
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
iin 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 ofxset 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
xitself asnums[0]. - Each subsequent integer
nums[i], where1 <= i < n, has to be greater thannums[i-1]but still needs to maintain the resultingxwhen 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
xthat 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
nand 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
minimumEndinginitializes with two parameters:len, which represents the length, andstart, an integer specifying the initial state. - The variable
finalis initially set tostartto hold the eventual result of the computation. - The
lenparameter is decremented by one at the beginning to adjust for the inclusion of thestartin the process. - A for-loop utilizes a
bitMaskinitialized 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 offinalwhich depends based on the least significant bit oflen(len & 1) multiplied bybitMask. - The value
lenis 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,
numberandbaseValue. ThefinalResultis initially set tobaseValue. - A loop then iterates for each bit position in
number. During each iteration:- The bit positions of the
baseValueare inspected. - If the current bit in
baseValueis unset (i.e., it is 0), and the corresponding lowest bit ofnumberis set (i.e., it is 1), the function sets the respective bit infinalResult. - The loop uses a
bitMaskwhich initially is 1 and left-shifts by 1 in each iteration, used to check individual bits. numberis right-shifted in each iteration to process the next bit in the subsequent iteration.
- The bit positions of the
- The output
finalResultis 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_valuewith the value ofinitial. - Decrement
lengthby one to adjust for zero-based indexing. - Establish
bit_maskinitialized at 1 to manipulate specific bits.
The main processing occurs within a while loop that continues until length reduces to 0:
- An
ifstatement checks if the least significant bit of theinitialis 0 using bitwise AND. - If true, it modifies
final_valueusing bitwise OR with the current least significant bit oflength. - The
lengthis shifted right to iterate through every bit, decreasing its value each loop iteration. - Simultaneously,
bit_maskis 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.