LeetCode #2767 — MEDIUM

Partition String Into Minimum Beautiful Substrings

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

Problem Statement

Given a binary string s, partition the string into one or more substrings such that each substring is beautiful.

A string is beautiful if:

It doesn't contain leading zeros.
It's the binary representation of a number that is a power of 5.

Return the minimum number of substrings in such partition. If it is impossible to partition the string s into beautiful substrings, return -1.

A substring is a contiguous sequence of characters in a string.

Example 1:

Input: s = "1011"
Output: 2
Explanation: We can paritition the given string into ["101", "1"].
- The string "101" does not contain leading zeros and is the binary representation of integer 5¹ = 5.
- The string "1" does not contain leading zeros and is the binary representation of integer 5⁰ = 1.
It can be shown that 2 is the minimum number of beautiful substrings that s can be partitioned into.

Example 2:

Input: s = "111"
Output: 3
Explanation: We can paritition the given string into ["1", "1", "1"].
- The string "1" does not contain leading zeros and is the binary representation of integer 5⁰ = 1.
It can be shown that 3 is the minimum number of beautiful substrings that s can be partitioned into.

Example 3:

Input: s = "0"
Output: -1
Explanation: We can not partition the given string into beautiful substrings.

Constraints:

1 <= s.length <= 15
s[i] is either '0' or '1'.

Step 01

Brute Force Baseline

Problem summary: Given a binary string s, partition the string into one or more substrings such that each substring is beautiful. A string is beautiful if: It doesn't contain leading zeros. It's the binary representation of a number that is a power of 5. Return the minimum number of substrings in such partition. If it is impossible to partition the string s into beautiful substrings, return -1. A substring is a contiguous sequence of characters in a string.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Dynamic Programming · Backtracking

Example 1

"1011"

Example 2

"111"

Example 3

"0"

Core Insight

What unlocks the optimal approach

To check if number x is a power of 5 or not, we will divide x by 5 while x > 1 and x mod 5 == 0. After iteration if x == 1, then it was a power of 5.
Since the constraint of s.length is small, we can use recursion to find all the partitions.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2767: Partition String Into Minimum Beautiful Substrings
class Solution {
    private Integer[] f;
    private String s;
    private Set<Long> ss = new HashSet<>();
    private int n;

    public int minimumBeautifulSubstrings(String s) {
        n = s.length();
        this.s = s;
        f = new Integer[n];
        long x = 1;
        for (int i = 0; i <= n; ++i) {
            ss.add(x);
            x *= 5;
        }
        int ans = dfs(0);
        return ans > n ? -1 : ans;
    }

    private int dfs(int i) {
        if (i >= n) {
            return 0;
        }
        if (s.charAt(i) == '0') {
            return n + 1;
        }
        if (f[i] != null) {
            return f[i];
        }
        long x = 0;
        int ans = n + 1;
        for (int j = i; j < n; ++j) {
            x = x << 1 | (s.charAt(j) - '0');
            if (ss.contains(x)) {
                ans = Math.min(ans, 1 + dfs(j + 1));
            }
        }
        return f[i] = ans;
    }
}

// Accepted solution for LeetCode #2767: Partition String Into Minimum Beautiful Substrings
func minimumBeautifulSubstrings(s string) int {
	ss := map[int]bool{}
	n := len(s)
	x := 1
	f := make([]int, n+1)
	for i := 0; i <= n; i++ {
		ss[x] = true
		x *= 5
		f[i] = -1
	}
	var dfs func(int) int
	dfs = func(i int) int {
		if i >= n {
			return 0
		}
		if s[i] == '0' {
			return n + 1
		}
		if f[i] != -1 {
			return f[i]
		}
		f[i] = n + 1
		x := 0
		for j := i; j < n; j++ {
			x = x<<1 | int(s[j]-'0')
			if ss[x] {
				f[i] = min(f[i], 1+dfs(j+1))
			}
		}
		return f[i]
	}
	ans := dfs(0)
	if ans > n {
		return -1
	}
	return ans
}

# Accepted solution for LeetCode #2767: Partition String Into Minimum Beautiful Substrings
class Solution:
    def minimumBeautifulSubstrings(self, s: str) -> int:
        @cache
        def dfs(i: int) -> int:
            if i >= n:
                return 0
            if s[i] == "0":
                return inf
            x = 0
            ans = inf
            for j in range(i, n):
                x = x << 1 | int(s[j])
                if x in ss:
                    ans = min(ans, 1 + dfs(j + 1))
            return ans

        n = len(s)
        x = 1
        ss = {x}
        for i in range(n):
            x *= 5
            ss.add(x)
        ans = dfs(0)
        return -1 if ans == inf else ans

// Accepted solution for LeetCode #2767: Partition String Into Minimum Beautiful Substrings
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2767: Partition String Into Minimum Beautiful Substrings
// class Solution {
//     private Integer[] f;
//     private String s;
//     private Set<Long> ss = new HashSet<>();
//     private int n;
// 
//     public int minimumBeautifulSubstrings(String s) {
//         n = s.length();
//         this.s = s;
//         f = new Integer[n];
//         long x = 1;
//         for (int i = 0; i <= n; ++i) {
//             ss.add(x);
//             x *= 5;
//         }
//         int ans = dfs(0);
//         return ans > n ? -1 : ans;
//     }
// 
//     private int dfs(int i) {
//         if (i >= n) {
//             return 0;
//         }
//         if (s.charAt(i) == '0') {
//             return n + 1;
//         }
//         if (f[i] != null) {
//             return f[i];
//         }
//         long x = 0;
//         int ans = n + 1;
//         for (int j = i; j < n; ++j) {
//             x = x << 1 | (s.charAt(j) - '0');
//             if (ss.contains(x)) {
//                 ans = Math.min(ans, 1 + dfs(j + 1));
//             }
//         }
//         return f[i] = ans;
//     }
// }

// Accepted solution for LeetCode #2767: Partition String Into Minimum Beautiful Substrings
function minimumBeautifulSubstrings(s: string): number {
    const ss: Set<number> = new Set();
    const n = s.length;
    const f: number[] = new Array(n).fill(-1);
    for (let i = 0, x = 1; i <= n; ++i) {
        ss.add(x);
        x *= 5;
    }
    const dfs = (i: number): number => {
        if (i === n) {
            return 0;
        }
        if (s[i] === '0') {
            return n + 1;
        }
        if (f[i] !== -1) {
            return f[i];
        }
        f[i] = n + 1;
        for (let j = i, x = 0; j < n; ++j) {
            x = (x << 1) | (s[j] === '1' ? 1 : 0);
            if (ss.has(x)) {
                f[i] = Math.min(f[i], 1 + dfs(j + 1));
            }
        }
        return f[i];
    };
    const ans = dfs(0);
    return ans > n ? -1 : ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.