First, let us ignore the second condition, that \(T_i\) does not end with *, before constructing \(T_i\). Initialize \(T_1,T_2,\dots,T_M\) with a length-\(N\) string consisting of *. Then, for each \(i\) and \(j\), assign the \(j\)-th character of \(S_i\) to the \((N-i+1)\)-th character of \(T_j\). Now, all but the second condition are satisfied.

To satisfy the second condition, for each \(T_i\), repeatedly remove the last character while it ends with *.

Sample code (Python)

N = int(input())
S = [input() for _ in range(N)]
M = max(map(len, S))
T = [["*"] * N for _ in range(M)]
for i in range(N):
    for j in range(len(S[i])):
        T[j][N - i - 1] = S[i][j]
for i in range(M):
    while T[i][-1] == "*":
        T[i].pop()
    print("".join(T[i]))

Sample code (C++)

#include <bits/stdc++.h>
using namespace std;

int main() {
    int N;
    cin >> N;
    vector<string> S(N);
    for (auto& s : S) cin >> s;
    int M = 0;
    for (auto& s : S) M = max(M, (int)s.size());
    vector<string> T(M, string(N, '*'));
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < (int)S[i].size(); ++j) {
            T[j][N - i - 1] = S[i][j];
        }
    }
    for (int i = 0; i < M; ++i) {
        while (T[i].back() == '*') {
            T[i].pop_back();
        }
        cout << T[i] << endl;
    }
}