CodingInterviewQuestions/Leetcode/Leetcode Problem 74 Search a 2D Matrix.txt at master · akuchotrani/CodingInterviewQuestions · GitHub

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74. Search a 2D Matrix
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.
Example 1:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 3
Output: true
Example 2:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 13
Output: false


Hint to solve: 2 methods
1st method: Use wrapping by division and modulo
2nd method: Use two binary search on row and cols

class Solution:

    #treat 2d matrix as a sorted list and generate mapping
    # row = mid/total_cols
    # col = mid%total_cols
    def method_1(self, matrix:List[List[int]], target:int) -> bool:
        if len(matrix) == 0 or len(matrix[0]) == 0:
            return False

        total_rows = len(matrix)
        total_cols = len(matrix[0])

        start = 0
        end = (total_rows * total_cols) - 1

        while(start <= end):
            mid = (int)((end+start)/2)
            mid_value = matrix[(int)(mid/total_cols)][mid%total_cols]
            #print(F"start:{start} end:{end} mid:{mid} midVal:{mid_value}")
            if mid_value == target:
                return True

            if mid_value < target:
                start = mid + 1
            elif mid_value > target:
                end = mid - 1

        return False

    #Perform 2 binary search
    #First binary search on the row
    #Second binary search on the col
    def method_2(self, matrix, target) -> bool:

        if len(matrix) == 0 or len(matrix[0]) == 0:
            return False
        total_rows = len(matrix)
        total_cols = len(matrix[0])

        #Binary search on row. To decide which row might contain the number
        start = 0
        end = total_rows - 1
        rowIndex = 0
        while(start <= end):
            mid = (int)((start+end)/2)
            rowIndex = mid
            #print(F"start:{start} end:{end} mid:{mid} midValue:{matrix[mid][0]}")
            if(matrix[mid][0] == target):
                return True

            if(matrix[mid][0] < target and matrix[mid][total_cols-1] >= target):
                break

            if matrix[mid][0] < target:
                start = mid + 1
            elif matrix[mid][0] > target:
                end = mid - 1


        #Binary search on column
        start = 0
        end = total_cols - 1

        while(start <= end):
            mid = (int)((start+end)/2)
            #print(F"start:{start} end:{end} mid:{mid} midValue:{matrix[rowIndex][mid]}")
            if(matrix[rowIndex][mid] == target):
                return True

            if matrix[rowIndex][mid] < target:
                start = mid + 1
            elif matrix[rowIndex][mid] > target:
                end = mid - 1

        return False

    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        # return self.method_1(matrix, target)
        return self.method_2(matrix, target)