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Problem solution · Java

Minimize OR of Remaining Elements Using Operations

Minimize OR of Remaining Elements Using Operations: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
34 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Minimize OR of Remaining Elements Using Operations, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 34 lines of Java from the credited upstream file 3022.java.
  • The implementation visibly relies on sequence storage.
  • 2 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimize OR of Remaining Elements Using Operations · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int minOrAfterOperations(int[] nums, int k) {    final int MAX_BIT = 30;    int ans = 0;    int prefixMask = 0; // Grows like: 10000 -> 11000 -> ... -> 11111.     for (int i = MAX_BIT; i >= 0; --i) {      // Add the i-th bit to `prefixMask` and attempt to "turn off" the      // currently added bit within k operations. If it's impossible, then we      // add the i-th bit to the answer.      prefixMask |= 1 << i;      if (getMergeOps(nums, prefixMask, ans) > k)        ans |= 1 << i;    }     return ans;  }   // Returns the number of merge operations to turn `prefixMask` to the target  // by ANDing `nums`.  private int getMergeOps(int[] nums, int prefixMask, int target) {    int mergeOps = 0;    int ands = prefixMask;    for (final int num : nums) {      ands &= num;      if ((ands | target) == target)        ands = prefixMask;      else        ++mergeOps; // Keep merging the next num.    }    return mergeOps;  }}

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