# Leftmost Column with atleast one 1 in a row-wise sorted binary matrix

Given a binary matrix mat[][] containing 0’s and 1’s. Each row of the matrix is sorted in the non-decreasing order, the task is to find the left-most column of the matrix with at least one 1 in it.

Note: If no such column exist return -1.

Examples:

```Input:
mat = { {0, 0},
{1, 1} }
Output: 1
Explanation:
The 1st column of the
matrix contains atleast a 1.

Input:
mat = {{0, 0},
{0, 0}}
Output: -1
Explanation:
There is no such column which
contains atleast one 1.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to iterate over all the rows of the matrix and Binary Search over them for the first occurrence of the 1 in the that row. The minimum of all the occurrence of the first 1 in the rows of the matrix will be the desired solution for the problem.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the ` `// Leftmost Column with atleast a  ` `// 1 in a sorted binary matrix ` ` `  `#include ` ` `  `#define N 3 ` `using` `namespace` `std; ` ` `  `// Function to search for the  ` `// leftmost column of the matrix ` `// with atleast a 1 in sorted  ` `// binary matrix ` `int` `search(``int` `mat[], ``int` `n, ``int` `m) ` `{ ` `    ``int` `i, a = INT_MAX; ` `     `  `    ``// Loop to iterate over all the  ` `    ``// rows of the matrix ` `    ``for` `(i = 0; i < n; i++) { ` `        ``int` `low = 0; ` `        ``int` `high = m - 1; ` `        ``int` `mid; ` `        ``int` `ans = INT_MAX; ` `         `  `        ``// Binary Search to find the  ` `        ``// leftmost occurence of the 1 ` `        ``while` `(low <= high) { ` `            ``mid = (low + high) / 2; ` `             `  `            ``// Condition if the column  ` `            ``// contains the 1 at this  ` `            ``// position of matrix ` `            ``if` `(mat[i][mid] == 1) { ` ` `  `                ``if` `(mid == 0) { ` `                    ``ans = 0; ` `                    ``break``; ` `                ``} ` `                ``else` `if` `(mat[i][mid - 1] == 0) { ` `                    ``ans = mid; ` `                    ``break``; ` `                ``} ` `            ``} ` `             `  `            ``if` `(mat[i][mid] == 1) ` `                ``high = mid - 1; ` `            ``else` `                ``low = mid + 1; ` `        ``} ` `         `  `        ``// If there is a better solution ` `        ``// then update the answer ` `        ``if` `(ans < a) ` `            ``a = ans; ` `    ``} ` `     `  `    ``// Condition if the solution  ` `    ``// doesn't exist in the matrix ` `    ``if` `(a == INT_MAX) ` `        ``return` `-1; ` `    ``return` `a+1; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `mat = { 0, 0, 0, ` `                    ``0, 0, 1, ` `                    ``0, 1, 1 }; ` `    ``cout << search(mat, 3, 3); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation to find the ` `// Leftmost Column with atleast a  ` `// 1 in a sorted binary matrix ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `     `  `static` `final` `int` `N = ``3``;  ` ` `  `// Function to search for the  ` `// leftmost column of the matrix ` `// with atleast a 1 in sorted  ` `// binary matrix ` `static` `int` `search(``int` `mat[][], ``int` `n, ``int` `m) ` `{ ` `    ``int` `i, a = Integer.MAX_VALUE; ` `     `  `    ``// Loop to iterate over all the  ` `    ``// rows of the matrix ` `    ``for``(i = ``0``; i < n; i++) ` `    ``{ ` `       ``int` `low = ``0``; ` `       ``int` `high = m - ``1``; ` `       ``int` `mid; ` `       ``int` `ans = Integer.MAX_VALUE; ` `        `  `       ``// Binary Search to find the  ` `       ``// leftmost occurence of the 1 ` `       ``while` `(low <= high) ` `       ``{ ` `           ``mid = (low + high) / ``2``; ` `            `  `           ``// Condition if the column  ` `           ``// contains the 1 at this  ` `           ``// position of matrix ` `           ``if` `(mat[i][mid] == ``1``) ` `           ``{ ` `               ``if` `(mid == ``0``) ` `               ``{ ` `                   ``ans = ``0``; ` `                   ``break``; ` `               ``} ` `               ``else` `if` `(mat[i][mid - ``1``] == ``0``)  ` `               ``{ ` `                   ``ans = mid; ` `                   ``break``; ` `               ``} ` `           ``} ` `           ``if` `(mat[i][mid] == ``1``) ` `               ``high = mid - ``1``; ` `           ``else` `               ``low = mid + ``1``; ` `       ``} ` `        `  `       ``// If there is a better solution ` `       ``// then update the answer ` `       ``if` `(ans < a) ` `           ``a = ans; ` `    ``} ` `     `  `    ``// Condition if the solution  ` `    ``// doesn't exist in the matrix ` `    ``if` `(a == Integer.MAX_VALUE) ` `        ``return` `-``1``; ` `    ``return` `a + ``1``; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `mat[][] = { { ``0``, ``0``, ``0` `}, ` `                    ``{ ``0``, ``0``, ``1` `}, ` `                    ``{ ``0``, ``1``, ``1` `} }; ` `                     `  `    ``System.out.print(search(mat, ``3``, ``3``)); ` `} ` `} ` ` `  `// This code is contributed by Amit Katiyar `

## C#

 `// C# implementation to find the leftmost   ` `// column with atleast a 1 in a sorted ` `// binary matrix ` `using` `System; ` ` `  `class` `GFG{ ` `     `  `//static readonly int N = 3;  ` ` `  `// Function to search for the leftmost   ` `// column of the matrix with atleast a ` `//  1 in sorted binary matrix ` `static` `int` `search(``int` `[,]mat, ``int` `n, ``int` `m) ` `{ ` `    ``int` `i, a = ``int``.MaxValue; ` `     `  `    ``// Loop to iterate over all the  ` `    ``// rows of the matrix ` `    ``for``(i = 0; i < n; i++) ` `    ``{ ` `       ``int` `low = 0; ` `       ``int` `high = m - 1; ` `       ``int` `mid; ` `       ``int` `ans = ``int``.MaxValue; ` `        `  `       ``// Binary Search to find the  ` `       ``// leftmost occurence of the 1 ` `       ``while` `(low <= high) ` `       ``{ ` `           ``mid = (low + high) / 2; ` `            `  `           ``// Condition if the column  ` `           ``// contains the 1 at this  ` `           ``// position of matrix ` `           ``if` `(mat[i, mid] == 1) ` `           ``{ ` `               ``if` `(mid == 0) ` `               ``{ ` `                   ``ans = 0; ` `                   ``break``; ` `               ``} ` `               ``else` `if` `(mat[i, mid - 1] == 0)  ` `               ``{ ` `                   ``ans = mid; ` `                   ``break``; ` `               ``} ` `           ``} ` `           ``if` `(mat[i, mid] == 1) ` `               ``high = mid - 1; ` `           ``else` `               ``low = mid + 1; ` `       ``} ` `        `  `       ``// If there is a better solution ` `       ``// then update the answer ` `       ``if` `(ans < a) ` `           ``a = ans; ` `    ``} ` `     `  `    ``// Condition if the solution  ` `    ``// doesn't exist in the matrix ` `    ``if` `(a == ``int``.MaxValue) ` `        ``return` `-1; ` `    ``return` `a + 1; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[,]mat = { { 0, 0, 0 }, ` `                   ``{ 0, 0, 1 }, ` `                   ``{ 0, 1, 1 } }; ` `                     `  `    ``Console.Write(search(mat, 3, 3)); ` `} ` `} ` ` `  `// This code is contributed by Amit Katiyar `

Output:

```2
```

Performance Analysis:

• Time Complexity: O(N*logN)
• Auxiliary Space: O(1)

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