brute force

Topics

• algorithmic paradigm

Try out all the possibilities to find the one that satisfies the problem’s criteria.

Let us illustrate this with the example problem of find all pairs of numbers abcde and fghij such that they are formed from digits 0 to 9 and they satisfy abcde/fghij = N for a given N, e.g. for N=62, we have: (79546, 01283), (94736, 01528), etc.

There are two ways of implementing brute force:

Approach 1: Generate and Check

1. Generate: We create all possible candidate solutions. In our example, this corresponds to generating all possible ways of assigning the digits 0−9 to the variables a, b, c, d, e, f, g, h, i, j. One way to do so it by using next_permutation STL to get a permutation and take first 5 digits and call it abcde and remaining 5 as fghij.
2. Check: Here, we test each candidate solution to see if it satisfies the given conditions. For our problem, we would check if the ratio abcde/fghij equals N and if all the digits assigned are unique.

This approach has a time complexity of $O(k!)$ where $k=10!$.

Approach 2: Loop and Check

1. Loop: This approach involves iterating through a range of possible values. For instance, we might loop through all 5-digit numbers for fghij. For each fghij, we calculate the corresponding abcde by multiplying fghij by N.
2. Check: Similar to the first approach, we then check if the generated abcde and fghij satisfy the problem’s conditions, namely if their ratio is N and if all digits are unique.

This approach has a time complexity of $O(m!)$ where $m=10^5$ since 98765 can be the largest 5 digit number that’s valid as per problem and 01234 the smallest.

Note

We can see from the number of operations that Approach 2 is faster. In brute force it’s crucial to understand which of the several approach is feasible.

void solve1(int n) {
  string s = "0123456789";
  do {
    int numer = stoi(s.substr(0, 5));
    int denom = stoi(s.substr(5, 5));
    if (numer % denom == 0 && numer / denom == n) {
      cout << numer << " " << denom << endl;
    }
  } while (next_permutation(s.begin(), s.end()));
}
 
void solve2(int n) {
  for (int fghij = 01234; fghij <= 98765 / n; fghij++) {
    int abcde = n * fghij;
    set < int > st;
 
    int temp = abcde;
    for (int i = 0; i < 5; i++) {
      st.insert(temp % 10);
      temp /= 10;
    }
 
    temp = fghij;
    for (int i = 0; i < 5; i++) {
      st.insert(temp % 10);
      temp /= 10;
    }
 
    if (st.size() == 10) {
      cout << abcde << " " << fghij << endl;
    }
  }
}

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