???? Problem Link: https://leetcode.com/problems/minimum-array-end/submissions/1447854076/?envType=daily-question&envId=2024-11-09
???? Solution: Pinned in the comments.
In today's LeetCode daily challenge, we solve the Minimum Array End problem. The solution is implemented in Java.
???? Problem Description:
Given an integer n (length of an array) and an integer x (starting element), the goal is to find the smallest possible last element in an array where each element is either equal to or a bitwise OR with the previous one. The task is to determine the minimum possible last element given these constraints.
???? Key Points:
Bitwise Operations: The solution heavily uses bitwise OR and checks individual bits.
Optimization with Shifts: Using bit shifts to modify x based on bits in remaining elements.
Iterative Updates: By updating each bit position conditionally, we achieve the minimum possible result.
???? Code Explanation:
Initial Setup: Initialize minLastElement with x and set the count of remaining elements.
Bit Manipulation Loop: For each bit, if it's zero in x, update minLastElement with OR operations based on elementsRemaining.
Return Result: Return the computed minimum last element after processing all bits.
???? Daily Solutions:
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???? Join the Community:
Share your thoughts on the problem in the comments. Discuss different approaches with fellow coders. Like, share, and subscribe for more daily coding challenges!
???? Solution: Pinned in the comments.
In today's LeetCode daily challenge, we solve the Minimum Array End problem. The solution is implemented in Java.
???? Problem Description:
Given an integer n (length of an array) and an integer x (starting element), the goal is to find the smallest possible last element in an array where each element is either equal to or a bitwise OR with the previous one. The task is to determine the minimum possible last element given these constraints.
???? Key Points:
Bitwise Operations: The solution heavily uses bitwise OR and checks individual bits.
Optimization with Shifts: Using bit shifts to modify x based on bits in remaining elements.
Iterative Updates: By updating each bit position conditionally, we achieve the minimum possible result.
???? Code Explanation:
Initial Setup: Initialize minLastElement with x and set the count of remaining elements.
Bit Manipulation Loop: For each bit, if it's zero in x, update minLastElement with OR operations based on elementsRemaining.
Return Result: Return the computed minimum last element after processing all bits.
???? Daily Solutions:
Subscribe for daily LeetCode solutions explained step-by-step in Java!
???? Join the Community:
Share your thoughts on the problem in the comments. Discuss different approaches with fellow coders. Like, share, and subscribe for more daily coding challenges!
- Категория
- Программирование на java
- Теги
- LeetCode, Coding, Programming
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