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Problem solution · Java

Minimize Manhattan Distances

Minimize Manhattan Distances: 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
55 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 Manhattan Distances, 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

  • 55 lines of Java from the credited upstream file 3102.java.
  • The implementation visibly relies on sequence storage.
  • 1 loop block 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 Manhattan Distances · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int minimumDistance(int[][] points) {    final int[] maxIndices = maxManhattanDistance(points, -1);    final int[] xiyi = maxManhattanDistance(points, maxIndices[0]);    final int[] xjyj = maxManhattanDistance(points, maxIndices[1]);    return Math.min(manhattan(points, xiyi[0], xiyi[1]), //                    manhattan(points, xjyj[0], xjyj[1]));  }   // Returns the pair of indices a and b where points[a] and points[b] have the  // maximum Manhattan distance and a != excludedIndex and b != excludedIndex.  private int[] maxManhattanDistance(int[][] points, int excludedIndex) {    int minSum = Integer.MAX_VALUE;    int maxSum = Integer.MIN_VALUE;    int minDiff = Integer.MAX_VALUE;    int maxDiff = Integer.MIN_VALUE;    int minSumIndex = -1;    int maxSumIndex = -1;    int minDiffIndex = -1;    int maxDiffIndex = -1;     for (int i = 0; i < points.length; ++i) {      if (i == excludedIndex)        continue;      final int x = points[i][0];      final int y = points[i][1];      final int sum = x + y;      final int diff = x - y;      if (sum < minSum) {        minSum = sum;        minSumIndex = i;      }      if (sum > maxSum) {        maxSum = sum;        maxSumIndex = i;      }      if (diff < minDiff) {        minDiff = diff;        minDiffIndex = i;      }      if (diff > maxDiff) {        maxDiff = diff;        maxDiffIndex = i;      }    }     return maxSum - minSum >= maxDiff - minDiff ? new int[] {minSumIndex, maxSumIndex}                                                : new int[] {minDiffIndex, maxDiffIndex};  }   private int manhattan(int[][] points, int i, int j) {    return Math.abs(points[i][0] - points[j][0]) + Math.abs(points[i][1] - points[j][1]);  }}

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