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

• Difficulty Level : Easy
• Last Updated : 26 May, 2021

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 exists 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.```

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 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`

## Python3

 `# Python3 implementation to find the``# Leftmost Column with atleast a``# 1 in a sorted binary matrix``import` `sys``N ``=` `3` `# Function to search for the``# leftmost column of the matrix``# with atleast a 1 in sorted``# binary matrix``def` `search(mat, n, m):` `    ``a ``=` `sys.maxsize``    ` `    ``# Loop to iterate over all the``    ``# rows of the matrix``    ``for` `i ``in` `range` `(n):``        ``low ``=` `0``        ``high ``=` `m ``-` `1``        ``ans ``=` `sys.maxsize``        ` `        ``# 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``                ` `                ``elif` `(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 ``=``=` `sys.maxsize):``        ``return` `-``1``    ``return` `a ``+` `1` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``  ` `    ``mat ``=` `[[``0``, ``0``, ``0``],``           ``[``0``, ``0``, ``1``],``           ``[``0``, ``1``, ``1``]]``    ``print``(search(mat, ``3``, ``3``))``  ` `# This code is contributed by Chitranayal`

## 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`

## Javascript

 ``
Output:
`2`

Performance Analysis:

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

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