let r = '';
    
    for(let i =0; i<words; i++) {
        for(let j = i; j < l; j += letters + d) {
            r += s[j];
        }
        r += ' '; // 1 space
    }
    
    return r;

Easy solution

include

using namespace std;

int main() { string st; cin>>st;

string noSpace="";

for(int i=0;i<st.length();i++){
    if(st[i]!=' '){
        noSpace+=st[i];
    }
}

int size=noSpace.length();

//cout<<size<<endl;

int fl=floor(sqrt(size));
int cl=ceil(sqrt(size));

int mul=fl*cl;

// cout<

    while(mul<size){
        fl+=1;
        mul=fl*cl;
    }
}

vector<vector<char>>ans;

int x=0;


    for(int i=0;i<fl;i++){
        vector<char>temp;
        for(int j=0;j<cl;j++){
            if(x<size){
                temp.push_back(noSpace[x]);
                x+=1;
            }else{
                temp.push_back('-');
            }
        }
        ans.push_back(temp);
    }


string ans1="";
for(int i=0;i<ans[0].size();i++){
    for(int j=0;j<ans.size();j++){
        if(ans[j][i]!='-'){
            ans1+=ans[j][i];
        }
    }
    ans1+=" ";

}

cout<<ans1<<endl;





return 0;

}

using nested loop in python:

def encryption(s): row, column = math.floor(math.sqrt(len(s))) , math.ceil(math.sqrt(len(s))) str = '' for x in range(column): for y in range(x, len(s), column): str = str + s[y] str = str + ' ' return str

Kotlin solution with O(n)

	var editedText = ""

        val breakerPoint = sqrt(s.length.toDouble()).toInt()
        val map = mutableMapOf<Int,String>()

        s.forEachIndexed { index, c ->
            map[index % (breakerPoint)] = (map[index % (breakerPoint)]?: "") + c
        }
        map.values.forEachIndexed { index, s ->
            editedText += s
            if (index != map.size - 1) {
                editedText += " "
            }
        }