LeetCode Problem 3864

Difficulty: 🔴 Hard
Topics: String, Divide and Conquer, Prefix Sum

Solution

LeetCode 3864 - Minimum Cost to Partition a Binary String

Problem Understanding

We are given a binary string s, where each character represents whether an element is sensitive:

• '1' means the element is sensitive.
• '0' means the element is not sensitive.

The entire string initially forms one segment. Every segment has a storage cost that depends on its length and the number of sensitive elements inside it.

For a segment with:

• Length L
• Number of sensitive elements X

the cost is:

• flatCost if X == 0
• L * X * encCost if X > 0

The interesting part of the problem is that a segment may be split only when its length is even. When a split is performed, the segment is divided into two contiguous halves of equal size. After splitting, each half may either remain as a final segment or be split again if its length is even.

The goal is to choose a valid sequence of splits that minimizes the total cost of all final segments.

The input size is large, up to 10^5 characters. This immediately tells us that any solution that explicitly explores all possible partition configurations will be far too slow. We need an approach that exploits the very restrictive splitting rule.

A few important observations:

• We can only split an interval exactly in half.
• Odd-length segments can never be split.
• Every valid partition corresponds to selecting some nodes in a recursive binary decomposition tree.
• Segment costs depend only on interval length and the number of '1' characters inside the interval.
• Prefix sums allow us to compute the number of sensitive elements in any interval in constant time.

Important edge cases include:

• Strings containing only '0', where keeping large segments may be cheaper than splitting.
• Strings containing only '1', where repeated splitting may dramatically reduce the cost.
• Lengths that are odd from the start, where no split is possible.
• Lengths that are powers of two, which generate the largest possible decomposition tree.

Approaches

Brute Force

A brute-force solution would recursively consider every possible decision:

• Keep the current segment as a final segment.
• If the segment length is even, split it into two halves and recursively solve both halves.

This approach is correct because it explores every valid partition obtainable through legal splits.

However, many identical subproblems are revisited repeatedly. More importantly, thinking about all possible partition configurations leads to exponential behavior in the size of the decomposition tree.

Given n ≤ 10^5, such an approach is completely infeasible.

Key Insight

The splitting rule is extremely restrictive.

A segment can only be split into its two equal halves. Therefore, every segment corresponds to a unique node in a binary decomposition tree.

For any interval:

• We can stop and pay its direct storage cost.
• Or, if its length is even, we can split it and pay the optimal costs of its two children.

This immediately gives a dynamic programming recurrence:

Let dp(interval) be the minimum cost achievable for that interval.

Then:

• If the interval length is odd:

dp = segmentCost

• If the interval length is even:

dp = min(segmentCost, dp(left) + dp(right))

Because every interval in the decomposition tree is evaluated exactly once, the problem becomes very efficient.

To compute segment costs quickly, we use a prefix sum array storing the cumulative number of '1' characters.

Approach Comparison

Approach	Time Complexity	Space Complexity	Notes
Brute Force	Exponential	Exponential	Explores every possible partition configuration
Optimal	O(n)	O(n)	Dynamic programming on the decomposition tree

Algorithm Walkthrough

1. Build a prefix sum array where prefix[i] stores the number of '1' characters in s[0:i].
2. Define a recursive function dfs(l, r) that computes the minimum cost for interval [l, r).
3. Compute the interval length:

length = r - l 4. Compute the number of sensitive elements:

ones = prefix[r] - prefix[l] 5. Compute the direct storage cost of this interval.

• If ones == 0, cost is flatCost.
• Otherwise, cost is length * ones * encCost.

6. If the interval length is odd, the interval cannot be split. Return the direct cost.
7. Otherwise, compute the midpoint:

mid = (l + r) // 2 8. Recursively compute:

• leftCost = dfs(l, mid)
• rightCost = dfs(mid, r)

9. The best achievable cost for this interval is:

min(directCost, leftCost + rightCost) 10. Return that value. 11. Invoke dfs(0, n) and return the result.

Why it works

Every valid partition is obtained by repeatedly applying the allowed equal-halves split operation. Therefore every possible segment that can appear corresponds to a node in the decomposition tree.

For any node, there are only two choices:

• Stop splitting and pay the segment's own cost.
• Split once and then optimally solve both children.

Since these are exactly the legal possibilities, and since the recurrence takes the minimum over them, the computed value is the true optimal cost for that interval. By induction on interval size, the result for the root interval is the minimum possible total cost.

Python Solution

from typing import List

class Solution:
    def minCost(self, s: str, encCost: int, flatCost: int) -> int:
        n = len(s)

        prefix = [0] * (n + 1)
        for i, ch in enumerate(s):
            prefix[i + 1] = prefix[i] + (ch == '1')

        def dfs(left: int, right: int) -> int:
            length = right - left
            ones = prefix[right] - prefix[left]

            if ones == 0:
                segment_cost = flatCost
            else:
                segment_cost = length * ones * encCost

            if length % 2 == 1:
                return segment_cost

            mid = (left + right) // 2

            split_cost = dfs(left, mid) + dfs(mid, right)

            return min(segment_cost, split_cost)

        return dfs(0, n)

Implementation Explanation

The prefix sum array allows the number of '1' characters in any interval to be computed in constant time.

The recursive function dfs(left, right) represents the minimum achievable cost for the interval [left, right).

For each interval, the implementation first computes its direct storage cost. If the interval length is odd, splitting is impossible, so that cost is immediately returned.

For even lengths, the interval can either remain intact or be split into two equal halves. The function recursively computes the optimal costs of both halves and returns the minimum between keeping the interval and splitting it.

Since every interval in the decomposition tree is visited only once, the solution is highly efficient.

Go Solution

func minCost(s string, encCost int, flatCost int) int64 {
	n := len(s)

	prefix := make([]int, n+1)
	for i := 0; i < n; i++ {
		prefix[i+1] = prefix[i]
		if s[i] == '1' {
			prefix[i+1]++
		}
	}

	var dfs func(int, int) int64

	dfs = func(left, right int) int64 {
		length := right - left
		ones := prefix[right] - prefix[left]

		var segmentCost int64
		if ones == 0 {
			segmentCost = int64(flatCost)
		} else {
			segmentCost = int64(length) * int64(ones) * int64(encCost)
		}

		if length%2 == 1 {
			return segmentCost
		}

		mid := (left + right) / 2

		splitCost := dfs(left, mid) + dfs(mid, right)

		if splitCost < segmentCost {
			return splitCost
		}
		return segmentCost
	}

	return dfs(0, n)
}

Go-Specific Notes

The maximum possible segment cost can reach:

100000 × 100000 × 100000 = 10^15

which does not fit into a 32-bit integer. Therefore the Go implementation uses int64 throughout all cost calculations.

The recursive depth is only the number of times a length can be halved, which is at most about 17 for n ≤ 100000, so recursion is completely safe.

Worked Examples

Example 1

Input:

s = "1010"
encCost = 2
flatCost = 1

Prefix sums:

Index	Prefix Ones
0	0
1	1
2	1
3	2
4	2

Root Interval [0,4)

Interval	Length	Ones	Direct Cost
[0,4)	4	2	4 × 2 × 2 = 16

Split into:

• [0,2)
• [2,4)

Interval [0,2)

Interval	Length	Ones	Direct Cost
[0,2)	2	1	4

Split again:

Interval	Cost
[0,1) = "1"	2
[1,2) = "0"	1

Split cost = 3

Optimal value = min(4, 3) = 3

Interval [2,4)

Exactly symmetric.

Optimal value = 3

Root Result

Choice	Cost
Keep whole segment	16
Split into halves	3 + 3 = 6

Answer:

6

Example 2

Input:

s = "1010"
encCost = 3
flatCost = 10

For each "10" segment:

Choice	Cost
Keep segment	2 × 1 × 3 = 6
Split into "1" and "0"	3 + 10 = 13

Keeping is better.

Root:

Choice	Cost
Whole string	24
Two halves	6 + 6 = 12

Answer:

12

Example 3

Input:

s = "00"
encCost = 1
flatCost = 2

Root interval:

Interval	Ones	Cost
"00"	0	2

Split result:

Interval	Cost
"0"	2
"0"	2

Total split cost = 4.

Therefore:

min(2, 4) = 2

Answer:

2

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Every interval in the decomposition tree is processed once
Space	O(n)	Prefix sum array dominates memory usage

The decomposition tree contains at most 2n - 1 nodes, which occurs when the string length is a power of two. Each node performs only constant-time work because interval counts are obtained from the prefix sum array. Therefore the total running time is linear in the size of the input.

Test Cases

sol = Solution()

assert sol.minCost("1010", 2, 1) == 6      # Example 1
assert sol.minCost("1010", 3, 10) == 12    # Example 2
assert sol.minCost("00", 1, 2) == 2        # Example 3

assert sol.minCost("0", 5, 7) == 7         # Single zero
assert sol.minCost("1", 5, 7) == 5         # Single one

assert sol.minCost("00", 100, 1) == 1      # All zeros, keep whole segment
assert sol.minCost("11", 1, 100) == 2      # Splitting helps

assert sol.minCost("0000", 1, 1) == 1      # Entire string contains no ones
assert sol.minCost("1111", 1, 100) == 4    # Split down to single characters

assert sol.minCost("101", 2, 1) == 12      # Odd length, no split possible

assert sol.minCost("1000", 10, 1) == 13    # Mixed values, selective splitting
assert sol.minCost("00000000", 5, 2) == 2  # Large zero-only segment

assert sol.minCost("1" * 16, 1, 1000) == 16  # Repeated splitting to leaves

Test Summary

Test	Why
"1010", 2, 1	Example 1
"1010", 3, 10	Example 2
"00", 1, 2	Example 3
"0"	Smallest zero-only input
"1"	Smallest sensitive input
"00", 100, 1	Entire segment cheaper than splitting
"11", 1, 100	Splitting improves cost
"0000"	All-zero interval
"1111"	All-one interval
"101"	Odd length root cannot split
"1000"	Mixed content
"00000000"	Larger zero-only segment
"1111111111111111"	Deepest useful splitting scenario

Edge Cases

Odd-Length Root Segment

If the original string length is odd, no split can ever occur at the root. A common mistake is to assume every interval can be recursively divided. The implementation explicitly checks whether the interval length is even before attempting any split. For odd lengths, it immediately returns the direct segment cost.

Segments Containing No Sensitive Elements

When a segment contains zero '1' characters, the cost is flatCost, not 0. Forgetting this special rule would produce incorrect answers. The implementation handles this case separately before applying the encrypted storage formula.

Very Large Costs

The maximum encrypted cost can reach approximately 10^15. Using 32-bit integers would overflow. The Go solution uses int64 for all cost calculations, and Python naturally supports arbitrary-precision integers.

All-Zero Strings

An all-zero string is particularly interesting because splitting never reduces the cost. Every segment would still cost flatCost, so the optimal answer is often to keep the largest possible segment. The dynamic programming recurrence naturally discovers this because the direct interval cost is compared against the sum of child costs at every step.

Power-of-Two Lengths

Lengths that are powers of two generate the largest decomposition trees and therefore stress the implementation the most. Even in this worst case, the number of tree nodes is at most 2n - 1, ensuring the algorithm remains linear and easily fits within the constraints.