Leftmost One

Similar Problems:

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- Tag: #binarysearch

Given a 2D array, and each line has only 0 and 1, the front part is 0, and the latter part is 1. Find the number of columns in the leftmost 1 of all the rows in the array.

Notice

- The number of rows in the array and the number of columns do not exceed 1000
- In order to limit the time complexity, your program will run 50000 times

Example

Given arr = [[0,0,0,1],[1,1,1,1]], return 0.

Explanation: Arr[1][0] is the leftmost 1 in all rows and its column is 0.

Given arr = [[0,0,0,1],[0,1,1,1]], return 1.

Explanation: Arr[1][1] is the leftmost 1 in all rows and its column is 1.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## https://code.dennyzhang.com/leftmost-one class Solution: """ @param arr: The 2-dimension array @return: Return the column the leftmost one is located """ def getColumn(self, arr): ## Basic Ideas: binary search ## Complexity: Time O(m*log(n)), Space O(1) row_count = len(arr) if row_count == 0: return 0 col_count = len(arr[0]) leftmost = row_count for i in range(row_count): # binarysearch l, r = 0, col_count-1 while l<r: # find first 1 m = int(l+(r-l)/2) if arr[i][m] == 1: r = m else: l = m + 1 if arr[i][r] == 1: leftmost = min(leftmost, r) else: leftmost = min(leftmost, r+1) return leftmost