Search a 2D Matrix (Java + Python)

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

• Integers in each row are sorted from left to right.
• The first integer of each row is greater than the last integer of the previous row.

For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

For an  O (rows * cols) solution is too simple to meet the interviewer’s expectation. So what is the better solution should be? Binary search should be a good tool when I face search problem, however, how BS could be apply  in matrix?   if we regard 0 as start point, total_cols*total_rows-1 as the end point to look the matrix as a array then we can apply BS. However, how could we convert the position to row_NO and col_NO?  We  know row_NO * total_cols+col_NO=position, So in programming  row_NO=postion/total_cols, col_NO=postion%total_cols. then we can adapt this feature in our program  to apply binary search to get O(log(cols+rows)) search

	/*
	Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
	Integers in each row are sorted from left to right.
	The first integer of each row is greater than the last integer of the previous row.
	For example,
	Consider the following matrix:
	[
	[1, 3, 5, 7],
	[10, 11, 16, 20],
	[23, 30, 34, 50]
	]
	Given target = 3, return true.
	*/
	public class Solution {
	public boolean searchMatrix(int[][] matrix, int target) {
	if (matrix==null || matrix.length==0){
	return false;
	}
	int start=0;
	int end=matrix.length*matrix[0].length-1;
	while (start<=end){
	int mid=start+(end-start)/2;
	int candidate=matrix[mid/matrix[0].length][mid%matrix[0].length];
	if (candidate==target){
	return true;
	}
	else if (candidate<target){
	start=mid+1;
	}
	else{
	end=mid-1;
	}
	}
	return false;
	}
	}

view raw SearchA2DMatrix.java hosted with ❤ by GitHub

	"""
	Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
	Integers in each row are sorted from left to right.
	The first integer of each row is greater than the last integer of the previous row.
	For example,
	Consider the following matrix:
	[
	[1, 3, 5, 7],
	[10, 11, 16, 20],
	[23, 30, 34, 50]
	]
	Given target = 3, return true.
	"""
	class Solution:
	# @param matrix, a list of lists of integers
	# @param target, an integer
	# @return a boolean
	def searchMatrix(self, matrix, target):
	if matrix==None or len(matrix)==0:
	return False
	start=0
	end=len(matrix)*len(matrix[0])-1
	while (start<=end):
	mid=start+(end-start)/2
	if matrix[mid/len(matrix[0])][mid%len(matrix[0])]==target:
	return True
	elif matrix[mid/len(matrix[0])][mid%len(matrix[0])]<target:
	start=mid+1
	else:
	end=mid-1
	return False

view raw SearchA2DMattix.py hosted with ❤ by GitHub