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1969. Minimum Non-Zero Product of the Array Elements

Description

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] in their binary representations. You are allowed to do 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. Corresponding bit refers to the bit that is in the same position in the other integer.

For example, if x = 1101 and y = 0011, after swapping the 2nd bit from the right, we have x = 1111 and y = 0001.

Find the minimum non-zero product of nums after performing the above operation any number of times. Return this product modulo 109 + 7.

Note: The answer should be the minimum product before the modulo operation is done.

 

Example 1:

Input: p = 1
Output: 1
Explanation: nums = [1].
There is only one element, so the product equals that element.

Example 2:

Input: p = 2
Output: 6
Explanation: nums = [01, 10, 11].
Any swap would either make the product 0 or stay the same.
Thus, the array product of 1 * 2 * 3 = 6 is already minimized.

Example 3:

Input: p = 3
Output: 1512
Explanation: nums = [001, 010, 011, 100, 101, 110, 111]
- In the first operation we can swap the leftmost bit of the second and fifth elements.
    - The resulting array is [001, 110, 011, 100, 001, 110, 111].
- In the second operation we can swap the middle bit of the third and fourth elements.
    - The resulting array is [001, 110, 001, 110, 001, 110, 111].
The array product is 1 * 6 * 1 * 6 * 1 * 6 * 7 = 1512, which is the minimum possible product.

 

Constraints:

  • 1 <= p <= 60

Solutions

  • class Solution {
    public int minNonZeroProduct(int p) {
        final int mod = (int) 1e9 + 7;
        long a = ((1L << p) - 1) % mod;
        long b = qpow(((1L << p) - 2) % mod, (1L << (p - 1)) - 1, mod);
        return (int) (a * b % mod);
    }

    private long qpow(long a, long n, int mod) {
        long ans = 1;
        for (; n > 0; n >>= 1) {
            if ((n & 1) == 1) {
                ans = ans * a % mod;
            }
            a = a * a % mod;
        }
        return ans;
    }
}

  • class Solution {
public:
    int minNonZeroProduct(int p) {
        using ll = long long;
        const int mod = 1e9 + 7;
        auto qpow = [](ll a, ll n) {
            ll ans = 1;
            for (; n; n >>= 1) {
                if (n & 1) {
                    ans = ans * a % mod;
                }
                a = a * a % mod;
            }
            return ans;
        };
        ll a = ((1LL << p) - 1) % mod;
        ll b = qpow(((1LL << p) - 2) % mod, (1L << (p - 1)) - 1);
        return a * b % mod;
    }
};

  • class Solution:
    def minNonZeroProduct(self, p: int) -> int:
        mod = 10**9 + 7
        return (2**p - 1) * pow(2**p - 2, 2 ** (p - 1) - 1, mod) % mod


  • func minNonZeroProduct(p int) int {
	const mod int = 1e9 + 7
	qpow := func(a, n int) int {
		ans := 1
		for ; n > 0; n >>= 1 {
			if n&1 == 1 {
				ans = ans * a % mod
			}
			a = a * a % mod
		}
		return ans
	}
	a := ((1 << p) - 1) % mod
	b := qpow(((1<<p)-2)%mod, (1<<(p-1))-1)
	return a * b % mod
}

  • function minNonZeroProduct(p: number): number {
    const mod = BigInt(1e9 + 7);

    const qpow = (a: bigint, n: bigint): bigint => {
        let ans = BigInt(1);
        for (; n; n >>= BigInt(1)) {
            if (n & BigInt(1)) {
                ans = (ans * a) % mod;
            }
            a = (a * a) % mod;
        }
        return ans;
    };
    const a = (2n ** BigInt(p) - 1n) % mod;
    const b = qpow((2n ** BigInt(p) - 2n) % mod, 2n ** (BigInt(p) - 1n) - 1n);
    return Number((a * b) % mod);
}

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