3133. Minimum Array End

[mah-shamim]

Topics: Bit Manipulation

You are given two integers n and x. You have to construct an array of positive integers nums of size n where for every 0 <= i < n - 1, nums[i + 1] is greater than nums[i], and the result of the bitwise AND operation between all elements of nums is x.

Return the minimum possible value of nums[n - 1].

Example 1:

• Input: n = 3, x = 4
• Output: 6
• Explanation: nums can be [4,5,6] and its last element is 6.

Example 2:

• Input: n = 2, x = 7
• Output: 15
• Explanation: nums can be [7,15] and its last element is 15.

Example 3:

• Input: cost = [[2, 5, 1], [3, 4, 7], [8, 1, 2], [6, 2, 4], [3, 8, 8]]
• Output: 10

Constraints:

• 1 <= n, x <= 108

Hint:

1. Each element of the array should be obtained by “merging” x and v where v = 0, 1, 2, …(n - 1).
2. To merge x with another number v, keep the set bits of x untouched, for all the other bits, fill the set bits of v from right to left in order one by one.
3. So the final answer is the “merge” of x and n - 1.

[mah-shamim]

We need to construct an array nums of positive integers of size n, where each successive element is greater than the previous. The bitwise AND of all elements in nums should yield x. We are asked to find the minimum possible value of nums[n-1].

Here’s the breakdown:

1. Bit Manipulation Insight: We can observe that nums[i] should be built by merging x with integers 0, 1, ..., n-1. This will help ensure the bitwise AND result yields x since we start with a base of x.

2. Building the Array Elements: Each element can be thought of as x merged with some integer, and we aim to keep x's bits intact. We fill in additional bits from the integer to get increasing numbers while maintaining the AND outcome as x.

3. Merging Strategy: To find the minimum nums[n-1], we only need to merge x with n-1. Merging in this context means that if any bit in x is 1, it remains 1. We use bits from n-1 to add any required additional bits without altering the bits set in x.

Let's implement this solution in PHP: 3133. Minimum Array End

<?php
/**
 * @param Integer $n
 * @param Integer $x
 * @return Integer
 */
function minEnd($n, $x) {
    // Calculate the maximum bit position to consider
    $kMaxBit = floor(log($n, 2)) + floor(log($x, 2)) + 2;
    $k = $n - 1;
    $ans = $x;
    $kBinaryIndex = 0;

    // Iterate through each bit position up to kMaxBit
    for ($i = 0; $i < $kMaxBit; ++$i) {
        // Check if the current bit in $ans is 0
        if ((($ans >> $i) & 1) == 0) {
            // Set the bit in $ans if the corresponding bit in $k is 1
            if (($k >> $kBinaryIndex) & 1) {
                $ans |= (1 << $i);
            }
            ++$kBinaryIndex;
        }
    }

    return $ans;
}

// Example 1
echo minimumArrayEnd(3, 4) . "\n";  // Output: 6

// Example 2
echo minimumArrayEnd(2, 7) . "\n";  // Output: 15
?>

Explanation:

1. Bit Checking and Setting:

   • We check each bit of ans (starting from x), and if a bit is 0 in ans, we look to the corresponding bit in k (which is n-1).
   • If that bit in k is 1, we set the bit in ans to 1. This process ensures the minimum increment in value while preserving bits set in x.

2. Loop Constraints:

   • We iterate through each bit position up to a calculated maximum (kMaxBit), ensuring that we cover the bits necessary from both x and n.

3. Result:

   • The final value of ans is the minimum possible value for nums[n-1] that satisfies the conditions.

Complexity:

• The algorithm operates in constant time since the number of bits in any integer in this range is bounded by 32, making this approach efficient even for large values of n and x.

This solution yields the desired minimum nums[n-1] while maintaining the required properties.

[kovatz]

Thanks to solve this problem

[mah-shamim]

Thanks