
Problem Statement
The challenge is to create an array of positive integers, nums
, of size n
which adheres to two primary conditions:
- Every integer in the array should progressively increase. This means for every integer
i
in the range0 <= i < n - 1
,nums[i + 1]
should be greater thannums[i]
. - The result of performing a bitwise AND operation over all integers in the array should equal a given integer
x
.
The ultimate goal is to calculate the smallest possible value for the last element in the array (nums[n - 1]
). This involves careful selection of each integer in nums
to both maintain ascending order and ensure that the bitwise AND of all the integers yields x
.
Examples
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.
Constraints
1 <= n, x <= 108
Approach and Intuition
To address this problem effectively, consider the properties of the bitwise AND operation and constraints of the problem:
Understanding bitwise AND requirements:
- For all integers to have a particular bitwise AND result of
x
, all integers must have the ON bits ofx
set in their binary representation. Additional bits can vary, provided they are the higher-order than the highest ON bit inx
.
- For all integers to have a particular bitwise AND result of
Constructing the array:
- Start the array with
x
itself asnums[0]
. - Each subsequent integer
nums[i]
, where1 <= i < n
, has to be greater thannums[i-1]
but still needs to maintain the resultingx
when ANDed with all previous numbers.
- Start the array with
Minimizing
nums[n - 1]
:- To meet the bitwise AND condition, each subsequent number in the array may need to increment in small amounts above the previous to preserve the binary structure.
- The minimal increments that maintain the AND condition depend on identifying the lowest unset bit positions in
x
that can be flipped without altering the critical bits determiningx
.
Example Walkthrough:
- For example, if
x = 4
(binary100
), the generated sequence must not alter the significant third bit (counting from zero). The incremental numbers can toggle bits lower than this bit to remain greater than their predecessor. - The specifics of the increment depend on the difference between
n
and the number of meaningful positions based on the binary length ofx
.
By considering these steps, we can iterate to build the required array nums
while ensuring both the constraints of gradual increment and the bitwise AND result are maintained correctly. The sequence's last element is carefully calculated to be as minimal as possible under these conditions.
Solutions
- C++
- Java
- Python
class Solution {
public:
long long minimumEnding(int len, int start) {
long long final = start, bitMask;
len -= 1; // Adjust `len` to account for `start`
// Iterate over each bit in `start` using bitMask initiated to 1
for (bitMask = 1; len > 0; bitMask <<= 1) {
if ((bitMask & start) == 0) { // Check if the bit is 0
final |= (len & 1) * bitMask; // Manage the bit position based on `len`
len >>= 1; // Move to the next bit
}
}
return final;
}
};
The solution in C++ defined in the class Solution
employs bit manipulation techniques to compute the minimum ending based on the len
and start
parameters.
- The function
minimumEnding
initializes with two parameters:len
, which represents the length, andstart
, an integer specifying the initial state. - The variable
final
is initially set tostart
to hold the eventual result of the computation. - The
len
parameter is decremented by one at the beginning to adjust for the inclusion of thestart
in the process. - A for-loop utilizes a
bitMask
initialized at 1, iteratively shifted left to examine each bit ofstart
. - Within the loop, it first checks if the current bit of
start
(usingbitMask & start
) is zero. If true, it involves altering the bit offinal
which depends based on the least significant bit oflen
(len & 1
) multiplied bybitMask
. - The value
len
is right-shifted, effectively halving it to move the significance to the next bit level. - The loop continues until all required bits are processed, taking into account changes in the bit positions as required by
len
.
The solution elegantly handles the given input by modifying final
according to the bits of start
and len
, ultimately returning the minimum ending value as a long long integer.
public class Solution {
public long minimumEnding(int number, int baseValue) {
long finalResult = baseValue;
long bitMask;
number--; // Decrementing to skip the baseValue
// Utilizing the bitmask to access each bit
for (bitMask = 1; number > 0; bitMask <<= 1) {
// Check if current bit in baseValue is unset
if ((bitMask & baseValue) == 0) {
// Set the current bit in finalResult if the lowest bit of number is set
finalResult |= (number & 1) * bitMask;
// Prepare for the next bit
number >>= 1;
}
}
return finalResult;
}
}
The provided Java code implements a function minimumEnding
within a class named Solution
. This function calculates a value based on manipulating bits of an integer. Here is a rundown of how the function executes:
- The function takes two integer parameters,
number
andbaseValue
. ThefinalResult
is initially set tobaseValue
. - A loop then iterates for each bit position in
number
. During each iteration:- The bit positions of the
baseValue
are inspected. - If the current bit in
baseValue
is unset (i.e., it is 0), and the corresponding lowest bit ofnumber
is set (i.e., it is 1), the function sets the respective bit infinalResult
. - The loop uses a
bitMask
which initially is 1 and left-shifts by 1 in each iteration, used to check individual bits. number
is right-shifted in each iteration to process the next bit in the subsequent iteration.
- The bit positions of the
- The output
finalResult
is produced after applying the necessary bit operations.
This function specifically adjusts the finalResult
based on a binary operation involving the position of bits in number
where the respective bit in baseValue
is unset, all directed towards minimizing the value set at the end. Furthermore, the nature of bit manipulation ensures that the operations are efficient and fit for scenarios where performance is critical.
class Solution:
def minimumEndValue(self, length: int, initial: int) -> int:
final_value = initial
length -= 1
bit_mask = 1
while length > 0:
if (bit_mask & initial) == 0:
final_value |= (length & 1) * bit_mask
length >>= 1
bit_mask <<= 1
return final_value
The code snippet provided outlines a Python function designed to determine the minimum end value for an array based on provided parameters. The function minimumEndValue
in the Solution
class accepts two arguments:
length
: an integer representing the array lengthinitial
: an integer indicating the initial value from which adjustments will be calculated
The goal is to manipulate and evaluate the bits of the initial value using bitwise operations. Here's a brief explanation of the operations within the function:
- Initialize
final_value
with the value ofinitial
. - Decrement
length
by one to adjust for zero-based indexing. - Establish
bit_mask
initialized at 1 to manipulate specific bits.
The main processing occurs within a while loop that continues until length
reduces to 0:
- An
if
statement checks if the least significant bit of theinitial
is 0 using bitwise AND. - If true, it modifies
final_value
using bitwise OR with the current least significant bit oflength
. - The
length
is shifted right to iterate through every bit, decreasing its value each loop iteration. - Simultaneously,
bit_mask
is shifted left to target the next bit in the sequence on the following iterations.
The function finally returns the final_value
, which represents the modified version of the initial
value after all iterations and tweaks as per the described behavior.
This succinct approach leverages bit manipulation to efficiently derive the resultant value based on the initial setup conditions.
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