SearchA2DMatrix.java

/* (C) 2024 YourCompanyName */
package divide_and_conquer;

/**
 * Created by gouthamvidyapradhan on 09/03/2017. Write an efficient algorithm that searches for a
 * value in an m x n matrix. This matrix has the following properties:
 *
 * <p>Integers in each row are sorted in ascending from left to right. Integers in each column are
 * sorted in ascending from top to bottom. For example,
 *
 * <p>Consider the following matrix:
 *
 * <p>[ [1, 4, 7, 11, 15], [2, 5, 8, 12, 19], [3, 6, 9, 16, 22], [10, 13, 14, 17, 24], [18, 21, 23,
 * 26, 30] ] Given target = 5, return true.
 *
 * <p>Given target = 20, return false.
 */
public class SearchA2DMatrix {
  /**
   * Main method
   *
   * @param args
   * @throws Exception
   */
  public static void main(String[] args) throws Exception {
    int[][] matrix = {
      {1, 3, 5, 7, 9}, // 1, 3, 5, 7, 9
      {2, 4, 6, 8, 10}, // 2, 4, 6, 8, 10
      {11, 13, 15, 17, 19}, // 11, 15, 17, 18, 19
      {12, 14, 16, 18, 20}, // 13, 20, 21, 22, 23
      {21, 22, 23, 24, 25} // 14, 25, 26, 27, 28
    };

    System.out.println(new SearchA2DMatrix().searchMatrix(matrix, 11));
  }

  private boolean searchMatrix(int[][] matrix, int target) {
    if (matrix.length == 0) return false;
    int M = matrix.length;
    int N = matrix[0].length;
    int r = 0, c = N - 1;
    while (r < M && c >= 0) {
      if (matrix[r][c] == target) return true;
      else if (target < matrix[r][c]) --c;
      else if (target > matrix[r][c]) r++;
    }
    return false;
  }
}