LeetCode 2767: Partition String Into Minimum Beautiful Substrings Solution

Master LeetCode problem 2767 (Partition String Into Minimum Beautiful Substrings), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2767. Partition String Into Minimum Beautiful Substrings

Medium

Hash Table String Dynamic Programming Backtracking

Problem Explanation

Explanation:

To solve this problem, we can iterate through the binary string s and try to find substrings that are beautiful. We can check if each substring is a beautiful number by converting it to an integer and checking if it is a power of 5.

Here is the algorithm:

Iterate through the binary string s and consider all possible substrings.
For each substring, convert it to an integer and check if it is a power of 5.
If a substring is a beautiful number, increment the count of beautiful substrings.
Return the minimum number of beautiful substrings found.

Time Complexity: O(n^2), where n is the length of the input string s. Space Complexity: O(n), to store the substrings.

Solution Code

class Solution {
    public int minPartitions(String s) {
        int n = s.length();
        int[] dp = new int[n+1];
        for(int i=1; i<=n; i++){
            dp[i] = Integer.MAX_VALUE;
            for(int j=i-1; j>=0; j--){
                if(s.charAt(j) == '0' && j != i-1){
                    continue;
                }
                String sub = s.substring(j, i);
                if(isPowerOfFive(sub)){
                    dp[i] = Math.min(dp[i], dp[j] + 1);
                }
            }
        }
        return dp[n] == Integer.MAX_VALUE ? -1 : dp[n];
    }
    
    private boolean isPowerOfFive(String str){
        if(str.charAt(0) == '0'){
            return false;
        }
        long num = Long.parseLong(str, 2);
        while(num > 1){
            if(num % 5 != 0){
                return false;
            }
            num /= 5;
        }
        return num == 1;
    }
}

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