Finding the number of palindromic sequences from a given set of numbers

Question

Suppose we are given a set of numbers like 20 40 20 60 80 60 It can be broken up into 2 palindromic sequences: 20 40 20, and 60 80 60. It can also be broken into 6 palindromic sequences each containing a single number.

How do we find the smallest number of palindromic sequences possible from a given set of numbers in c++?

PS- This is not my homework. Genuine question.

Answer 1

A straightforward approach begins by looking at each of the O(n³) subsequences and checking to see if it's a palindrome. Once we know which subsequences are palindromic, we can do dynamic programming in O(n²) time to find the minimal number of consecutive subsequences that cover the whole sequence.

For the input 20 40 20 60 80 60, the C++ implementation below prints [20 40 20] [60 80 60].

#include <cstdio>
#include <vector>
using namespace std;

int main() {
  // Read the data from standard input.
  vector<int> data;
  int x;
  while (scanf("%d", &x) != EOF) {
    data.push_back(x);
  }
  int n = data.size();

  // Look at every subsequence and determine if it's a palindrome.
  vector<vector<bool> > is_palindrome(n);
  for (int i = 0; i < n; ++i) {
    is_palindrome[i] = vector<bool>(n);
    for (int j = i; j < n; ++j) {
      bool okay = true;
      for (int left = i, right = j; left < right; ++left, --right) {
        if (data[left] != data[right]) { 
          okay = false;
          break; 
        }
      }
      is_palindrome[i][j] = okay;
    }
  }

  // Dynamic programming to find the minimal number of subsequences.
  vector<pair<int,int> > best(n);
  for (int i = 0; i < n; ++i) {
    // Check for the easy case consisting of one subsequence.
    if (is_palindrome[0][i]) {
      best[i] = make_pair(1, -1);
      continue;
    }
    // Otherwise, make an initial guess from the last computed value.
    best[i] = make_pair(best[i-1].first + 1, i-1);
    // Look at earlier values to see if we can improve our guess.
    for (int j = i-2; j >= 0; --j) {
      if (is_palindrome[j+1][i]) {
        if (best[j].first + 1 < best[i].first) {
          best[i].first = best[j].first + 1;
          best[i].second = j;
        }
      }
    }
  }

  printf("Minimal partition: %d sequences\n", best[n-1].first);
  vector<int> indices;
  int pos = n-1;
  while (pos >= 0) {
    indices.push_back(pos);
    pos = best[pos].second;
  }
  pos = 0;
  while (!indices.empty()) {
    printf("[%d", data[pos]);
    for (int i = pos+1; i <= indices.back(); ++i) {
      printf(" %d", data[i]);
    }
    printf("] ");
    pos = indices.back()+1;
    indices.pop_back();
  }
  printf("\n");

  return 0;
}