# Binary Matrix Leftmost One - Facebook Top Interview Questions

### Problem Statement :

You are given a two-dimensional list of integers matrix which contains 1s and 0s.

Given that each row is sorted in ascending order with 0s coming before 1s, return the leftmost column index with the value of 1.

If there's no row with a 1, return -1.

Can you solve it faster than \mathcal{O}(nm)O(nm).

Constraints

n, m ≤ 250 where n and m are the number of rows and columns in matrix

Example 1

Input

matrix = [

[0, 0, 0, 0],

[0, 0, 1, 1],

[0, 0, 0, 1],

[0, 1, 1, 1]

]

Output

1

Explanation

The last row contains the leftmost column with a one at index 1.

### Solution :

                        Solution in C++ :

int solve(vector<vector<int>>& mat) {
int res = INT_MAX;
for (int i = 0; i < mat.size(); i++) {
int pos = -1;
int l = 0, r = mat[0].size() - 1;
while (l <= r) {
int mid = l + (r - l) / 2;
if (mat[i][mid] == 1) {
pos = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
if (pos != -1) res = min(res, pos);
}

return res == INT_MAX ? -1 : res;
}


                        Solution in Java :

import java.util.*;

class Solution {
public int solve(int[][] matrix) {
if (matrix.length == 0 || matrix[0].length == 0) {
return -1;
}

int n = matrix.length;
int m = matrix[0].length;

int row = n - 1;
int col = m - 1;

int ans = m;

while (col >= 0 && row >= 0) {
if (matrix[row][col] == 1) {
ans = col;
col--;
} else {
row--;
}
}

return ans == m ? -1 : ans;
}
}


                        Solution in Python :

class Solution:
def solve(self, matrix):
if not matrix:
return -1
ans = float("inf")
COLS = len(matrix[0])
for row in matrix:
l, r = 0, COLS - 1
while l < r:
mid = l + (r - l) // 2
if row[mid] == 1:
r = mid
else:
l = mid + 1
if row[l] == 1:
ans = min(ans, l)
return ans if ans != float("inf") else -1


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