ZCO15001 - Editorial

PROBLEM LINK

ZCO15001

DIFFICULTY

Easy

PREREQUISITES

Dynamic Programming

PROBLEM IN SHORT

Output the smallest n such that the given array can be broken into n contiguous sub-arrays which are palindromes.

SOLUTION

This problem has a dynamic programming solution. Simply stated, the constraints are small enough for a $O(n^3)$ solution. If we can somehow find the solution for the sub-array from index $i$ to $(n - 1)$, and then use it to find subsequent answers - we’ll have a solution.

The simplest way to do this, is to define an array $dp$, which stores the answers for each $i$, from $0$ to $(n - 1)$. Also, $dp[i] = min(dp[i], dp[j + 1] + 1)$, if the sequence from $i$ to $j$ is a palindrome, for all $j$, from $i$ to $(n - 1)$.

This is because, if the sequence from i to j is a palindrome, that’s one palindromic sequence and thus we add one to the optimal solution for the array from $(j + 1)$ to $(n - 1)$. Finally, $dp[0]$ is the solution, because it’ll be the optimal solution for the array $0$ till $(n - 1)$.

TIME COMPLEXITY

$O(n^3)$ for building the $dp$ array. It can be further optimised to $O(n^2)$ if you find out if the array from $i$ to $j$ is a palindrome using a 2D dynamic programming approach.

EDITORIALIST's SOLUTION

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

int n, a[307];
int dp[307];

int isPalin(int i, int j) { // is the sequence from i to j a palindrome - including both i and j
    int length = j - i + 1;
    for (int it = 0; it < length/2; it++) {
        if (a[i + it] != a[j - it])
            return 0;
    }
    return 1;
}

int recur(int cur) {
    if (dp[cur] != -1)
        return dp[cur];
    else {
        int ans = INT_MAX;
        for (int i = cur; i < n; i++) {
            if ( isPalin(cur, i) )
                ans = min(ans, recur(i + 1) + 1);
        }
        return dp[cur] = ans;
    }
}

int main() {
    memset(dp, -1, sizeof(dp));
    scanf("%d", &n);
    dp[n] = 0;
    dp[n - 1] = 1;
    for (int i = 0; i < n; i++)
        scanf("%d", &a[i]);
    printf("%d", recur(0));
    //for (int i = 0; i < n; i++)
    //  printf("%d ", recur(i));
    //printf("\n %d", isPalin(3, 4));
    return 0;
}