LeetCode Problem 1969

Difficulty: 🟡 Medium
Topics: Math, Greedy, Recursion

Solution

LeetCode 1969, Minimum Non-Zero Product of the Array Elements

Problem Understanding

The problem gives us an integer p and defines an array containing every number from 1 to 2^p - 1. Each number is represented in binary using exactly p bits.

We are allowed to repeatedly perform a special operation. In one operation, we choose two numbers and swap one bit position between them. The swap must occur at the same bit index in both numbers.

The goal is to make the product of all numbers as small as possible, while ensuring the product is still non-zero. Since the numbers can become extremely large, we return the answer modulo 10^9 + 7.

The important detail is that the modulo is applied only at the end. We must minimize the actual mathematical product first.

The constraint 1 <= p <= 60 is extremely important. The array size is 2^p - 1, which becomes astronomically large even for moderate values of p. For example:

• p = 20 gives more than one million numbers
• p = 60 gives more than 10^18 numbers

This immediately tells us that we cannot explicitly build or manipulate the array. The solution must rely on mathematical reasoning.

Another key observation is that swapping bits preserves the total count of set bits at each bit position across the entire array. We cannot arbitrarily change the binary representation of the collection, only redistribute bits among numbers.

The most important edge cases are:

• p = 1, where the array contains only [1]
• Very large p, where direct computation would overflow normal integer ranges
• Ensuring the product never becomes zero, since introducing a 0 invalidates the result

Approaches

Brute Force Approach

A brute force solution would attempt to simulate all possible swaps and search for the minimum product arrangement.

The array contains 2^p - 1 elements, and each element has p bits. Every operation can choose:

• Any pair of numbers
• Any bit position

The number of reachable states grows exponentially. Even for small p, the search space becomes impossibly large.

Additionally, evaluating the product for each state is expensive because the numbers themselves become huge.

This approach is theoretically correct because it explores all possible configurations, but it is computationally infeasible.

Key Insight

The crucial mathematical insight is that the minimum non-zero product occurs when:

• One number remains as large as possible, namely 2^p - 1

• The remaining numbers are paired so that most become either:

• 1

• 2^p - 2

The optimal arrangement becomes:

• One copy of 2^p - 1
• (2^(p-1) - 1) copies of 2^p - 2
• (2^(p-1) - 1) copies of 1

This leads to the formula:

$$(2^p - 1) \times (2^p - 2)^{(2^{p-1} - 1)}$$

The reason this works comes from the AM-GM style intuition that, for a fixed bit distribution, the product becomes smaller when values are pushed toward extremes rather than kept balanced.

We must compute this efficiently using modular exponentiation.

The final formula is:

$(2^p-1)\cdot(2^p-2)^{(2^{p-1}-1)}$

Approach Comparison

Approach	Time Complexity	Space Complexity	Notes
Brute Force	Exponential	Exponential	Explores all possible bit swap configurations
Optimal	O(log(2^p))	O(1)	Uses mathematical derivation and fast exponentiation

Algorithm Walkthrough

1. Compute the maximum value in the array.

The largest number is:

$$2^p - 1$$

This is the number whose binary representation contains all 1s. 2. Compute the second-largest value.

The value:

$$2^p - 2$$

has all bits set except the least significant bit. 3. Determine how many times this value appears.

The exponent becomes:

$$2^{p-1} - 1$$

This comes from pairing numbers in the optimal redistribution pattern. 4. Use modular exponentiation.

Directly computing the power is impossible because the exponent can exceed 10^18.

Instead, use fast exponentiation:

• Repeatedly square the base
• Halve the exponent each step
• Multiply into the result when the exponent bit is set

5. Multiply the final components.

The answer is:

$$(2^p - 1) \times (2^p - 2)^{(2^{p-1}-1)} \mod (10^9+7)$$

Why it works

The allowed swaps preserve the total number of set bits at every position. To minimize the non-zero product, we want as many numbers as possible to become small, while concentrating remaining bits into fewer large numbers.

The optimal configuration pushes the values to the extremes:

• Many numbers become 1
• Many numbers become 2^p - 2
• One number remains 2^p - 1

This arrangement satisfies the bit-count constraints while producing the smallest possible non-zero product.

Python Solution

class Solution:
    def minNonZeroProduct(self, p: int) -> int:
        MOD = 10**9 + 7

        max_value = (1 << p) - 1
        second_max = max_value - 1
        exponent = (1 << (p - 1)) - 1

        return (max_value * pow(second_max, exponent, MOD)) % MOD

The implementation directly follows the mathematical formula.

First, we compute max_value = 2^p - 1 using bit shifting. Since 1 << p equals 2^p, subtracting one produces the all-ones binary number.

Next, second_max becomes 2^p - 2.

The exponent is calculated as 2^(p-1) - 1.

Python provides a built-in modular exponentiation function:

pow(base, exponent, mod)

This performs fast exponentiation internally in logarithmic time, which is essential for handling very large exponents efficiently.

Finally, we multiply the remaining factor max_value and take the result modulo 10^9 + 7.

Go Solution

func minNonZeroProduct(p int) int {
	const MOD int64 = 1_000_000_007

	maxValue := (int64(1) << p) - 1
	secondMax := maxValue - 1
	exponent := (int64(1) << (p - 1)) - 1

	power := modPow(secondMax, exponent, MOD)

	return int((maxValue * power) % MOD)
}

func modPow(base, exponent, mod int64) int64 {
	result := int64(1)

	base %= mod

	for exponent > 0 {
		if exponent&1 == 1 {
			result = (result * base) % mod
		}

		base = (base * base) % mod
		exponent >>= 1
	}

	return result
}

The Go solution uses a custom modular exponentiation helper because Go does not provide a built-in equivalent to Python's three-argument pow.

The implementation uses int64 throughout because values such as 2^60 exceed the range of standard 32-bit integers.

The exponentiation loop repeatedly squares the base and halves the exponent, ensuring logarithmic running time.

Worked Examples

Example 1

Input

p = 1

Step 1: Compute values

Variable	Value
max_value	1
second_max	0
exponent	0

Step 2: Compute power

$$0^0 = 1$$

Python's modular exponentiation correctly returns 1.

Step 3: Final answer

$$1 \times 1 = 1$$

Output

1

Example 2

Input

p = 2

Step 1: Compute values

Variable	Value
max_value	3
second_max	2
exponent	1

Step 2: Compute power

$$2^1 = 2$$

Step 3: Final answer

$$3 \times 2 = 6$$

Output

6

Example 3

Input

p = 3

Step 1: Compute values

Variable	Value
max_value	7
second_max	6
exponent	3

Step 2: Compute power

$$6^3 = 216$$

Step 3: Final answer

$$7 \times 216 = 1512$$

Output

1512

Complexity Analysis

Measure	Complexity	Explanation
Time	O(log(2^p)) = O(p)	Fast exponentiation halves the exponent each iteration
Space	O(1)	Only a few integer variables are used

The dominant operation is modular exponentiation. Since exponentiation by squaring processes one bit of the exponent at a time, the number of iterations is proportional to the number of bits in the exponent, which is O(p).

No auxiliary data structures proportional to the input size are required.

Test Cases

solution = Solution()

assert solution.minNonZeroProduct(1) == 1       # Smallest possible input
assert solution.minNonZeroProduct(2) == 6       # Example from statement
assert solution.minNonZeroProduct(3) == 1512    # Example from statement

assert solution.minNonZeroProduct(4) == 581202553  # Larger intermediate powers
assert solution.minNonZeroProduct(5) == 202795991  # Tests modular exponentiation

assert solution.minNonZeroProduct(10) > 0       # Medium-sized exponent
assert solution.minNonZeroProduct(20) > 0       # Large exponent handling
assert solution.minNonZeroProduct(60) > 0       # Maximum constraint

# Verify result always stays within modulo range
MOD = 10**9 + 7
assert 0 <= solution.minNonZeroProduct(60) < MOD

Test Summary

Test	Why
p = 1	Validates smallest edge case
p = 2	Verifies basic formula behavior
p = 3	Matches detailed sample explanation
p = 4	Tests larger exponentiation
p = 5	Verifies modular arithmetic correctness
p = 10	Ensures scalability beyond small inputs
p = 20	Confirms efficient logarithmic exponentiation
p = 60	Validates maximum constraint handling

Edge Cases

Edge Case 1, p = 1

This is the smallest possible input. The array contains only [1], so no swaps are possible. A common bug is mishandling the exponent:

$$2^{p-1} - 1 = 0$$

The implementation correctly computes pow(0, 0, MOD) as 1, giving the correct final answer of 1.

Edge Case 2, Very Large p

When p = 60, values like 2^60 exceed 32-bit integer limits. A naive implementation using regular integers may overflow.

The Python solution naturally supports arbitrary precision integers. The Go solution explicitly uses int64 to safely handle these values.

Edge Case 3, Avoiding Zero Product

If swaps produced even one zero value, the total product would become zero, which is forbidden by the problem.

The mathematical derivation guarantees that the optimal arrangement keeps every number positive. The smallest values become 1, never 0, ensuring the product remains non-zero.