Leetcode 1969: Minimum Non-Zero Product of the Array Elements
You are given a positive integer
p. Consider an array nums (1-indexed) that consists of the integers in the inclusive range [1, 2p - 1] represented in binary form. You are allowed to perform the following operation any number of times: Choose two elements x and y from nums. Choose a bit in x and swap it with its corresponding bit in y. The goal is to find the minimum non-zero product of the elements in nums after performing the above operation any number of times. Return this minimum product modulo (10^9 + 7).Problem
Approach
Steps
Complexity
Input: The input consists of a single integer `p`.
Example: p = 3
Constraints:
• 1 <= p <= 60
Output: The output should be the minimum non-zero product of the elements in `nums` modulo (10^9 + 7).
Example: Output: 1512
Constraints:
• The product should be returned modulo (10^9 + 7).
Goal: The goal is to minimize the product of the elements in `nums` by swapping bits between elements.
Steps:
• Step 1: Determine the maximum value in the range `[1, 2^p - 1]`.
• Step 2: Calculate the minimum non-zero product of the array by applying the bit-swapping operations as needed.
• Step 3: Return the product modulo (10^9 + 7).
Goal: The constraints ensure that the input is manageable within the given bounds.
Steps:
• The integer `p` will always be between 1 and 60.
Assumptions:
• The integer `p` is always a positive integer and within the given bounds.
• Input: Input: p = 1, Output: 1
• Explanation: In this case, the array `nums` contains only one element [1]. Therefore, the product is simply 1.
• Input: Input: p = 2, Output: 6
• Explanation: The array `nums` contains [1, 2, 3]. The product of these numbers is 1 * 2 * 3 = 6. No further bit-swapping is needed as the product is already minimized.
• Input: Input: p = 3, Output: 1512
• Explanation: In this case, the array `nums` consists of [1, 2, 3, 4, 5, 6, 7]. After performing the bit-swapping operations, we achieve a minimized product of 1512.
Approach: To minimize the product of the numbers in `nums`, we need to efficiently calculate the product of the numbers while applying the bit-swapping operation. The key is to identify the largest possible numbers and optimize the product using modular arithmetic.
Observations:
• We need to minimize the product of numbers using bit-swapping, which suggests an approach involving modular arithmetic.
• The key idea is to reduce the product as much as possible by strategically applying bit-swapping, ensuring the result is not zero.
Steps:
• Step 1: Calculate the largest number `2^p - 1`.
• Step 2: Compute the minimum non-zero product using the power of the largest number and apply the bit-swapping operations.
• Step 3: Return the product modulo (10^9 + 7).
Empty Inputs:
• There are no empty inputs as the integer `p` is always provided.
Large Inputs:
• For large values of `p`, up to 60, the solution should handle large arrays efficiently.
Special Values:
• When `p = 1`, the array contains only one element, which simplifies the problem.
Constraints:
• Ensure that the solution works within the time and space limits for `p` values up to 60.
long long mod = 1000000007;
long long expn(long long n, long long p) {
if(p == 0) return 1;
if(p == 1) return n%mod;
if(p % 2 == 0) {
long long ans = expn(n, p / 2);
return ((ans ) * (ans )) % mod;
} else {
long long ans = expn(n, p / 2);
ans = ((ans ) * (ans )) % mod;
return (ans * (n % mod)) % mod;
}
return 0;
}
int minNonZeroProduct(int p) {
long long val = pow(2, p);
val--;
long long ans = expn(val-1, val/2);
return ans * (val %mod)% mod;
}
1 : Declaration
long long mod = 1000000007;
This line declares a constant `mod` with the value 1000000007, used as a modulus for all calculations to prevent overflow in large integer computations.
2 : Function Definition
long long expn(long long n, long long p) {
Defines the function `expn` which calculates the exponentiation of `n` raised to the power `p` modulo `mod`. It uses recursion to efficiently compute the result.
3 : Base Case
if(p == 0) return 1;
Base case for recursion: when the exponent `p` is 0, return 1, as any number raised to the power of 0 is 1.
4 : Base Case
if(p == 1) return n%mod;
Base case: when the exponent `p` is 1, return `n` modulo `mod`.
5 : Recursive Case
if(p % 2 == 0) {
Checks if the exponent `p` is even. If it is, it recursively calculates `n^(p/2)`.
6 : Recursive Case
long long ans = expn(n, p / 2);
Recursively calculates `n^(p/2)`, stores the result in `ans`.
7 : Recursive Case
return ((ans ) * (ans )) % mod;
Returns the square of `ans` modulo `mod` to compute `n^p` when `p` is even.
8 : Recursive Case
} else {
Handles the case when `p` is odd.
9 : Recursive Case
long long ans = expn(n, p / 2);
Recursively calculates `n^(p/2)` for odd `p`.
10 : Recursive Case
ans = ((ans ) * (ans )) % mod;
Squares `ans` and takes modulo `mod` to compute the result of `n^p`.
11 : Recursive Case
return (ans * (n % mod)) % mod;
Since `p` is odd, multiply `ans` with `n % mod` and return the result modulo `mod`.
12 : End of Function
return 0;
If the exponentiation process does not resolve correctly (though this should never happen), return 0 as a fallback.
13 : Function Definition
int minNonZeroProduct(int p) {
Defines the function `minNonZeroProduct` which computes a specific result using the `expn` function.
14 : Calculation
long long val = pow(2, p);
Calculates 2 raised to the power `p`.
15 : Calculation
val--;
Decreases `val` by 1.
16 : Calculation
long long ans = expn(val-1, val/2);
Calls the `expn` function to calculate `(val-1)^(val/2) % mod`.
17 : Return Statement
return ans * (val %mod)% mod;
Returns the result of multiplying `ans` by `val % mod` and taking modulo `mod`.
Best Case: O(log p)
Average Case: O(log p)
Worst Case: O(log p)
Description: The time complexity is dominated by the exponentiation step, which is O(log p) due to the binary exponentiation algorithm.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is constant, O(1), as we only need a few variables for the computation.
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